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Calculations In Chemistry
Modules 13 to 16
A Note to the Student
The focus of these lessons is to provide methods will help you to solve numeric calculations
in first-year chemistry. This is only one part of a course in chemistry, but it is often the
most challenging.
Problem Notebook: The purchase of a spiral problem notebook is suggested as a place to
write your work when solving the problems in these lessons. A notebook that has graphpaper as its pages will be especially helpful.
Choosing a Calculator: As you do problems in these lessons (and assigned homework)
that require a calculator, use the same calculator that you will be allowed to use during
quizzes and tests. Calculators have many different labels and placements of keys. It is
advisable to practice the rules and keys for a calculator before quizzes and tests.
Many courses will not allow the use of a graphing calculator or other types of calculators
with extensive memory during tests. If a type of calculator is specified for your course, buy
two if possible. When one becomes broken or lost, you will have a familiar backup if the
bookstore is sold out later in the term.
If no type of calculator is specified for your course, any inexpensive calculator with a 1/x
or x-1 , yx or ^ , log or 10x , and ln functions will be sufficient for most calculations
in introductory chemistry courses.
When to Do the Lessons: You will receive the maximum benefit from these lessons by
completing each topic before it is addressed in your class.
Where to Start and Lesson Sequence: The order of these lessons may not always match
the order in which topics are covered in your course. If you are using these modules as
part of a course, you should do the lessons in the order in which they are assigned by your
instructor. If you are using these lessons on your own to assist with a course, begin by
•
Determining the topics that will be covered on your next graded assignment:
problem set, quiz, or test.
•
Find that topic in the Table of Contents.
•
Download the modules that precede and include the topics.
•
Find the prerequisite lessons for the topic, listed at the beginning of the module or
lesson. Print the needed lessons. Do the prerequisites, then the topics related to
your next graded assignments.
•
Follow the instructions on “How to Use These Lessons” on page 1.
If you begin these lessons after the start of your course, when time permits, review prior
topics in these lessons as needed, starting with Module 1. You will need all of these
introductory modules for later topics -- and for your final exam.
Check back for updates at www.ChemReview.Net .
©2008 ChemReview.net v. f9
Page i
Table of Contents — Volume 1
How to Use These Lessons ............................................................................................... 1
Module 1 – Scientific Notation ........................................................................................ 2
Lesson 1A:
Lesson 1B:
Lesson 1C:
Moving the Decimal ..............................................................................................3
Calculations Using Exponential Notation ..........................................................7
Tips for Complex Calculations...........................................................................12
Module 2 – The Metric System ......................................................................................19
Lesson 2A:
Lesson 2B:
Lesson 2C:
Lesson 2D:
Metric Fundamentals...........................................................................................20
Calculations With Units ......................................................................................25
Metric Prefix Formats ..........................................................................................27
Cognitive Science -- and Flashcards ..................................................................30
Module 3 – Significant Figures...................................................................................... 35
Lesson 3A:
Lesson 3B:
Lesson 3C:
Rules for Significant Figures...............................................................................35
Sig Figs -- Special Cases.......................................................................................37
Sig Fig Summary and Practice............................................................................41
Module 4 – Conversion Factors......................................................................................44
Lesson 4A:
Lesson 4B:
Lesson 4C:
Lesson 4D:
Lesson 4E:
Conversion Factor Basics ....................................................................................44
Single Step Conversions......................................................................................48
Multi-Step Conversions.......................................................................................51
English/Metric Conversions ..............................................................................53
Ratio Unit Conversions .......................................................................................57
Module 5 – Word Problems ............................................................................................62
Lesson 5A:
Lesson 5B:
Lesson 5C:
Lesson 5D:
Lesson 5E:
Lesson 5F:
Lesson 5G:
Answer Units -- Single Or Ratio?.......................................................................62
Mining The DATA ...............................................................................................64
Solving For Single Units......................................................................................67
Finding the Given .................................................................................................72
Some Chemistry Practice ....................................................................................75
Area and Volume Conversions ..........................................................................77
Densities of Solids: Solving Equations .............................................................82
Module 6 – Atoms, Ions, and Periodicity..................................................................... 89
Lesson 6A:
Lesson 6B:
Lesson 6C:
Lesson 6D:
Lesson 6E:
Atoms.....................................................................................................................89
The Nucleus, Isotopes, and Atomic Mass.........................................................94
Elements, Compounds, and Formulas ............................................................101
The Periodic Table..............................................................................................106
A Flashcard Review System .............................................................................110
Module 7 – Writing Names and Formulas.................................................................113
Lesson 7A:
Lesson 7B:
Lesson 7C:
Naming Elements and Covalent Compounds ...............................................113
Naming Ions .......................................................................................................118
Names and Formulas for Ionic Compounds..................................................128
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Module 8 – Grams and Moles...................................................................................... 142
Lesson 8A:
Lesson 8B:
Lesson 8C:
Lesson 8D:
The Mole ............................................................................................................. 142
Grams Per Mole (Molar Mass)......................................................................... 144
Converting Between Grams and Moles ......................................................... 146
Converting Particles, Moles, and Grams........................................................ 150
Module 9 – Mole Applications .................................................................................... 155
Lesson 9A:
Lesson 9B:
Lesson 9C:
Lesson 9D:
Fractions and Percentages................................................................................ 155
Empirical Formulas........................................................................................... 157
Empirical Formulas from Mass or % Mass .................................................... 158
Mass Fraction, Mass Percent, Percent Composition..................................... 163
Module 10 – Balanced Equations and Stoichiometry.............................................. 172
Lesson 10A:
Lesson 10B:
Lesson 10C:
Lesson 10D:
Lesson 10E:
Lesson 10F:
Chemical Reactions and Equations................................................................. 172
Balancing Equations.......................................................................................... 175
Using Coefficients -- Molecules to Molecules ............................................... 180
Mole to Mole Conversions ............................................................................... 182
Stoichiometry ..................................................................................................... 185
Limiting Reactants and Reactants in Excess .................................................. 192
Module 11 – Molarity .................................................................................................... 200
Lesson 11A:
Lesson 11B:
Lesson 11C:
Lesson 11D:
Lesson 11E:
Lesson 11F:
Lesson 11G:
Ratio Unit Review ............................................................................................. 200
Word Problems With Ratio Answers ............................................................. 201
Molarity .............................................................................................................. 207
Conversions and Careers ................................................................................. 213
Units and Dimensions ..................................................................................... 216
Ratios Versus Two Related Amounts ............................................................ 221
Solving Problems With Parts .......................................................................... 228
Module 12 – Molarity Applications............................................................................ 238
Lesson 12A:
Lesson 12B:
Lesson 12C:
Dilution .............................................................................................................. 238
Ion Concentrations ............................................................................................ 247
Solution Stoichiometry ..................................................................................... 254
Module 13 – Ionic Equations and Precipitates ........................................................ 265
Lesson 13A:
Lesson 13B:
Lesson 13C:
Lesson 13D:
Predicting Solubility for Ionic Compounds .................................................. 265
Total and Net Ionic Equations ......................................................................... 269
Predicting Precipitation.................................................................................... 273
Precipitate and Gravimetric Calculations ...................................................... 280
Module 14 – Acid-Base Neutralization ...................................................................... 287
Lesson 14A:
Lesson 14B:
Lesson 14C:
Lesson 14D:
Lesson 14E:
Ions in Acid-Base Neutralization .................................................................... 287
Balancing Hydroxide Neutralization ............................................................. 291
Acid-Hydroxide Neutralization Calculations ............................................... 298
Neutralization Calculations in Parts............................................................... 304
Carbonate Neutralization................................................................................. 311
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Module 15 – Redox Reactions ......................................................................................319
Lesson 15A:
Lesson 15B:
Lesson 15C:
Lesson 15D:
Lesson 15E:
Oxidation Numbers ...........................................................................................319
Balancing Charge ...............................................................................................324
Balancing Redox Using Oxidation Numbers .................................................328
Oxidizing and Reducing Agents..................................................................... 330
Redox Stoichiometry......................................................................................... 334
Module 16 – Half-Reaction Balancing........................................................................338
Lesson 16A:
Lesson 16B:
Lesson 16C:
Lesson 16D:
Constructing Half-Reactions – The CA-WHe! Method ................................338
Balancing By Adding Half-Reactions..............................................................344
Separating Redox Into Half-Reactions ............................................................347
Balancing Redox With Spectators Present......................................................350
Volume 2
Module 17 – Ideal Gases ...............................................................................................357
Lesson 17A:
Lesson 17B:
Lesson 17C:
Lesson 17D:
Lesson 17E:
Lesson 17F:
Lesson 17G:
Gas Fundamentals..............................................................................................357
Gases at STP........................................................................................................361
Complex Unit Cancellation...............................................................................366
The Ideal Gas Law and Solving Equations.....................................................371
Density, Molar Mass, and Choosing Equations.............................................375
Using the Combined Equation .........................................................................382
Gas Law Summary and Practice ......................................................................388
Module 18 – Gas Labs, Gas Reactions ........................................................................392
Lesson 18A:
Lesson 18B:
Lesson 18C:
Lesson 18D:
Charles’ Law; Graphing Direct Proportions...................................................392
Boyle’s Law; Graphs of Inverse Proportions..................................................399
Avogadro’s Hypothesis; Gas Stoichiometry ..................................................403
Dalton’s Law of Partial Pressures ....................................................................410
Module 19 – Graphing...................................................................................................417
Lesson 19A:
Lesson 19B:
Lesson 19C:
Lesson 19D:
Lesson 19E:
Lesson 19F:
Graphing Fundamentals ...................................................................................417
The Specific Equation for a Line ......................................................................428
Graphing Experimental Data ...........................................................................435
Deriving Equations From Linear Data ............................................................445
Linear Equations Not Directly Proportional ..................................................456
Graphing Inverse Proportions..........................................................................463
Module 20 – Phases Changes and Energy..................................................................472
Lesson 20A:
Lesson 20B:
Lesson 20C:
Lesson 20D:
Lesson 20E:
Phases and Phase Changes ...............................................................................472
Specific Heat Capacity and Equations ............................................................485
Water, Energy, and Consistent Units ..............................................................492
Calculating Joules Using Unit Cancellation ...................................................496
Calorimetry .........................................................................................................502
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Module 21 – Heats Of Reaction (ΔH) ......................................................................... 510
Lesson 21A:
Lesson 21B:
Lesson 21C:
Exo- And Endothermic Reactions .................................................................. 510
Heats of Formation and Element Formulas................................................... 514
Adding ΔH Equations (Hess’s Law)............................................................... 521
Module 22 – Light and Spectra ................................................................................... 531
Lesson 22A:
Lesson 22B:
Lesson 22C:
Lesson 22D:
Lesson 22E:
Waves ................................................................................................................. 531
Planck's Constant ............................................................................................. 536
DeBroglie’s Wavelength .................................................................................. 544
The Hydrogen Atom Spectrum ....................................................................... 549
Quantum Mechanics ........................................................................................ 555
Module 23 – Electron Configuration .......................................................................... 560
Lesson 23A:
Lesson 23B:
Lesson 23C:
Lesson 23D:
The Multi-Electron Atom ................................................................................. 560
Abbreviated Electron Configurations............................................................. 564
The Periodic Table and Electron Configuration ........................................... 570
Electron Configurations: Exceptions and Ions ............................................ 574
Module 24 – Bonding ................................................................................................... 579
Lesson 24A:
Lesson 24B:
Lesson 24C:
Lesson 24D:
Lesson 24E:
Lesson 24F:
Lesson 24G:
Lesson 24H:
Covalent Bonds.................................................................................................. 579
Molecular Shapes and Bond Angles ............................................................... 584
Electronegativity................................................................................................ 592
Molecular Polarity............................................................................................. 597
Solubility............................................................................................................. 602
Double and Triple Bonds ................................................................................. 606
Ion Dot Diagrams .............................................................................................. 612
Orbital Models for Bonding ............................................................................. 614
Volume 3
Module 25 – Kinetics ..................................................................................................... 619
Lesson 25A:
Lesson 25B:
Lesson 25C:
Lesson 25D:
Lesson 25E:
Lesson 25F:
Lesson 25G:
Lesson 25H:
Kinetics Fundamentals .................................................................................... 619
Rate Laws ........................................................................................................... 623
Integrated Rate Law --Zero Order .................................................................. 633
Logarithms ........................................................................................................ 641
Integrated Rate Law -- First Order.................................................................. 656
Reciprocal Math................................................................................................. 666
Integrated Rate Law -- Second Order ............................................................. 671
Half-Life.............................................................................................................. 678
Module 26 – Equilibrium.............................................................................................. 687
Lesson 26A:
Lesson 26B:
Lesson 26C:
Lesson 26D:
Lesson 26E:
Le Châtelier’s Principle..................................................................................... 688
Powers and Roots of Exponential Notation................................................... 700
Equilibrium Constants...................................................................................... 710
K Values ............................................................................................................. 716
Kp Calculations .................................................................................................. 720
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Lesson 26F:
Lesson 26G:
Lesson 26H:
Lesson 26I:
Lesson 26J:
The Rice Moles Table.........................................................................................726
K Calculations From Initial Concentrations ...................................................734
Q: The Reaction Quotient..................................................................................740
Calculations Using K and Q..............................................................................743
Solving Quadratic Equations............................................................................749
Module 27 – Acid-Base Fundamentals .......................................................................760
Lesson 27A:
Lesson 27D:
Acid-Base Math Review ....................................................................................760
Kw Calculations: H+ and OH─ .......................................................................763
Strong Acid Solutions........................................................................................767
The [OH─] in Strong Acid Solutions...............................................................770
Lesson 27E:
Strong Base Solutions ........................................................................................774
Lesson 27F:
The pH System ...................................................................................................776
Lesson 27B:
Lesson 27C:
Module 28 – Weak Acids and Bases............................................................................786
Lesson 28A:
Lesson 28B:
Lesson 28C:
Lesson 28D:
Lesson 28E:
Lesson 28F:
Lesson 28G:
Lesson 28H:
Weak Acids and Ka Expressions ......................................................................786
Ka Math................................................................................................................791
Ka Calculations ...................................................................................................794
Solving Ka Using the Quadratic Formula .......................................................805
Weak Bases and Kb Calculations......................................................................808
Polyprotic Acid Calculations............................................................................819
Brønsted-Lowry Definitions .............................................................................824
Which Acid-Base Reactions Go? ......................................................................830
Module 29 – Salts and Buffers .....................................................................................838
Lesson 29A:
Lesson 29B:
Lesson 29C:
Lesson 29D:
Lesson 29E:
Lesson 29F:
The Acid-Base Behavior of Salts.......................................................................838
Calculating the pH of a Salt Solution .............................................................846
Salts That Contain Amphoteric Ions ...............................................................860
Acid-Base Common Ions, Buffers ....................................................................854
Buffer Equations.................................................................................................860
Buffer Calculations.............................................................................................865
Additional topics and resources are available online at www.ChemReview.Net .
•••••
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Page vi
Module 13 — Ionic Equations and Precipitates
Module 13 — Ionic Equations and Precipitates
Pretests: On any of these lessons, if you think you know the topic, try the last problem in
the lesson. If you can do that problem, you may skip the lesson.
Lesson 13A: Predicting Solubility for Ionic Compounds
Prerequisites: If you have any problems translating between the names, solid formulas,
and separated formulas for ionic compounds during Module 13, review your Lesson 7C
and your ion name and formula flashcards from Lesson 7B.
* * * * *
Solubility Terminology
All ionic compounds dissolve to some extent in water, and some dissolve to a substantial
extent.
Solubility for ionic compounds is temperature dependent. Some dissolve to a greater
extent in warmer solvents, some dissolve less, and some ionic compounds have borderline
solubility at room temperature.
Those messy realities aside, most ion combinations can generally be classified as soluble or
insoluble in water. A commonly accepted definition is
•
if 0.10 moles or more of a solid dissolve per liter of solution at room temperature,
the solid is termed soluble;
•
if the solid dissolves less than 0.10 moles per liter, it is considered either slightly
soluble or insoluble.
Most introductory chemistry courses ask that you memorize a set of solubility rules that
will allow you to predict the solubility of many ion combinations. Although there are some
patterns to solubility based on the Periodic Table, there are limited simple rules. The
solubility scheme on the following page will cover most ions customarily assigned.
However,
•
if your course requires that you memorize a particular chart, memorize it instead of
this one.
•
If you are allowed to consult a particular solubility chart on quizzes and tests, use it
in place of this one.
•
If the following chart conflicts in its predictions from a chart you are assigned in
your class, use the one that you are assigned.
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Module 13 — Ionic Equations and Precipitates
A Solubility Scheme
The scheme below is hierarchical: higher rules take precedence over those beneath them. To
predict the solubility of ion combinations, apply the rules in order from the top.
If only one ion of the two in a compound is in this table, you may presume that the
compound will follow the rule for the ion that is shown. (This will not always be accurate,
but is a best guess. With solubility, there are exceptions.)
Positive Ions
Negative Ions
─
─
─
─
1. (alkali metals)+ , NH4+ NO3 , CH3COO , ClO3 ; ClO4
Soluble
(nitrate, acetate, chlorate, perchlorate)
2. Pb2+ , Hg22+ , Ag+
CO32─ , PO43─ , S2─ , CrO42─
Insoluble
3.
Cl─ , Br─ , I─ (except Cu+)
Soluble
4. Ba2+ (except OH─)
OH─ (except Ba2+ , Sr2+, Ca2+)
Insoluble
5.
SO42─(except Sr2+, Ca2+)
Soluble
Exceptions include: Column 2 sulfides and aluminum sulfide decompose in water.
CH3COOAg, Ag2SO4 and Hg2SO4 are moderately soluble.
Memorization Tips
1. Most people best recall what they hear repeatedly (especially if it rhymes). You may
want to recite the ion names in a series, as well as write them.
2. Notice the patterns in the last column, and in the empty boxes.
3. Focus especially on the rules near the top. In this scheme, the higher the rule, the more
often it will be used.
Using the Scheme to Make Predictions
In using any solubility scheme, if you are unsure of the ions in a compound, you should
write out the separated formula that shows the ion charges. Atoms that have two possible
charges, such as Cu+ and Cu2+, can have differing solubilities.
Let’s try a few predictions.
Q.
For the questions below, use the solubility scheme of your choice. Write your
answer and your reason.
1. Is Ba(NO3)2 soluble?
2. Is PbCl2 soluble or insoluble?
3. Is Hg(NO3)2 soluble or insoluble?
* * * * *
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Module 13 — Ionic Equations and Precipitates
Answers
1. Yes. Ba(NO3)2 Æ Ba2+ + 2 NO3─ , and all nitrates are soluble.
2. PbCl2 Æ Pb2+ + 2 Cl─ . Pb2+ ion by Rule 2 is insoluble. Chloride ion in Rule 3 is
soluble, but rule 2 takes precedence. PbCl2 is insoluble.
3. Hg(NO3)2 Æ Hg2+ + 2 NO3─ . The above solubility scheme makes no specific
predictions for Hg2+, the mercuric [mercury (II)] ion, but based on Rule 1 that all
nitrates are soluble, predict soluble. If only one ion in a pair is in the table, base your
prediction on the rule for that ion.
Practice
If you are required to memorize solubility rules, do so before you do the practice below. To
be useful, practice should be treated as a test (but one that you grade).
Whenever there are rules you need to memorize for an assignment, quiz, or test, try
writing them at the top of your assignment at the beginning.
1. Write your solubility scheme from memory.
2. Label the ion combinations as soluble or insoluble, and state a reason for your
prediction. Check your answers after each one or two parts.
a. K+ + Br─
b. Sr2+ + Cl─
c. Ca2+ + CO32─
d. Ag+ + CrO42─
3. Write the separated ions for these combinations, then label the ion combination as
soluble or insoluble, and state a reason for your prediction. (If you are using a different
scheme, your reason may differ, but your answer will probably be the same.)
Do every other part, and check your answers frequently. Do more if you need more
practice. If you have trouble writing the separated ions, redo Lessons 7B and 7C.
a. Lead (II) Bromide Æ Pb2+ + 2 Br─ ; Insoluble -- Rule 2 for Pb2+ (example)
b. Barium Carbonate Æ
c. Sodium Hydroxide Æ
d. SrBr2 Æ
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Module 13 — Ionic Equations and Precipitates
e. Silver nitrate Æ
f. Ammonium hydroxide Æ
g. Fe3(PO4)2 Æ
h. Pb(CH3COO)2 Æ
i. NiCl2 Æ
j. BaSO4 Æ
k. RbBr Æ
l. Fe2S3 Æ
4. Which ions in combination with Pb2+ are soluble compounds?
5. Are all nitrates soluble?
Are all phosphates insoluble?
Answers
These answers are based on the solubility scheme above. You will have different rules if you use a different
scheme, but you should have the same answer in most cases, and a similar reason.
2. a. K+ + Br─
Soluble. Rule 1, first column metal ions are always soluble
b. Sr2+ + Cl─ Soluble. Rule 3, chloride
c. Ca2+ + CO32─ Insoluble. Rule 2, carbonates (if only one ion is in table, use it)
d. Ag+ + CrO42─ Insoluble. Rule 2, chromate ion
3. b. Barium Carbonate Æ Ba2+ + CO32─ Insoluble by Rule 2 for carbonates
c. Sodium Hydroxide Æ Na+ + OH─ Soluble by Rule 1 for alkali-metal ions
d. SrBr2 Æ Sr2+ + 2 Br─ Soluble by Rule 3 for bromides
e. Silver nitrate Æ Ag+ + NO3─ Soluble by Rule 1 for nitrates
f. Ammonium hydroxide Æ NH4+ + OH─
Soluble by Rule 1 for ammonium ion
g. Fe3(PO4)2 Æ 3 Fe2+ + 2 PO43─ Insoluble by Rule 2 for phosphate ions
h. Pb(CH3COO)2 Æ Pb2+ + 2 CH3COO─ Soluble by Rule 1 for acetates
i. NiCl2 Æ Ni2+ + 2 Cl─
Soluble by Rule 3 for chlorides
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Module 13 — Ionic Equations and Precipitates
BaSO4 Æ Ba2+ + SO42─ Insoluble by Rule 4 for barium ions
k. RbBr Æ Rb+ + Br─
Soluble by Rule 1 for alkali-metal ions
j.
l. Fe2S3 Æ 2 Fe3+ + 3 S2─
Insoluble by Rule 2 for sulfide ions
4. Which ions in combination with Pb2+ would make soluble compounds? Nitrate, acetate, chlorate, and
perchlorate.
5. Are all nitrates soluble? YES. That is a rule used frequently.
Are all phosphates insoluble? NO. Alkali-metal and ammonium-ion phosphates are soluble.
*
*
* * *
Lesson 13B: Total and Net Ionic Equations
Prerequisites: Lessons 7C, 10B, and 13A.
* * * * *
Mixing Ions
When solutions of different ions are mixed, chemical reactions can occur.
One type of reaction is precipitation. When solutions of different soluble ionic compounds
are mixed, the ions may “trade partners:” new combinations of positive and negative ions
are possible. If a new combination is possible that is insoluble in water, it will precipitate:
it will form a cloud of solid particles in the solution that sink to the bottom.
For example, when a 0.1 M solution of sodium chloride (table salt) is mixed with a 0.1
M solution of silver nitrate, a bright white cloud immediately forms. Solid particles
from the cloud slowly settle to the bottom of the solution. Over time, in light, the
surface of the solid slowly turns grey.
The precipitate formed is silver chloride (AgCl), a compound that reacts with light.
The total ionic equation for this reaction is
Na+(aq) + Cl─(aq) + Ag+(aq) + NO3─(aq) Æ AgCl(s) + Na+(aq) + NO3─(aq)
The (aq) is an abbreviation for aqueous, which means dissolved in water. The (s)
abbreviates solid.
Note that the sodium and nitrate ions did not change in the reaction. Ions that are present,
but do not change during a reaction, are called spectator ions. In total ionic equations, the
spectator ions are included on both sides, and they will be the same on both sides. Including
the spectators helps us to see all of the reactant and product compounds.
However, as in math equations, terms that are the same on both sides of an equation can be
cancelled. Canceling the spectators gives the net ionic equation, which shows only the
particles that changed in the reaction.
For the reaction above, cancel the spectators and write the net ionic equation.
* * * * *
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Module 13 — Ionic Equations and Precipitates
Ag+(aq) + Cl─(aq)
Practice A:
Æ AgCl(s)
Do Part a on each of these. If you need more practice, do Part b.
1. For these total ionic equations, circle the precipitate, cross out the spectators, and write
the net ionic equation.
a. Pb2+(aq) + 2 NO3― (aq) + Cu2+(aq) + 2 Cl―(aq) Æ Cu2+(aq) + 2 NO3―(aq) + PbCl2(s)
b. 6 Na+(aq) + 2 PO43―(aq) + 3 Mg2+(aq) + 3 SO42―(aq) Æ Mg3(PO4)2(s) + 6 Na+(aq) + 3 SO42―(aq)
2. Balance these precipitation equations. (For rules, see Lesson 10B).
a.
Fe(NO3)3(aq) +
NaOH(aq) Æ
Fe(OH)3(s) +
NaNO3(aq)
b.
Ca(NO3)2(aq) +
K3PO4(aq) Æ
KNO3(aq) +
Ca3(PO4)2(s)
3. Formulas for “aqueous solids,” such as in Problem 2 above, are often written to
represent ionic solids dissolved in a solution. In reality, however, if an ionic solid
dissolves in water, its ions will separate in the solution. Total and net ionic equations
show separated-ion formulas -- unless the ions precipitate.
Re-write the balanced equations for 2a and 2b using separated-ion formulas. This will
be the total ionic equation for each reaction.
4. Write the net ionic equations for the reactions in 2a and 2b.
Balancing Total Ionic Equations
For precipitation reactions, you will often be asked to balance a total ionic equation. To
balance properly, you must first balance each of the reactant and product formulas for
charge, then balance again to account for ratios of reaction.
In the following equation, the brackets show the original ion combinations on the left and
the new possible combinations on the right.
[
Ca2+ +
NO3― ] + [
K+ +
PO43―] Æ Ca3(PO4)2(s) + [
K+ + NO3― ]
To balance, first add coefficients inside the brackets so that the charges are balanced for each
substance formula, and then check your answer below. (For reactions conducted in
aqueous solutions, if no state is shown, assume (aq).)
* * * * *
[ 1 Ca2+ + 2 NO3― ] + [ 3 K+ + 1 PO43―] Æ Ca3(PO4)2(s) + [ 1 K++ 1 NO3― ]
Now add coefficients in front of the brackets so that all of the atoms balance.
* * * * *
To form the precipitate, 3 calcium atoms and 2 phosphate ions are needed, so:
3 [1 Ca2+ + 2 NO3―] + 2 [3 K+ + 1 PO43―] Æ 1 Ca3(PO4)2(s) + 6 [1 K+ + 1 NO3―]
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Module 13 — Ionic Equations and Precipitates
Write the total ionic equation by removing the brackets. Check that the total ionic equation is
balanced for atoms. In total ionic equations, check also that the charges on each side total
to zero as well.
* * * * *
3 Ca2+ + 6 NO3― + 6 K+ + 2 PO43― Æ 1 Ca3(PO4)2(s) + 6 K+ + 6 NO3―
Check: 3 Ca, 6 N, 6K, 2 P, 26 O on both sides. Zero net charge on both sides. Balanced.
Practice B:
Re-write these equations as balanced total ionic equations.
SO42― ] +
1.
[
K+ +
2.
[
Fe2+ +
Br ― ] + [
[
Sr2+ + NO3― ] Æ
Na+ + PO43― ] Æ [
SrSO4(s) +
Na+ +
[ K+ +
Br― ] +
NO3― ]
Fe3(PO4)2(s)
Balancing Equations That Omit Some Spectator Ions
Until this point, most of the equations we have encountered have contained either
•
formulas for neutral compounds, or
•
separated ions, with a total charge of zero on both sides of the equation.
In reactions that involve ions, equations may also be written that omit some but not all of
the spectator ions. This can result in an equation that is balanced, but has a net charge on
both sides.
For example, the reaction of silver nitrate and magnesium chloride solutions can be
represented with ionic solid and molecular formulas as
2 AgNO3(aq) + MgCl2(aq)
(1)
Æ 2 AgCl(s) + Mg(NO3)2(aq)
However, separated-ion formulas better represent soluble substances dissolved in water.
The total ionic equation is
2 Ag+ + 2 NO3─ + Mg2+ + 2 Cl─
Æ 2 AgCl(s) + Mg2+ + 2 NO3─
(2)
A third way to write the reaction is to leave out some of the spectators. For example, the
above reaction can be written as
2 AgNO3 +
2 Cl─
Æ 2 AgCl(s) + 2 NO3─
(3)
In this equation, the spectator magnesium ions have been left out. The reaction might be
written in this format to emphasize that silver nitrate solution will form a precipitate when
mixed with any solution that contains chloride ions.
Note that this partial ionic equation (3), unlike the total ionic equation, does not have a zero
charge on both sides. However, it does have a balanced charge on both sides: the total
charge on both sides is positive two. This is an important rule:
Equations are considered balanced if they have the same number and kind of atoms and
the same net charge on both sides.
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In balancing equations, if the atoms are balanced, and the total charge is the same on both
sides, the equation is considered balanced even if some of the atoms and charges that must
be present have been left out. In a balanced equation, the total charge on both sides does
not need to be zero, but it must be the same on both sides.
By definition, in total and net ionic equations, the charges must add up to zero on both sides,
but in the partial ionic equations that are also frequently written, the atom counts and total
charge must simply be the same on both sides.
Leaving Out the Spectators
Leaving out the spectators, in both net and in partial ionic equations, provides and
equation that is quicker to write than the total ionic equation. The equation focuses on the
important particles: those that change in the reaction.
In a similar shortcut, in chemistry labs a container may be labeled “Ag+” or “ OH─.”
However, if any container is labeled as containing a single ion, it does not mean that only
one kind of ion is present. The spectator ions are present, but they have been left out. All
stable matter must be composed of balanced positive and negative charges.
“Leaving out the spectators” on container labels and in equations is simply a shortcut used
to emphasize the ions that react.
Practice C:
Balance these partial ionic equations. Check answers as you go. Do every
other letter, and more if you need more practice.
1.
Co(NO3)2
2.
AgNO3 +
3.
CO32─ +
4.
HCO3─ +
5.
Al3+ +
OH─ Æ
+
CrO42─ Æ
HCl
Æ
OH─
NO +
NO3─
Co(OH)2(s) +
NO3─
Ag2CrO4(s) +
H2CO3 +
Æ
H2O Æ
H2O +
Al +
Cl―
CO32―
NO3
─
+
H+
ANSWERS
Practice A: Assume (aq) if no state is shown.
1a. Pb2+(aq) + 2 NO3― (aq) + Cu2+(aq) + 2 Cl―(aq) Æ Cu2+(aq) + 2 NO3―(aq) + PbCl2(s)
Net ionic equation: Pb2+(aq) + 2 Cl―(aq) Æ PbCl2(s)
1b.. 6 Na+ + 2 PO43―+ 3 Mg2+ + 3 SO42― Æ
Net ionic equation:
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Mg3(PO4)2(s)
+ 6 Na+ + 3 SO42―
3 Mg2+(aq) + 2 PO43―(aq) Æ Mg3(PO4)2(s)
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Module 13 — Ionic Equations and Precipitates
2a.
1 Fe(NO3)3(aq) + 3 NaOH(aq) Æ 1 Fe(OH)3(s) + 3 NaNO3(aq)
2b.
3 Ca(NO3)2 + 2 K3PO4 Æ 6 KNO3 + 1 Ca3(PO4)2(s)
1 Fe3+ + 3 NO3― + 3 Na+ + 3 OH― Æ 1 Fe(OH)3(s) + 3 Na+ + 3 NO3―
3 Ca2+ + 6 NO3― + 6 K+ + 2 PO43― Æ 6 K+ + 6 NO3― + 1 Ca3(PO4)2(s)
Net ionic equation: 1 Fe3+(aq) + 3 OH―(aq) Æ 1 Fe(OH)3(s)
Net ionic equation: 3 Ca2+(aq) + 2 PO43―(aq) Æ 1 Ca3(PO4)2(s)
3a.
3b.
4a.
4b.
Practice B: Assume (aq) if no state is shown.
1. 2 K+ + 1 SO42― + 1 Sr2+ + 2 NO3― Æ 1 SrSO4(s) + 2 K+ + 2 NO3―
2. 3 Fe2+ + 6 Br ― + 6 Na+ + 2 PO43― Æ 6 Na+ + 6 Br― + Fe3(PO4)2(s)
Practice C: (coefficients of 1 may be omitted).
2 OH─ Æ
1 Co(OH)2(s) +
2 NO3─
2 AgNO3 +
1 CrO42─ Æ
1 Ag2CrO4(s) +
2 NO3─
3.
1 CO32─ +
2 HCl
4.
1 HCO3─ +
1 OH─ Æ
1.
1 Co(NO3)2 +
2.
Æ
1 H2CO3 +
1 Al3+ + 1 NO + 2 H2O Æ
* * * * *
5.
2 Cl―
1 CO32―
─
1 Al + 1 NO3 + 4 H+
1 H2O +
Lesson 13C: Predicting Precipitation
Prerequisites: Lessons 7C, 10B, 13A, and 13B.
Timing: Do this lesson when you are asked to predict whether a precipitate will form, or to
write a formula for a precipitate which forms, when ions are mixed.
* * * * *
Steps in Predicting Precipitation
This key rule for predicting precipitation must be memorized.
When two solutions of soluble ions are mixed,
•
if a new combination is possible that is insoluble, it will precipitate.
•
If a new possible combination is soluble, its ions will remain separated and
dissolved in the solution.
To predict whether the mixing of soluble ions will produce a precipitate, you must analyze
the solubility of the new ionic combinations that can be formed. Let’s walk through that
prediction process one step at a time. Answer these questions in your problem notebook.
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Q. Two separate aqueous solutions are prepared, one by dissolving Ca(NO3)2 and the
other by dissolving Na2CO3.
a. Write balanced equations showing the ions formed when each of these two ionic
solids dissolves in water, then check your answer below.
* * * * *
a. Ca(NO3)2(s) Æ Ca2+(aq) + 2 NO3―(aq)
Na2CO3(s) Æ 2 Na+(aq) + CO32―(aq)
* * * * *
b. When the two solutions above are mixed, two new combinations of positive and
negative ions are possible. Each positive ion can attract either its original negative
partner or the new negative ion with which it has been mixed. Fill in the blanks
below to show the new possible ion combinations, then add coefficients to balance
each new combination for charge.
1
Ca2+ + ____ _______ Æ
___ _____ + ____
NO3― Æ
In solution reactions, if no state is shown, assume (aq).
* * * * *
b.
1 Ca2+ + 1 CO32― Æ
1 Na+ + 1 NO3― Æ
* * * * *
c. After the arrows, write solid formulas for these new combinations.
d. Based on ion solubility rules, label each of the solids as soluble or insoluble.
e. Apply the rule: When two solutions of soluble ions are mixed,
•
if a new combination is possible that is insoluble, it will precipitate.
•
If the new combination is soluble, its ions remain separated.
Will there be a precipitate in this case? If so, what is its solid formula?
Write answers for the above three steps, and then check your answers below.
* * * * *
c. 1 Ca2+ + 1 CO32―Æ CaCO3
1 Na+ + 1 NO3― Å NaNO3
Insoluble by Rule 2. Will precipitate.
Soluble by Rule 1. Ions stay separated.
e. The precipitate is solid CaCO3 .
* * * * *
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Module 13 — Ionic Equations and Precipitates
f.
To write the total ionic equation, write the 4 original reactant ions. Then, for
products, write the formula for the solid precipitate, plus the separated ions for the
new combination that did not precipitate.
Fill in the blanks to give the total ionic equation for the reaction above.
1 Ca2+ +___ ________ +___ ______ + 1 CO32― Æ ___ _______(s) + ___ ______ + ___ ______
g. In your total ionic equation, cross out the spectator ions that are the same on both
sides. These ions did not react. Rewrite the equation without the spectators. This is
the net ionic equation.
* * * * *
f.
Ca2+ + 2 NO3― + 2 Na+ + CO32― Æ CaCO3(s) + 2 Na+ + 2 NO3―
g. Net ionic equation:
Ca2+(aq) + CO32―(aq) Æ CaCO3(s)
In net ionic equations, spectator ions are omitted that can cancel in the total ionic
equation. Net ionic equations show only the particles that change in the reaction.
* * * * *
Predicting Precipitation – The Chart Method
The chart method below can help in predicting whether a precipitate will form. Copy the
following chart into your notebook.
Solid
Formulas
Dissolved Ions
before mixing
Mixed – Possible
new ion combinations
Possible
Precipitates
1.
2.
Let’s use an example to see how the chart is used.
Q2. When solutions of Pb(NO3)2 and KI are mixed, a yellow precipitate forms. Using
the steps on the previous page, fill in the chart above and write the formula for the
precipitate.
If you need a hint, read a part of the answer below and then try again.
* * * * *
Answer
Ionic compounds in solution are often represented using solid formulas, but when soluble
ionic solids dissolve, the ions separate. In these two solutions are these ions:
Solid
Formulas
Dissolved Ions
before mixing
1. Pb(NO3)2 Æ
2.
KI
Æ
Pb2+ + 2 NO3─
K+ + I─
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Module 13 — Ionic Equations and Precipitates
After they are mixed, the ions can “trade partners.” Two new combinations of positive and
negative ions are possible.
Solid
Formulas
Dissolved Ions
before mixing
Pb(NO3)2 Æ Pb2+ + 2 NO3─
KI
Æ
K+ + I─
Mixed – Possible
new ion combinations
Pb2+ + I─
K+ + NO3─
These new combinations may or may not form solid precipitates. Add the possible solid
formulas.
Solid
Formulas
Dissolved Ions
before mixing
Pb(NO3)2 Æ Pb2+ + 2 NO3─
KI
Æ
K+ + I─
Mixed – Possible
new ion combinations
Possible
Precipitates
Pb2+ + 2 I─
Æ
PbI2
K+ + NO3─
Æ
KNO3
Note that the coefficients are not the same before mixing and after. You must adjust the
coefficients so that each new ion combination is balanced for atoms and charge.
Will a precipitate form? If a new combination is possible that is insoluble, it will precipitate.
Label each of the possible precipitates in the chart as soluble or insoluble.
* * * * *
PbCl2 is insoluble by rule 2 (Pb2+); KNO3 is soluble by rule 1. The result is
Solid
Formulas
Dissolved Ions
before mixing
Pb(NO3)2 Æ Pb2+ + 2 NO3─
KI
Æ
K+ + I─
Mixed – Possible
new combinations
Possible
Precipitates
Pb2+ + 2 I─
PbI2 (Insoluble, forms ppt.)
Æ
K+ + NO3─ Å
KNO3 (Soluble, ions stay dissolved)
* * * * *
Q3. For the above reaction, write the total and then the net ionic equation.
* * * * *
Answer
The chart shows the ions before mixing (column 2) and the products after mixing (column 3
or 4). To write the total ionic equation, copy all of the column two ions on a line on your
paper, followed by a reaction arrow. To the right of the arrow, write the new products that
form, shown in column 3 or column 4. Adjust the coefficients so that both each ion
combination and the overall equation are balanced for atoms and charge.
* * * * *
To make the precipitate, we need 2 I─. That means we must have 2 K+ to start).
Pb2+ + 2 NO3─ + 2K+ + 2 I─ Æ PbI2(s) + 2 K+ + 2 NO3─
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Module 13 — Ionic Equations and Precipitates
That is the balanced total ionic equation. Now write the net ionic equation.
* * * * *
Pb2+(aq) + 2 I─(aq) Æ PbI2(s)
Practice A:
Net ionic equations show the ions that change.
Answers are at the end of this lesson.
1. When AgNO3 solution is mixed with Na2CrO4 solution, a blood red precipitate
appears. Fill in a chart like the one above for this reaction, and then write the total and
net ionic equations.
2. When potassium hydroxide and cobalt nitrate solutions are mixed, an intense blue
precipitate forms. Do the chart, then write the total and net ionic equations.
The 3-Line Method
The chart method can be used for as long as needed to gain confidence that you are
arriving at correct answers. At that point, you can try a quicker “3 line method” (draft,
total ionic, and net ionic equations) that handles more of the steps in your head. Let’s learn
the method with the following example.
Q4. Write the total and net ionic equation identifying the white precipitate that forms
when solutions of barium chloride and potassium sulfate are mixed.
Write then check your answers to the steps below.
1. On one line, write the separated-ion formulas for the ions being mixed.
* * * * *
{ Ba2+ + 2 Cl─ } + { 2 K+ + SO42─ } Æ
2. After the arrow, write the new possible ion combinations.
* * * * *
{ Ba2+ + 2 Cl─ } + { 2 K+ + SO42─ } Æ { Ba2+ + SO42─ } + { K+ + Cl─ }
3. Use solubility rules to label each new combination as soluble or insoluble.
4. If a new combination of ions is possible that is insoluble, it will precipitate. Write the
solid formula for any precipitate that would form.
* * * * *
{ Ba2+ + 2 Cl─ } + { 2 K+ + SO42─ } Æ { Ba2+ + SO42─ } + { K+ + Cl─ }
INsoluble
Soluble
BaSO4(s)
5. That’s the rough draft. Copy the draft result without the brackets, adjusting the
coefficients as needed, to write the total ionic equation.
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Module 13 — Ionic Equations and Precipitates
* * * * *
Ba2+ + 2 Cl─ + 2 K+ + SO42─ Æ BaSO4(s) + 2 K+ + 2 Cl─
6. Write the net ionic equation.
* * * * *
Ba2+(aq) + SO42─(aq) Æ BaSO4(s)
* * * * *
Flashcards: Add these to your collection. Run each until perfect for 3 days.
But remember, you must also memorize a) a detailed solubility scheme and b) a method to
predict whether mixed ions will precipitate, and what the precipitate formula will be.
One-way cards (with notch)
Back Side -- Answers
Chemical equations must be balanced for
Atoms and charge
Net ionic equations leave out
Spectator ions
When will mixing two soluble ionic compounds
produce a precipitate?
If a new possible combination is insoluble
When will mixing two soluble ionic compounds
not result in a precipitate?
If both new possible combinations are soluble
Two-way cards (without notch):
Ion combinations with these positive ions will
always be soluble
(alkali metals)+ , NH4+
Ion combinations with these negative ions will
always be soluble
nitrate, acetate, chlorate, perchlorate
Practice B
1. For these, write total and net ionic equations. Try using the 3-line method.
a. Ferric nitrate and rubidium hydroxide Æ
b. MgSO4(aq) and BaCl2(aq) Æ
2. Combining solutions of magnesium chloride and sodium sulfate,
a. what will be the names of the two new combinations that are possible?
b. Which of the new combinations are soluble in aqueous solutions?
c. Which of the new combinations will precipitate?
3. When these solutions are mixed:
NiBr2(aq) + K3PO4(aq)
a. What are the solid formulas for the two new combinations that are possible?
b. Will a precipitate form in the mixture? If so, what is its name and formula?
c. Write the total and the net ionic equations for the reaction.
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Module 13 — Ionic Equations and Precipitates
ANSWERS
( Assume the state is (aq) if not shown.)
Practice A
Problem 1
Solid
Formula
1. AgNO3
Dissolved Ions
before mixing
Æ
Ag+ + NO3─
2. Na2CrO4 Æ 2 Na+ + CrO42─
Mixed – Possible
new combinations
Possible
Precipitates
2 Ag+ + CrO42─ Æ Ag2CrO4 (Insoluble, ppt.)
Na+ + NO3─
Å NaNO3 (Soluble, does not ppt.)
Total ionic equation:
2 Ag+ + 2 NO3─ + 2Na+ + CrO42─ Æ Ag2CrO4(s) + 2 Na+ + 2 NO3─
2 Ag+(aq) + CrO42─(aq) Æ Ag2CrO4(s)
Net ionic equation:
Problem 2
Solid
Formula
Dissolved Ions
before mixing
Potassium Hydroxide Æ K+ + OH─
Cobalt Nitrate
Æ
Co2+ + 2 NO3─
Mixed – Possible
new combinations
Possible
Precipitates
K+ + NO3─ Å KNO3 (Soluble, does not ppt.)
Co2+ + 2 OH─ Æ Co(OH)2 (Insoluble, ppt.)
Total ionic equation:
2 K+ + 2 OH─ + Co2+ + 2 NO3─ Æ Co(OH)2(s) + 2 K+ + 2 NO3─
Net ionic equation:
Co2+(aq) + 2 OH─(aq) Æ Co(OH)2(s)
Practice B
1a.
{ Fe3+ + 3 NO3─}
+
{ Rb+ + OH─ } Æ { Fe3+ + 3 OH─ } + { Rb+ + NO3─ }
Insoluble
Soluble
Fe(OH)3(s)
(That’s the rough draft. To write total and net ionic equations, remove brackets and adjust coefficients so
that all atoms and charges balance.)
Total ionic: Fe3+ + 3 NO3─ + 3 Rb+ + 3 OH─ Æ Fe(OH)3(s) + 3 Rb+ + 3 NO3─
Net ionic equation: Fe3+(aq) + 3 OH─(aq) Æ Fe(OH)3(s)
1b.
{ Mg2+ + SO42─ } +
{ Ba2+ + 2 Cl─} Æ { Mg2+ + 2 Cl─ }
Soluble
+
{ Ba2+ + SO42─ }
Insoluble
BaSO4(s)
Total ionic: Mg2+ + SO4─ + Ba2+ + 2 Cl─ Æ Mg2+ + 2 Cl─ + BaSO4(s)
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Module 13 — Ionic Equations and Precipitates
NET ionic equation:
Ba2+(aq) + SO4─(aq) Æ BaSO4(s)
2. a. Magnesium sulfate and sodium chloride.
3.
b. Both
c. Neither
{ Ni2+ + 2 Br─ } + { 3 K+ + PO43─} Æ { 3 Ni+2 + 2 PO43─} + { K+ + Br─ }
Insoluble
Soluble
Ni3(PO4)2(s)
a. KBr and Ni3(PO4)2
b. Yes. Ni3(PO4)2 (nickel phosphate)
c. Total: 3 Ni2+ + 6 Br─ + 6 K+ + 2 PO43─ Æ 1 Ni3(PO4)2(s) + 6 K+ + 6 Br─
Net ionic equation: 3 Ni2+(aq) + 2 PO43─ (aq) Æ
Ni3(PO4)2(s)
(Note how the coefficients are balanced above. When writing each combination at each stage, adjust
coefficients to balance the combination for atoms and charge. Then, for the total ionic equation, adjust the
coefficients again to balance each ion combination plus all of the atoms and charges on both sides.
If the coefficients do not balance easily, it may indicate an error, and you should double-check your work.)
* * * * *
Lesson 13D: Precipitation and Gravimetric Calculations
Timing: If there is a gap between the time that you are asked to predict which ion
combinations precipitate, and when you are assigned calculations involving precipitates,
delay this lesson until precipitation calculations are assigned.
* * * * *
Gravimetric Calculations
Stoichiometry answers how much of the reactants are used up and products are formed in
chemical reactions. Once the moles of any one reaction component are determined, moles of
other components can be solved using simple-whole-number ratios: the coefficients of the
balanced equation.
In some precipitation calculations, volume and concentration are used to solve for moles.
In others, precipitates are dried and weighed to determine moles; a process termed
gravimetric because the moles of precipitate are determined by its weight, which is the
attraction of its mass by gravity.
Precipitation/Gravimetric Stoichiometry
Precipitation is a reaction. In most respects, precipitation/gravimetric calculations follow
the same rules for reaction stoichiometry that you used in Lessons 12C, but in two ways
they often differ.
1. In most previous equations to balance, you were supplied with the formulas of the
reactants and products.
In precipitation calculations, you will often be asked to both predict all of the product
formulas and identify products that precipitate.
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Module 13 — Ionic Equations and Precipitates
2. Precipitation calculations can involve measurements for both a solid formula and the
ions formed when the solid dissolves in water.
To handle these differences for precipitation reactions, we will add these special steps to
our stoichiometry process.
In precipitation calculations, do stoichiometry steps 3 and 4 as follows.
3. Balance the equation.
a. Write and balance a precipitation equation in three parts:
Reactants in solid formulas Æ Reactants as separated ions Æ Products
b. As you write the product formulas, decide whether a new combination will be
soluble and separated, or insoluble and precipitate. Use one of the methods
for predicting product formulas and precipitates in Lesson 13C.
In the products, write solid formulas for insoluble precipitates and separatedions for combinations that are soluble.
c. Once the precipitate formula is identified, in the WANTED and/or DATA,
replace ppt. with the formula for the precipitate. If this results in grams of a
formula in WANTED or DATA, add the molar mass of the ppt. to the DATA.
4: Write the mol-mol bridge.
In the mol-mol bridge, include all chemical formulas that are written after units in
the WANTED and DATA. This may result in three or more terms with
coefficients, moles, and chemical formulas that are equal, instead of two.
Cover the answer below, then apply the steps to the following problem. If you get stuck,
read part of the answer below, then try again.
Q. If 14.0 mL of Pb(NO3)2 solution is reacted with excess MgCl2 solution, and the
resulting precipitate when dried weighs 0.314 grams, what was the original [Pb2+]?
* * * * *
Answer
1. WANT:
=
? mol Pb2+
L Pb2+ soln.
(a ratio unit is WANTED)
Concentration is a ratio that is constant in a well-mixed solution. The ratio for any
sample will be the same as the ratio for the whole solution. We can easily measure the
liters in a sample, but we cannot count the moles of ions in a sample directly.
However, if we precipitate the ions, weigh the precipitate, and identify the precipitate
formula, we can find the moles of ions in the original sample. We can then divide to
find the mol/L ratio for the sample, and that is the WANTED unit. Since weighing the
dried precipitate is one step, this is a gravimetric calculation.
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Module 13 — Ionic Equations and Precipitates
2. DATA:
14.0 mL Pb(NO3)2 soln. = 0.314 g ppt.
(Equivalent: 2SUA-R)
In the calculation of a WANTED ratio based on a reaction, two DATA amounts will be
equivalent in the reaction (Lesson 12C).
The MgCl2 is in excess. Amounts of reactants in excess can be ignored in the DATA,
since they do not determine the amounts of reactants used up or products formed
(Lesson 10F).
3. Balance. Since this is a precipitation reaction, apply the special steps above.
a. Write the balanced equation for precipitation in 3 parts: Reactants as solid
formulas, reactants in separated-ions, and products as either precipitates or as
separated, soluble spectator ions, depending on their state in the products.
Solid Reactants:
1 Pb(NO3)2 + 1 MgCl2 Æ
Separated Reactants:
Æ 1 Pb2+ + 2 NO3─ + 1 Mg2+ + 2 Cl─ Æ
Products:
Æ 1 PbCl2(s) + 1 Mg2+ + 2 NO3─
Adjust coefficients so that all three parts balance for atoms and charge.
b. Decide the precipitate formula.
Upon mixing the two solutions, two new ion combinations can form.
•
•
•
One new combination is lead ion with chloride ion. By the solubility rules in
Lesson 13A, the combination of Pb2+and Cl─ is insoluble. A possible new
combination that is insoluble will precipitate, so PbCl2 will precipitate.
The other new possible combination is Mg2+and NO3─ ions. All
combinations that include nitrate ion are soluble. Write these ions as
separated on the products side.
The only precipitate is therefore PbCl2. The ions that do not precipitate are
written as separated ions in the products.
c. In the DATA, replace ppt. with its solid formula:
DATA:
14.0 mL Pb(NO3)2 soln. = 0.314 g ppt. PbCl2(s)
(Equivalent)
Now that the DATA includes grams of a formula, what can be added to the DATA
that will likely be needed to solve?
* * * * *
278.1 g PbCl2 = 1 mol PbCl2
(grams prompt)
4. Bridge conversion. In precipitation reactions, write mole ratios between all of the
chemical formulas which are written after units in the WANTED or DATA.
In this problem, the WANTED and DATA involve measurements of three chemical
formulas: Pb2+, Pb(NO3)2, and PbCl2. Write the bridge ratios for those three formulas
using the coefficients in the balanced equations above.
* * * * *
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Module 13 — Ionic Equations and Precipitates
Bridge: 1 mol Pb2+(aq) = 1 mol Pb(NO3)2(aq) = 1 mol PbCl2(s)
Any two of those three terms can be used as a bridge conversion.
* * * * *
5. SOLVE. For reaction calculations in which the answer unit is a ratio, our strategy is to
find a value for the top and bottom WANTED units separately, then divide (see Lesson
12C). The steps are
a. Label the DATA equality that has two amounts that are equivalent at the
endpoint as “Equivalent” or “2SUA-R.”
b. Solve for the WANTED top or bottom unit that is not moles first, using as a given
the side of the equivalency that measures the WANTED formula or sample.
c. Solve for moles WANTED using single-unit stoichiometry. Start with the other
half of the equivalency as the given.
c. Divide the two WANTED amounts to find the WANTED ratio.
* * * * *
The WANTED unit that is not moles is L Pb2+. Pick one of the terms in the equivalency
to convert to Pb2+. Which one?
In stoichiometry, pick the term in the equivalency that measures the WANTED
substance or sample. Neither of the two equivalent amounts has the formula Pb2+.
However, the labels Pb(NO3)2 and Pb2+ are simply two ways to identify the same
solution: one label has the nitrate spectator ions included, and the other does not. The
mL volume of Pb(NO3)2 is in the equivalency, so use that term as your given.
? L Pb2+ soln. = 14.0 mL Pb(NO3)2 soln. = 0.0140 L Pb(NO3)2 = 0.0140 L Pb2+ soln.
An good general rule is:
If you get stuck, add more complete labels to the WANTED and DATA.
In this case, by labeling all of the volume units as soln., it is a indication that they can be
related. Now solve for the other WANTED unit and then check below.
* * * * *
Starting from the other half of the equivalency as given,
? mol Pb2+ = 0.314 g PbCl2 ● 1 mol PbCl2 ● 1 mol Pb2+ = 1.129 x 10─3 mol Pb2+
278.1 g PbCl2 1 mol PbCl2
Finish the problem, and then check below.
* * * * *
The sample of Pb2+ solution was found to have 1.129 x 10─3 mol Pb2+ in 0.0140 L Pb2+
soln. Knowing those amounts, we can find the WANTED ratio.
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Module 13 — Ionic Equations and Precipitates
─3
2+
? mol Pb2+
= 1.129 x 10 mol Pb
=
2+
2+
L Pb soln.
0.0140 L Pb soln.
0.0806 mol Pb2+
L Pb2+ soln.
* * * * *
Flashcards: Add these to your collection. Run each until perfect for 3 days, then move
them to stack 2 (see Lesson 6E).
One-way cards (with notch)
Back Side -- Answers
In gravimetric or precipitation calculations,
what special rule applies to the balance step?
Write solid reactants Æ separated reactants
Æ products
In gravimetric or precipitation calculations,
when you identify the ppt. formula…
If grams ppt is in WANTED or DATA,
add the molar mass to the DATA
Practice:
Memorize the special rules for precipitate stoichiometry, then apply them to
these problems. If you get stuck, read a part of the answer and try again.
1. A 25.0 mL sample of K2SO4 solution is reacted with an excess amount of BaCl2 solution.
Assuming that all of the sulfate ions precipitate, if the mass of the dried precipitate is
1.167 grams, what was the original [K2SO4]?
2. 50.0 mL of a solution containing Pb2+ requires 12.0 mL of a 0.200 M NaCl solution to
precipitate all of the lead ions. What was the original [Pb2+]?
ANSWERS
1. 1.
WANTED:
2. DATA:
? mol K2SO4
L K2SO4 soln.
25.0 mL K2SO4 soln. = 1.167 g ppt.
(Equivalent – 2SUA-R)
(The problem is about a chemical reaction. The WANTED and DATA involve 2 substance (the ppt.
being one.) That’s the stoichiometry prompt. All stoichiometry begins with the same 4 steps.)
3. Balance.
a. (In precipitation, balance in 3 parts)
Solid Reactants:
Separated Reactants:
Products:
1 K2SO4(aq) + 1 BaCl2(aq) Æ
Æ 2 K+(aq) + 1 SO42─(aq) + 1 Ba2+(aq) + 2 Cl─(aq) Æ
Æ 1 BaSO4(s) + 2 K+(aq) + 2 Cl─ (aq)
b. (Identify the precipitate. Since KCl is soluble, the ppt. must be BaSO4.
c. Adjust the DATA.)
DATA:
25.0 mL K2SO4 soln. = 1.167 g ppt. BaSO4(s)
(equivalent – 2SUA-R)
(What additional key information can now be added to the DATA?
* * * * *
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Module 13 — Ionic Equations and Precipitates
233.4 g BaSO4 = 1 mol BaSO4
4. Bridge.
(grams of a formula = g prompt)
1 mole K2SO4 = 1 mole BaSO4
(K2SO4 and BaSO4 are the only formulas that are written after units in the WANTED or DATA.
(If needed, adjust your work and finish from here.)
* * * * *
5. SOLVE. (Since the answer unit is a ratio, find amounts for the two WANTED units separately.
a. Label the equivalency in the DATA.
b. First find the WANTED unit that is not moles, choosing as a given the side of the equivalency that
measures the formula WANTED.)
? L K2SO4 = 25.0 mL K2SO4 = 0.0250 L K2SO4
c. (Find moles WANTED using the other half of the equivalency as a given.)
* * * * *
? mol K2SO4 = 1.167 g BaSO4 ● 1 mol BaSO4 ● 1 mol K2SO4 = 5.000 x 10─3 mol K2SO4
233.4 g BaSO4 1 mol BaSO4
d. (Find the final WANTED unit.)
* * * * *
─3
? mol K2SO4 = 5.000 x 10 mol K2SO4
L K2SO4
0.0250 L K2SO4 solution
2. 1. WANT:
2. DATA:
=
0.200 mol K2SO4
L K2SO4
? mol Pb2+ =
L Pb2+ soln.
50.0 mL Pb2+ soln = 12.0 mL NaCl soln.
0.200 mol NaCl = 1 L NaCl soln.
(equivalent – 2SUA-R)
(M prompt – a ratio)
3. Balance.
a. (Write the balanced equation for precipitate formation in three parts: solid, separated, products.)
1 Pb2+ + 2 NaCl Æ 1 Pb2+ + 2 Na+ + 2 Cl─ Æ 1 PbCl2(s) + 2 Na+
(The anion that was initially combined with lead ion is not known, but it is not needed as long as
the above equations are balanced for atoms and charge.
b. Identify: The precipitate must be PbCl2.
c. Adjust: Since the ppt. amount is not measured in the DATA, there is no DATA to adjust.)
4. Bridge. (The WANTED and DATA include measurements (units and formulas) for only 2 substance
formulas: Pb2+ and NaCl.)
Bridge: 1 mol Pb2+ = 2 mol NaCl
(If needed, adjust your work and solve from here.)
* * * * *
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Module 13 — Ionic Equations and Precipitates
5. SOLVE. (Since the answer unit is a ratio, find amounts for the two WANTED units separately.
a. Label the equivalency in the DATA.
b. First find the WANTED unit not moles, choosing as a given the side of the equivalency that
measures the formula WANTED.)
? L Pb2+ soln. = 50.0 mL Pb2+ = 5.000 x 10─2 L Pb2+
(mL to L, divide by 1,000)
c. (Using other half of equivalency as given, find moles WANTED using stoichiometry steps.)
? mol Pb2+ = 12.0 mL NaCl ● 1 L ● 0.200 mol NaCl ● 1 mol Pb2+ = 1.200 x 10─3
103 mL
1 L NaCl
2 mol NaCl
mol Pb2+
d. (Find the final WANTED unit.
Above, in the sample of the WANTED solution, you found 1.20 x 10─3 mol Pb2+ was in 0.050 L Pb2+.)
? mol Pb2+
L Pb2+
=
1.200 x 10─3 mole Pb2+
5.000 x 10─2 L Pb2+
= 0.0240 M Pb2+
* * * * *
Summary: Solubility and Precipitation
1. Most ionic solids can be characterized as soluble or insoluble in water. The rules for
solubility must be memorized, and exceptions occur. Some frequently used rules are
a. compounds containing alkali-metal atoms, and NH4+, NO3─ , CH3COO─, ClO3─ ,
and ClO4─ ions are nearly always soluble in water;
b. except in the above cases, compounds containing Pb2+ , Hg22+, Ag+, CO32─,
PO43─ , S2─, or CrO42─ ions will generally be insoluble in water.
2. When two solutions of soluble ions are mixed,
•
if a new combination is possible that is insoluble, it will precipitate;
•
if a new combination is soluble, its ions will remain separated.
3. Total ionic equations show all of the ions and precipitates present when two soluble
solutions are mixed. Net ionic equations include the ions and solids that change in a
reaction, and omit spectator ions that do not change.
4. In precipitation calculations:
a. Write a balanced equation for precipitate formation in 3 parts.
Reactants as solid formulas Æ Reactants as separated ions Æ Products
b. Use solubility rules to determine which ions precipitate.
c. Use stoichiometry steps to solve.
5. In general, to understand the reactions of ionic compounds,
•
Re-write the reactants in their separated-ions format;
•
Look for reactions among the new possible ion combinations.
# # # #
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Module 14 — Acid-Base Neutralization
Module 14 — Acid-Base Neutralization
Timing: This module covers acid-base neutralization. Other acid-base topics, including
pH, weak acids and bases and buffers are covered in Modules 27 - 29.
There are two types of acid-base neutralization calculations that may be assigned in your
course.
•
If all of the neutralization calculations that you are assigned supply chemical
formulas for the products in the reactions, you do not need to do Module 14. This
type of calculation was covered in Lesson 12C.
•
If your course assigns in which you must predict the product formulas for acid-base
neutralization, complete Module 14.
* * * * *
Lesson 14A: Ions In Acid-Base Neutralization
Prerequisites: Lessons 7B and 7C.
Pretest: If you can do all of the following problems, you may skip Lessons 14A and 14B.
Check answers at the end of this lesson.
Assuming complete neutralization of all protons and basic ions, write product formulas
and balance these.
a.
KHC2O4 +
KOH Æ
b.
H3Unk +
Mg(OH)2 Æ
c.
HCO3─ +
OH─ Æ
* * * * *
Terminology
Many substances can be classified as acids or bases (and some can be both). There are a
variety of definitions for acids and bases, with each definition helpful in certain kinds of
reactions.
In an acid-base neutralization reaction, an acid and a base are the reactants. When they
are mixed together, they react, and both are used up. Both the acid and the base are said
to be neutralized.
For the limited purpose of studying neutralization, we will define
•
an acid as a substance that creates H+ ions when dissolved in water, and
•
a base as a substance that can react with (use up) H+ ions.
Because strong acids and bases can react with many substances, they are often termed
corrosive: they may damage a surface, metal, or fabric. When an acid or base is neutralized,
its most reactive ions are used up, and its corrosive power is reduced.
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Module 14 — Acid-Base Neutralization
Understanding Neutralization
As substances, acids and bases may be ionic or covalent compounds. However, when
dissolved in water, however, acids and bases react with the water to form ions. A key to
understanding neutralization is to write the acid and base reactants as separated ions.
Acids
When they are dissolved in water, acids form H+ ions plus other ions. The ions formed
will always be present in ratios that guarantee electrical neutrality.
A strong acid is one that separates essentially 100% into ions when it dissolves in water.
Example – Nitric acid: 1 HNO3
H2O
Æ
Other frequently encountered strong acids include
1 H+ + 1 NO3─
(goes ~100%)
•
HCl , hydrochloric acid, ionizes in water to form an H+ ion and a chloride ion (Cl─).
•
H2SO4 , sulfuric acid, is used in many car batteries. It is termed a diprotic acid
because each neutral H2SO4 molecule can release two H+ ions.
Strong acids can neutralize both strong and weak bases.
Weak acids ionize only slightly in water. An example is CH3COOH (acetic acid), the active
ingredient in vinegar. Acetic acid ionizes slightly in water to form one H+ ion and one
acetate anion (CH3COO─).
The four italicized acid names above are encountered frequently. Their names and formulas
should be memorized.
Bases
Strong bases can neutralize both strong and weak acids. Strong bases often contain
hydroxide ions. Examples include
•
NaOH (sodium hydroxide);
•
KOH (potassium hydroxide); and
•
Ca(OH)2 (calcium hydroxide), which ionizes to form two hydroxide ions.
Compounds that contain carbonate ions (CO32─) are also bases. Carbonates are weaker
bases than hydroxides, but carbonate ions are strong enough as bases to neutralize both
strong and moderately weak acids. Examples of carbonates include
•
Na2CO3 (sodium carbonate) and
•
CaCO3 (calcium carbonate).
Hydrogen carbonate ions (HCO3─, also called bicarbonate ion) can also react as a base, but
the HCO3─ ion is a weaker base than carbonate and hydroxide ions, and it reacts only with
only acids that are relatively strong.
Many other molecules and ions can act as bases, but those containing OH─, CO32─, or
HCO3─ ions are the bases most frequently used in acid-base neutralization.
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Module 14 — Acid-Base Neutralization
Electrolytes
Because acids and bases form ions when they dissolve in water, they are termed
electrolytes: their solutions can conduct electricity. Compounds that ionize close to 100%
are called strong electrolytes. Strong acids and strong bases are strong electrolytes.
The Structure of H+: A Proton
Neutral hydrogen atoms contain one proton and one electron. (About 1% of H atoms also
contain one or two neutrons, but the neutrons have no impact on the types of reactions in
which hydrogen participates.)
An H+ ion is a hydrogen atom without an electron, so H+ ions behave chemically as a
single proton. The terms H+ ion and proton are often used interchangeably to describe the
active particle in an acid.
Memorize: acid ion = H+ = proton.
Identifying Acids and Bases
From a chemical formula, how can you tell whether a substance will be an acid or a base?
y Compounds that ionize in water to form H+ ions are acids.
y Compounds that contain OH─, CO32─, or HCO3─ ions can act as bases.
To predict whether a particle formula will be an acid or a base, write the equation for
the formula separating into ions. Most of the ions should be familiar, and the equation
must be balanced for atoms and charge.
Practice A:
For the reaction in which the compounds below ionize in water, write
balanced equations showing the ions formed, then label each initial reactant as an acid or a
base. Answers are at the end of the lesson. For help, review Lesson 7C.
a. LiOH
H2O
b. HBrO4
Æ
Æ
c. HCN Æ
d. Na2CO3 Æ
e. Ca(OH)2 Æ
f. NH4OH Æ
Identifying Acidic and Non-Acidic Hydrogens
Hydrogen atoms in compounds can be divided into two types.
•
Acidic hydrogens are generally defined as those that react with hydroxide ions, and
•
non-acidic hydrogens are those that do not.
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Module 14 — Acid-Base Neutralization
Compounds often contain both acidic and non-acidic hydrogens.
For example, in CH3COOH (acetic acid), the H atom at the end of the formula reacts
with NaOH, but the other three H atoms do not. The H at the end is the acidic
hydrogen. The other H atoms are not acidic hydrogens.
In a substance formula, how can you predict which hydrogens in a compound will be
acidic, and which will not? In general, the rules are:
•
If one or more H atoms is written toward the front of a formula, while others are
not, the H atoms at the front will be acidic, the others will not.
Example: Acetic acid is often written as HC2H3O2. Only the H in front is
acidic.
•
The H at the end of a ─COOH (also written ─CO2H) functional group is acidic.
Examples:
In C6H5COOH, only the H at the end is acidic.
In C3H7CO2H, only the H at the end is acidic.
•
If H is the second atom in the formula, written after a metal atom and before other
atoms, it is acidic.
Examples:
In KHC8H4O4, the first H (and only the first) is acidic.
In NaH2PO4, the two H’s after the metal are acidic.
•
If a substance with only one H reacts with hydroxide ion, the H is acidic.
Practice B
1. Draw an arrow toward and count the acidic hydrogens in these compounds.
a. NaH2PO4
d. H3AsO4
b. C6H5COOH
e. KHC8H4O4
c. H2C4H4O6
f. NaHSO4
2. Fill in the blanks to show the number of protons formed when these compounds ionize
in water.
a. C3H7CO2H Æ ___ H+
b. HC2H3O2 Æ ___ H+
c. NaH2(C3H5O(COO)3) Æ ___ H+
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Module 14 — Acid-Base Neutralization
ANSWERS
Pretest: Coefficients of 1 may be omitted as understood.
a. 1 KHC2O4 + 1 KOH Æ 1 H2O + 1 K2C2O4
b. 2 H3Unk + 3 Mg(OH)2 Æ 6 H2O + 1 Mg3(Unk)2
c. HCO3─ + OH─ Æ H2O + CO32─
Practice A
a. LiOH Æ Li + + OH─ Base
b. HBrO4 Æ H+ + BrO4─ Acid
c. HCN Æ H+ + CN─
d. Na2CO3 Æ 2 Na+ + CO32─ Base
Acid
e. Ca(OH)2 Æ Ca2+ + 2 OH─
f. NH4OH Æ NH4+ + OH─ Base
Base
Practice B
1. a. NaH2PO4 Two
d. H3AsO4
2
b. C6H5COOH
Three
e. KHC8H4O4
a. C3H7CO2H Æ 1 H+
One
One
c. H2C4H4O6 Two
One
f. NaHSO4
b. HC2H3O2 Æ 1 H+
c. NaH2(C3H5O(COO)3) Æ 2 H+
* *
*
* *
Lesson 14B: Balancing Hydroxide Neutralization
Prerequisites: Lesson 14A.
* * * * *
Hydroxide Neutralization
In the reaction of an acid and a hydroxide, the neutralization reaction forms water (which
can be written as H─OH or HOH or H2O).
H+ + OH─ Æ
H-OH(l)
The reaction of an acid and a hydroxide is driven to completion by the formation of water: a
low potential energy molecule. Whenever a physical or chemical system can go to lower
potential energy, there is a strong tendency to do so.
The reaction of acids with compounds containing hydroxide ions can be represented by the
general equation
An acid + a base containing OH─ Æ
H2O + a salt
(1)
A typical acid-hydroxide neutralization is the reaction of hydrochloric acid with sodium
hydroxide to form sodium chloride (also known as table salt).
HCl(aq) + NaOH(aq)
©2008 ChemReview.net v. f9
Æ
H-OH(l) + NaCl(aq)
(2)
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Module 14 — Acid-Base Neutralization
The state (aq) means aqueous (dissolved in water). Most (but not all) acid-base reactions are
carried out in water. In the acid-base reaction equations in these lessons, you may assume
that the water is liquid and the formulas for other compounds and ions are aqueous (aq)
unless otherwise noted. Writing water in the form H─OH form helps to emphasize the
reaction that occurs between acids and hydroxide ions.
Equation (2) above is one way that this reaction is represented. However, both HCl and
NaOH, when dissolved in water, separate completely into ions. The table salt in the
solution after the reaction also exists as separate ions of Na+ and Cl─. Re-writing the
equation to show the separated ions that actually exist in the solution, the reaction is
H+ + Cl─ + Na+ + OH─ Æ
H─OH(l) + Na+ + Cl─
(3)
Note in reaction (3) that the sodium and chloride ions are spectators: they do not change
during the reaction. The term salt in reaction (2) refers to the combination of spectator ions
that are the products, in addition to water, when the acid and hydroxide ion react.
Balancing Neutralization
If formulas are supplied for all the reactants and products, neutralization equations can be
balanced by trial and error using the methods in Lesson 10B.
However, in acid-base neutralization problems, often the product formulas are not
supplied. In these problems, you may be asked to
•
predict formulas for all of the products of the reaction, or
•
add the coefficients for the acid and base without knowing the products.
The following rules will solve these problems.
Predicting All Products of Hydroxide Neutralization
In neutralization, by separating the solid (molecular) reactant formulas into ions, you can
predict the products and balancing the equation.
In hydroxide neutralization, the key rule is
If an acid reacts with OH─, one of the products is H─OH.
Knowing H─OH is one product, you can often predict the other product formula(s).
Complete the steps below in your notebook.
Q. Write a balanced equation for the complete neutralization of H3PO4 (phosphoric
acid) by Ca(OH)2 (calcium hydroxide).
Steps:
1. Write the acid and base reactants in their solid (molecular) formulas.
2. On the line below, re-write each reactant inside parentheses ( ) in its separated-ion
format.
3. After the reactants, add “ Æ H─OH + ______ + ______”.
* * * * *
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Module 14 — Acid-Base Neutralization
Solid:
H3PO4 +
Ca(OH)2 Æ
Separated: ( 3 H+ + PO43─ ) +
( Ca2+ + 2 OH─ )
Æ H-OH + ______ + ______
4. Add lowest-whole-number coefficients in front of the parentheses ( ) to balance the H+,
OH─, and H-OH.
* * * * *
Separated: 2 ( 3 H+ + PO43─ ) + 3 ( Ca2+ + 2 OH─ ) Æ 6 H-OH + ____ + ____
^
^
^
[ ( 6 H+ )
( 6 OH─ ) Æ ( 6 H-OH ) ]
The total H+ ions must equal the total OH─ ions must equal the total H2O.
5. Add in the products side blanks the formulas and coefficient totals for the ions on the
left that are not H+and OH─.
6. Move the left side coefficients that are in front of the ( ), plus the coefficient of the water,
up to the top line.
7. Finish adding coefficients to the top equation. To write a molecular formula for the
remaining product, use the rules for writing ionic solid formulas in Lesson 7C.
*
* * * *
Solid:
2 H3PO4 + 3 Ca(OH)2
Æ
6 H2O
+ 1 Ca3(PO4)2(s)
Separated: 2 ( 3 H+ + PO43─ ) + 3 ( Ca2+ + 2 OH─ ) Æ 6 H2O + 2 PO43─ + 3 Ca2+
In most neutralization reactions, the ions that are not H+ and OH─ will simply remain
as aqueous “spectator ions,” dissolved in the final solution. In this problem, however,
calcium and phosphate ions are an insoluble combination. If a combination can form
that is insoluble, it will precipitate. In this reaction, we have both a neutralization and a
precipitation.
8. Check that the final equation is balanced.
Practice A.
Balance by inspection or the methods above. Assume that the acids and
bases are completely neutralized. Do every other part, and more if you need more practice.
1. Write the product formulas in the solid (molecular) format and balance the equation.
a.
HNO3 +
KOH Æ
b.
KOH +
H2SO4
c.
H2SO4 +
Al(OH)3 Æ
Æ
2. Write reactant and product formulas, then balance the equation. Write final formulas in
the ionic-solid (molecular) format.
a. Barium hydroxide plus sulfuric acid
b. Hydrochloric acid plus magnesium hydroxide
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Module 14 — Acid-Base Neutralization
Balancing Neutralization with Spectators Omitted
In all stable substances and mixtures, the total of all of the positive and negative charges
must add up to zero. However, as a shortcut, spectator ions are often omitted when
representing substances or writing reaction equations. In such cases, the written charges
may not add up to zero.
For example, in the lab, solutions of acids and bases may be labeled as simply H+ or
OH─ to emphasize the reactive particles. In those solutions, however, there must also
be other ions that balance the overall charge.
Similarly, in writing neutralization equations, partial ionic equations may be written that
leave out spectators. In these equations, if some or all spectators are omitted, the total
charge may not be neutral on both sides, but that is considered as an acceptable shortcut
when writing balanced equations.
For a written reaction equation to be considered balanced, what is required is that the
atoms and the total charge be the same on both sides (but not necessarily zero).
To balance a neutralization equation in which some spectators have been left out, use the
same steps as above: break the formulas into familiar ions, one reactant including H+ and
the other including a basic ion, then balance the atoms and total charge on each side.
Try this example.
Q. Write the products and balance:
HSO4─ +
OH─ Æ
* * * * *
• Inside parentheses, break the reactants into familiar ions. Form water on the right.
•
Add coefficients that balance the particles that form water.
•
Balance the particles that don’t react.
•
Make sure that the atoms balance and the total charge is the same in all three parts.
1 HSO4─ + 1 OH─
Æ
1 ( 1 H+ + 1 SO42─ ) + 1 OH─
Æ
1 H-OH + 1 SO42─
Atoms balance, and in all three parts, the total charge is ─2. Balanced!
Practice B.
Check answers as you go. Do every other letter, and more for more practice.
1. Write product formulas and balance the equation. Assume all acids and bases are
totally neutralized. Products may be molecules or ions.
a.
H+ +
b.
HSO4─ +
KOH Æ
c.
KHSO4 +
OH─
d.
CH3COOH +
©2008 ChemReview.net v. f9
Al(OH)3 Æ
Æ
OH─
Æ
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Module 14 — Acid-Base Neutralization
Partial Balancing
To solve most neutralization stoichiometry, we will need only the coefficients for the
reactants: the acid and the base. This partial balancing can be completed if know the
formula for one product, and in acid-hydroxide neutralization, we do. One product is
always H-OH.
For this problem, write H-OH as a product, balance the left side, and then check below.
Q. What is the ratio of reaction for
H2SO4 +
Al(OH)3 Æ
* * * * *
Either balance by inspection (total H+ = total OH─) or by using these steps.
1. Below the two reactants, write the number of acidic and basic ions in the reactants.
2. On the right, add one product of the reaction.
H2SO4
+
( 2 H+ ) +
Al(OH)3
Æ
H─OH + …
(3 OH─ )
Æ
H─OH + …
3. Add the coefficients to balance the atoms and charge.
H2SO4
+
3 ( 2 H+ ) +
Al(OH)3
Æ
2 (3 OH─ )
H─OH + …
Æ 6 H─OH
+ …
4. Transfer the coefficients to the original reaction equation.
3 H2SO4
+ 2 Al(OH)3 Æ 6 H─OH + …
3 ( 2 H+ ) + 2 (3 OH─ ) Æ 6 H─OH + …
The reactant ratios will be all that is needed to solve most neutralization stoichiometry.
Practice C.
Assume all reactants are completely neutralized. Balance by inspection or
using the steps above. Do every other problem, and more if you need more practice.
1. Balance the coefficients of the two reactants.
a.
CsOH +
H2SO4
b.
Ca(OH)2 +
nitric acid Æ
c.
HSO4─ +
OH─
Æ
Æ
2. Add ratios of reaction for these reactants.
a.
CH3COOH
b.
Calcium hydroxide +
c.
NaHC2O4 +
d.
Sodium hydroxide +
©2008 ChemReview.net v. f9
+
NaOH Æ
HC2H3O2 Æ
Al(OH)3 Æ
KHC8H4O4 Æ
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Module 14 — Acid-Base Neutralization
Balancing with an Unknown Formula
In some neutralization calculations, the formula for an acid or the base is not supplied, but
the number of protons in the acid, or basic ions in the base, is provided. Often this
information is all that you need to balance the acid-base ratio. Try this example.
Q. A solid acid has an unknown formula but is known to contain three acidic hydrogens.
What will be the ratio for the reaction of this acid with NaOH?
Check part of the answer below if you need a hint.
* * * * *
Answer
One way to balance is to write the acid formula as H3Unk , where the Unk stands for
unknown, and H3 represents three acidic hydrogens.
The two reactants can then be written as:
H3Unk +
NaOH Æ
Complete the reactant balancing.
* * * * *
H3Unk + NaOH Æ 1 ( 3 H+ ) + 3 ( OH─ ) Æ 3 H─OH + ….
The acid-base ratio must be
1 H3Unk + 3 NaOH Æ
A formula could be written for the other products, but it would be speculation. The acid
anion could remain intact, or it could decay in some fashion. However, for most
neutralization stoichiometry, formulas for the other products are not required. The acidbase ratio on the left side of the equation will be all that is needed to solve.
* * * * *
Summary: Neutralization rules so far.
1. Acid-base neutralization is an ionic reaction. To understand ionic reactions, write
the separated-ion formulas.
2. Ions: Acids contain H+. The reacting particle in acids = H+ = a proton.
Bases include compounds with hydroxide (OH─), carbonate (CO32─), and
hydrogen carbonate (HCO3─) ions.
3. Products: For acids + OH─, one product is water: H-OH.
H+ + OH─ Æ H-OH
4. Balancing: To predict the products and balance the equations,
•
Write the separated-ion formulas in ( ).
•
Write one product.
•
Finish by balancing atoms and charge.
5. To balance when a formula is unknown,
•
If a substance has 2 acidic hydrogens, write its formula as H2Unk.
•
If a base has 3 hydroxides, write Unk(OH)3.
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Practice D: Learn the rules above, then do these problems. Assume that all reactants
are completely neutralized.
1. Supply the ratios of reaction for the two reactants.
a.
H2Unk +
Sr(OH)2 Æ
b.
H2SO4 +
Unk(OH)3 Æ
c.
HCl +
UnkOH Æ
2. Add coefficients to balance the two reactants.
a. An unknown acid with two acidic hydrogens is totally neutralized by potassium
hydroxide.
b. An unknown base with three hydroxide ions is totally neutralized by nitric acid.
ANSWERS
Practice A
Coefficients of one may be omitted. Any coefficient ratios which are the same as these are not incorrect, but
lowest whole-number ratios are preferred.
1
1 HNO3 +
b.
2 KOH + 1 H2SO4
c.
3 H2SO4 + 2 Al(OH)3 Æ
2. a.
b.
1 KOH Æ
1 H2O + 1 KNO3
a.
2 H2O + 1 K2SO4
Æ
1 Ba(OH)2 + 1 H2SO4 Æ
1 Al2(SO4)3 + 6 H2O
2 H2O + 1 BaSO4
2 HCl + Mg(OH)2 Æ 2 H2O + 1 MgCl2
Practice B
1. a. 3 H+ + 1 Al(OH)3 Æ 3 H-OH + Al3+
+ 1 KOH Æ 1 H-OH + 1 K+ + 1 SO42─
c. 1 KHSO4 + 1 OH─ Æ 1 H-OH + 1 K+ + 1 SO42─
d. 1 CH3COOH + 1 OH─ Æ 1 H-OH + 1 CH3COO─
b. 1 HSO4─
Practice C To balance the reactants, write one product of the reaction.
1. a. 2 CsOH + 1 H2SO4 Æ 2 H-OH + …
b. 1 Ca(OH)2 + 2 HNO3 Æ 2 H-OH + …
c. 1 HSO4─ +
1 OH─
©2008 ChemReview.net v. f9
Æ 1 H-OH + …
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Module 14 — Acid-Base Neutralization
2. a. 1 CH3COOH + 1 NaOH
b. 1 Ca(OH)2 + 2 HC2H3O2
Æ 1 H2O + …
Æ
2 H2O + …
c. 3 NaHC2O4 + 1 Al(OH)3 Æ 3 H2O + …
d. 1 NaOH + 1 KHC8H4O4
Æ 1 H2O + …
Practice D
1. a. 1 H2Unk + 1 Sr(OH)2 Æ 2 H-OH + …
b. 3 H2SO4 + 2 Unk(OH)3 Æ 6 H2O + …
c. 1 HCl + 1 UnkOH Æ H-OH + …
2. a. 1 H2Unk + 2 KOH Æ 2 H2O + …
b. 3 HNO3 + 1 Unk(OH)3 Æ 3 H2O + …
* * * * *
Lesson 14C: Acid-Hydroxide Neutralization Calculations
Prerequisites: Lessons 7C, 10E, 11D, 12C, 14A+B.
* * * * *
Terminology
Titration is an experimental technique used to gather the information needed for
stoichiometry calculations. Calibrated burets are used to precisely measure the amount of
solution added as a chemical reaction takes place.
Indicators are dyes used in titration that change color at the instant the moles of two
reacting particles are equal or have reacted in a simple-whole-number ratio. When this
equivalence point is reached, a change in indicator color signals the endpoint of the
titration.
Acid-base titration is simply neutralization in which the amounts of acid and base are
carefully measured.
In an acid-base neutralization, an acid and base are mixed. If one of the two reactants is a
strong acid or base, the reaction will proceed until the limiting reactant is essentially
completely used up. Standard solution stoichiometry methods can then be used to
calculate the amounts of reactants used up and products that form.
In the special case of a titration of an acid with a base that contain hydroxide ions, when the
equivalence point is reached, the moles of H+ ions from the original solution that have
reacted equal the moles of OH─ ions in the original base solution that have reacted.
At an acid-hydroxide endpoint, moles H+ reacted = moles OH─ reacted
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Module 14 — Acid-Base Neutralization
When an acid and a hydroxide ion are mixed, one reaction is always
1 H+ + 1 OH─ Æ
1 H-OH(l)
When titrating a weak acid or base, the opposite solution must be strong, and a careful
selection of the indicator dye will be required to show a sharp endpoint. However, for all
acids and bases, whether one or both is strong, stoichiometry calculations are done using
the same steps.
Neutralization Calculations Based on Partial Balancing
To solve stoichiometry, you do not always need a complete balanced equation. What you
need is the bridge conversion: a numeric ratio between the WANTED and given substances
in the reaction.
In most neutralization calculations, the WANTED and given are the acid and the base
reactants. Their coefficients can be determined by the partial balancing practiced in Lesson
14B. This special rule is:
In neutralization stoichiometry, use partial balancing at Step 3 to write the Step 4 bridge.
Once partial balancing supplies the acid and base coefficients, neutralization calculations
can be solved using the same stoichiometry steps that were practiced in Lesson 12C.
Try this example.
Q.
If 19.29 mL of a 0.120 M KOH solution is needed to completely neutralize 0.1484
grams of an unknown acid, given that acid has three acidic protons, what is molar
mass of the acid?
If you get stuck, read a portion of the answer until you are unstuck, then try again.
* * * * *
Answer: Your paper should look like this.
1. WANT:
? g acid
mol acid
2. DATA:
0.1484 g acid = 19.29 mL KOH
(write the unit WANTED)
(2SUA-R: equivalent at endpt.)
0.120 mol KOH = 1 L KOH
3. Balance.
(For acid-base neutralization, use partial balancing)
1 H3Unk + 3 KOH Æ 3 H2O + …
4. Bridge.
1 mol H3Unk = 3 mol KOH
Rules used above include
•
Reaction calculations with data for two substances are solved by stoichiometry.
•
All stoichiometry starts with the same 4 steps: WDBB.
•
Solving for a ratio unit, all of the DATA will be in equalities.
•
In stoichiometry for a ratio unit, two amounts will be equivalent at the endpoint.
•
In neutralization, use partial balancing at Step 3 to write the Step 4 bridge.
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Module 14 — Acid-Base Neutralization
•
An unknown acid with 3 acidic protons can be written as H3Unk when balancing.
If needed, adjust your work and finish from here.
* * * * *
5. SOLVE.
In stoichiometry, when a ratio unit is WANTED, solve separately for the top and
bottom WANTED amounts in a sample, then divide.
a. Label the DATA equality that has the two amounts that are equivalent at the
endpoint as “Equivalent”or “2SUA-R.”
b. Solve first for the easier unit (the one not moles), using as a given the side of the
equivalency that contains the WANTED substance formula.
In this problem, the WANTED amount not moles is g acid. The grams of acid in the
sample are supplied in the equivalency. That was easy.
? g acid in sample = 0.1484 g acid
c. To find the moles of acid WANTED, start with the other half of the equivalency as
the given.
* * * * *
In this problem, the acid may be written as acid or as H3Unk .
Start:
? mol acid = 19.29 mL KOH (other half of equivalency)
Use single-unit stoichiometry. Convert to the moles of the given that reacted,
? mol WANTED =
unit given
>>
? mol acid = 19.29 mL KOH •
mol given
1 L • 0.120 mol KOH •
1 L KOH
103 mL
* * * * *
Cross the mol/mol bridge to reach the moles of WANTED substance.
? mol WANTED = unit given >>
mol given >> mol WANTED (stop)
? mol acid = 19.29 mL KOH • 1 L • 0.120 mol KOH • 1 mol acid =
1 L KOH
3 mol KOH
103 mL
= 7.716 x 10─4 mol H3Unk in the sample.
* * * * *
Solve for the final WANTED ratio.
? g acid
mol acid
=
g acid in sample
mol acid in sample
=
0.1484 g acid
=
─4
7.716 x 10 mol acid
192 g acid
mol
The molar mass of a substance is constant: a ratio that is characteristic. The concentration
of a well-mixed solution is a ratio that is constant. In cases where a ratio WANTED is
constant, if the amounts of the top and bottom WANTED units in a sample can be found,
that ratio will be true for all samples.
* * * * *
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Module 14 — Acid-Base Neutralization
Stoichiometry and CSI
Consider what we were able to determine in the above calculation.
Many molecules that are important in chemical and biological systems, including legal and
illegal drugs, are acids or bases. In the above problem, we did not know the chemical
formula for the acid. However, by finding the ratio of the grams of acid per mole of the acid
(its molar mass), we can get close to identifying the acid.
The acid can be weighed to find the grams in a sample, but nearly always we do not have a
way to count the number of particles (the moles) in a sample directly. However, for acids
and bases, we can determine the moles in a sample by titrating with an acid or base
solution of a known concentration and applying stoichiometry.
In titration, when the indicator changes color, the moles of two reacting particles are either
equal or in a whole-number ratio.
Knowing the grams and moles in a sample, we can find the grams to moles ratio: the molar
mass. From the molar mass of a substance and its melting point or other physical
constants, tables in chemistry reference media will identify many unknown substances
with near certainty.
The ability to identify an unknown acid or base is a skill that is marketable in forensic
criminology, medicine, and other rewarding careers.
Practice: If needed, review the steps for solving solution stoichiometry in Lesson 12C.
Answers are at the end of this lesson. If you get stuck, read a portion of the answer until
unstuck, then try again.
1. If 124 milliliters of a barium hydroxide solution is needed to completely neutralize 2.34
grams of arsenic acid (H3AsO4), what is the original [Ba(OH)2]?
2. A 2.00 M H3PO4 solution is completely neutralized by 1.50 liters of 0.500 M OH─. How
many mL of the H3PO4 solution were neutralized?
3. Oxalic acid (90.03 g•mol─1) is a solid at room temperature. Each oxalic acid particle
contains two acidic hydrogens. If 44.43 mL of an NaOH solution is needed to
completely neutralize 0.100 g of oxalic acid crystals, find the [NaOH].
4. Review Lessons 14A-C and prepare flashcards that cover the fundamentals.
Take the Paper You Need
In many problems, a methodical approach will require using an entire sheet of paper to
solve each question. That’s OK. Paper recycles. Trees grow. Your understanding will
also grow if you take the time and paper that is needed for careful work.
If you are solving in a graph-paper notebook (a good idea), try working with the paper
in the landscape mode (turned sideways). This provides more room for the long string
of conversions that stoichiometry often requires.
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Module 14 — Acid-Base Neutralization
ANSWERS
1. 1. WANTED:
2. DATA:
? mol Ba(OH)2
L soln
(the unit WANTED is mol/L)
124 mL Ba(OH)2 soln. = 2.34 g H3AsO4
(equivalent at end point)
141.9 g H3AsO4 = 1 mol H3AsO4
(g prompt)
(When stoichiometry WANTS a ratio unit, all of the data will be in equalities, and one equality will be
two amounts that are equivalent at the endpoint.
When reaction calculations involve two 2 substances, that’s the stoichiometry prompt. All
stoichiometry begins with the same 4 steps: WDBB.)
3. Balance. (Using the rules in Lesson 14B,)
2 H3AsO4 + 3 Ba(OH)2 Æ 6 H―OH + …
4. Bridge.
(use partial balancing)
2 mol H3AsO4 = 3 mol Ba(OH)2
5. SOLVE. (Since a ratio unit is WANTED, solve for the top and bottom WANTED amounts separately.
a. Label the equivalency in the data.
b. Solve for the easier WANTED amount first (usually the one not moles), using as a given the side
of the equivalency that measures the WANTED substance.)
? L Ba(OH)2 = 124 mL Ba(OH)2 • 1 L = 0.1240 L Ba(OH)2
103 mL
(move decimal 3 times)
c. (To find the other WANTED unit (usually moles), use single-unit stoichiometry steps. Start with
the other half of the equivalency as your given.)
? mol Ba(OH)2 = 2.34 g H3AsO4 • 1 mol H3AsO4 • 3 mol Ba(OH)2 = 0.02473 mol Ba(OH)2
141.9 g H3AsO4 2 mol H3AsO4
(Carry an extra sf until the final step. Finish, then check below.
* * * * *
d. Solve for the final WANTED unit using the two amounts found above.)
? mol Ba(OH)2 = mol Ba(OH)2 in sample = 0.02473 mol Ba(OH)2 =
L soln
L Ba(OH)2 in sample
0.1240 L Ba(OH)2 soln.
2.
0.200 mol Ba(OH)2
L Ba(OH)2
(For reaction calculations, use stoichiometry steps. Stoichiometry starts with WDBB.)
WANT: ? mL H3PO4
DATA: 2.00 mol H3PO4 = 1 L H3PO4
1.50 L OH─ soln.
0.500 mol OH─ = 1 L OH─
(you want a single unit)
(M prompt)
(the single unit given)
(M prompt)
(Since you have solutions of two different substances, include formulas after all units to identify the
substance that each number and unit measures.
Since all volumes in the problem are for aqueous solutions, you may label one volume as “soln.” but
omit the other “solution” labels as “understood.”)
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Module 14 — Acid-Base Neutralization
Balance:
1 H3PO4 + 3 OH─ Æ 3 H2O + PO43─
Bridge.
1 mol H3PO4 = 3 mol OH─
SOLVE: Since a single unit is WANTED, use single-unit stoichiometry, chaining conversions.)
? mL H3PO4 = 1.50 L OH─ • 0.500 mol OH─ • 1 mol H3PO4 • 1 L H3PO4
• 103 mL =
1 L OH─
3 mol OH─
2.00 mol H3PO4 1 L
= 125 mL H3PO4
3.
WANTED:
? mol NaOH
L NaOH soln
DATA:
44.43 mL NaOH = 0.100 g oxalic acid
(Equivalent at endpoint)
90.03 g oxalic acid = 1 mol oxalic acid
Balance:
Let oxalic acid be H2Unk.
Bridge.
1 H2Unk + 2 NaOH Æ 2 H2O + ….
1 mol H2Unk = 2 mol NaOH
SOLVE.
(Not moles): ? L NaOH = 44.43 mL NaOH = 0.04443 L NaOH
(by inspection)
(moles): ? mol NaOH = 0.100 g H2Unk • 1 mol acid • 2 mol NaOH = 2.221 x 10─3 mol
90.03 g acid
1 mol acid
NaOH
? = mol NaOH =
L NaOH
2.221 x 10─3 mol NaOH =
0.04443 L NaOH soln.
0.0500 mol NaOH
L NaOH
4. Your flashcards might include the following.
One-way cards (with notch)
Names and formulas for 3 strong acids
Back Side -- Answers
Hydrochloric: HCl, sulfuric: H2SO4,
nitric: HNO3
Ion symbol for a proton
H+
At the endpoint of acid-hydroxide neutralization
Moles H+ = moles OH─
Formula for an unknown base with 3 hydroxides
Unk(OH)3
Formula for an unknown with 2 acidic protons
H2Unk
Two-way cards (without notch):
An acid
A substance that produces H+ in water
A base
A substance that neutralizes H+
3 moderate to strong basic ions used in
neutralization
OH─, CO32─, HCO3─
* * * * *
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Lesson 14D: Neutralization Calculations In Parts
Timing: Do this lesson if you are assigned neutralization calculations that include either
measurements for three particles or questions with multiple parts.
* * * * *
Calculations Involving Both Ionized and Un-Ionized Acids or Base Formulas
For neutralization calculations that include data for both un-ionized acid or base formulas
and their reacting ions, you will need to write a relationship between the ionized and unionized particles. Use this special rule.
1. In stoichiometry, when the WANTED and DATA include units that measure both
an solid formula and its ions,
a. write the balanced reaction equation in three parts:
Reactants in molecular formulas Æ Reactants as separated ions Æ Products
b. In the bridge conversion, write all formulas in the WANTED and DATA that
have units attached. (This may result in three or more terms that are equal in
the bridge conversion, instead of two.)
If a unit with a chemical formula attached is written in the WANTED or DATA, it
represents a measured quantity. To solve, mole to mole relationships are needed among all
of the measured quantities. We find those relationships from the coefficients of the
balanced equation.
There are other ways to track the relationship between particles and their ions, but the
above rules have the advantage of being consistent with both the rule for precipitation
reactions and our fundamental rule for reactions that involve ions:
To understand the reactions of ionic compounds, write the reactants as separated ions,
then look for possible reactions in the new mixture of ions.
Use Rule 1 to solve the following.
Q. A sample of 15.0 mL of Ba(OH)2 solution is neutralized by 28.14 mL of
0.100 M HCl. What was the [OH─] in the original base solution?
Write only the first four stoichiometry steps, then check your answer below.
* * * * *
Answer: Your paper should look like this.
1. WANT:
? mol OH─
L OH─
2. DATA:
15.0 mL Ba(OH)2 = 28.14 mL HCl
0.100 mol HCl
©2008 ChemReview.net v. f9
=
(Equivalent at endpt. -- 2SUA-R)
1 L HCl
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Module 14 — Acid-Base Neutralization
3. Balance.
Reactants – molecular: 1 Ba(OH)2 + 2 HCl Æ
Reactants as ions:
Æ 1 Ba2+ + 2 OH─ + 2 H+ + 2 Cl─ Æ
Products:
4. Bridge:
Æ 2 H2O + ….
1 mol Ba(OH)2 = 2 mol HCl = 2 mol OH─
* * * * *
Rules that apply above include
•
In stoichiometry for a ratio unit, two amounts will be equivalent at the endpoint.
•
In acid-base stoichiometry, the WANTED and DATA are nearly always
measurements of the two reactants. You will only need to know one product
formula to find the coefficients for the reactants.
•
In the WANTED and DATA there are three chemical formulas with units in front:
OH─, Ba(OH)2, and HCl. This is a prompt to balance the equation in 3 parts.
An additional prompt is that the OH─ ion must come from the Ba(OH)2. When the
WANTED and DATA include both an un-ionized formula and one or more of its
ions, write the balanced equation in 3 parts: solid formulas, separated ions,
products (in neutralization, one product).
•
In step 4, include in the bridge conversion the moles of all the chemical formulas in
the WANTED and DATA that were listed after units of measure. This problem has
three such formulas.
Now solve for the WANTED unit.
* * * * *
To solve a reaction calculation for a ratio, solve for top and bottom WANTED units
separately, then divide.
a. Label the DATA equality that has two amounts equivalent at the endpoint as
“Equivalent”or “2SUA-R.”
b. Solve first for the easier unit (the one not moles), using as a given the side of the
equivalency that measures the WANTED chemical formula.
In this problem, the WANTED unit not moles is L OH─. In the DATA equivalency
is mL of Ba(OH)2 solution. Since the solution labeled Ba(OH)2 is the solution that
contains the OH─ ions, the Ba(OH)2 and OH─ solutions are the same solution. The
moles of Ba(OH)2 and OH─ differ, but the volume of the solution in which those
chemical formulas are found in the same.
? L OH─ soln. = 15.0 mL Ba(OH)2 soln. = 0.0150 L Ba(OH)2 = 0.01500 L OH─
If needed, adjust your work and finish.
* * * * *
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Module 14 — Acid-Base Neutralization
c. Solve for the other WANTED unit using single-unit stoichiometry. Use the other
half of the equivalency as your given. Use the bridge to go from given to WANTED.
? mol OH─ = 24.18 mL HCl ● 1 L HCl ● 0.100 mol HCl ● 1 mol OH─
1 L HCl
2 mol HCl
103 mL
= 1.209 x 10─3 mol OH─
=
(carry an extra sf until the final step)
d. Solve for the final WANTED unit.
* * * * *
? mol OH─ = 1.209 x 10─3 mol OH─ =
L OH─
0.01500 L OH─
0.0806 mol OH─
L OH─
* * * * *
Neutralization Calculations in Parts
When problems contain more than one question about a common set of DATA, the general
rules (from Lesson 11F) included
•
First list the common DATA, then write the WANTED unit for each part. As parts
are completed, consider those answers to be additional DATA.
Similarly, for reaction calculations that have multiple questions, use special rule 2.
2. When stoichiometry contains more than one question about a common set of data,
•
First do stoichiometry steps 2, 3, and 4 (DATA, balance, bridge),
•
then do step 1(WANTED) for each part.
Stoichiometry that has multiple questions will also rely on these previous rules.
•
To solve for single units when all of the data is in equalities, label the DATA as
SUA, R, or 2SUA-R (Lesson 11F). Use single-unit amounts (SUA or 2SUA-R), not
ratios, as a given to solve for single units.
•
To solve for a single unit, try to select a given from the SUA DATA that solves in the
fewest steps. Include as DATA answers from previous steps.
A Shortcut: Moles H+ = Moles OH─
When a stoichiometry problem asks questions in multiple parts, each part can be treated as
a separate new problem in which you “start over” to solve. However, the parts of a
problem can often be solved more quickly by using answers to previous parts plus
fundamental relationships.
A fundamental equality that can be used in acid-hydroxide reactions is special rule 3.
3. In hydroxide neutralization, some steps may be solved by inspection using this rule:
At the equivalence point, the moles of H+ in the original acid sample equal the moles
of OH─ in the original base sample.
At the endpoint of acid-OH─ titration, mol H+ reacted = mol OH─ reacted
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Module 14 — Acid-Base Neutralization
Use special rules 1, 2, and 3 to solve the following problem.
Q. KH(IO3)2 (potassium hydrogen iodate – 389.9 g/mol) is a primary standard solid
acid that can be used to determine the concentration of base solutions. In such a
titration, the indicator changes color when 19.29 mL of a KOH solution has been
added to 0.338 grams of KH(IO3)2.
A. How many moles of H+ ions were in the acid sample?
B.
C.
* * * * *
Answer
How many moles of hydroxide ions were in the added base sample?
Find the [KOH].
2. DATA:
389.9 g KH(IO3)2 = 1 mol KH(IO3)2
19.29 mL KOH = 0.338 g KH(IO3)2
(R)
(2SUA-R - equivalent at endpt.)
3. Balance:
Solid formulas:
1 KH(IO3)2 + 1 KOH Æ
Separated formulas:
Æ 1 K+ + 1 H+ + 2 IO3─ + 1 K+ + 1 OH─ Æ
Products:
4. Bridge:
Æ 1 H2O + ….
1 mol KH(IO3)2 = 1 mol KOH = 1 mol H+ = 1 mol OH─
The above uses
Partial balancing with one product will find the neutralization bridge conversions.
Rule 3: For stoichiometry that asks several questions, first complete steps 2, 3 and 4, then
do step 1 for each part.
Rule 1: This problem and its parts includes measurements for four particles: KH(IO3)2 ,
KOH, H+, and OH─. When the WANTED and DATA includes both solid/molecular
formulas and ions, write the balanced equation in three parts: reactants in molecular
(solid) formulas, reactants as separated-ions, then products using formulas for the
substances and ions actually present in the products.
When solving a problem with parts, you may not catch that both ions and molecular
formulas are both included until reaching a later part of a problem. If that happens,
add the 3-part balancing step to the DATA at that time.
In the bridge conversion, include all of the particle formulas that label units in the
DATA and WANTED terms, including WANTED units found in later parts of the
problem.
Solve Part A.
* * * * *
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Module 14 — Acid-Base Neutralization
Part A:
WANT:
? mol H+
Strategy: WANT a single unit? Start with a single unit. Since all of the DATA is in
equalities, pick one of the single-unit amounts, from the two that are equivalent,
as your given. Pick the amount that you can most easily convert to the
WANTED mol H+.
* * * * *
The H+ ions are formed from KH(IO3)2 in this reaction:
1 mol KH(IO3)2 Æ 1 mol H+ + …..
Since this is a reaction, try stoichiometry steps using the known amount of
KH(IO3)2 as your given.
* * * * *
amount
known bridge relationship
? mol H+= 0.338 g KH(IO3)2• 1 mol KH(IO3)2 •
1 mol H+ =
389.9 g KH(IO3)2 1 mol KH(IO3)2
8.669 x 10─4 mol H+
Carry an extra sf until the last part. Add this answer to the DATA.
Using the Part A answer, try solving Part B by inspection.
* * * * *
Part B: WANT: ? mol OH─
In problems with parts, watch for easy conversions from answers to earlier parts. At the
endpoint of an acid-hydroxide titration, when the indicator changes color:
moles H+ reacted = moles OH─ reacted.
Using the Part A answer: ? mol OH─ = 8.669 x 10─4 mol H+ = 8.669 x 10─4 mol OH─
Once you know the moles of either H+ or OH─, you know the moles of the other in the
sample.
Part C:
WANT:
? mol KOH
L KOH soln.
We solve stoichiometry for a ratio in three steps: find the top amount, bottom amount, and
divide. Try solving for the first two of those three steps by inspection.
* * * * *
You could start at the beginning and do complete ratio-unit stoichiometry, but using
answers found so far will save steps.
For the top WANTED unit: 1 mole KOH Æ 1 mole OH─ , so from Part B,
8.669 x 10─4 moles OH─ =
8.669 x 10─4 mol KOH
in the sample.
* * * * *
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Module 14 — Acid-Base Neutralization
For the bottom unit (L KOH in the sample), the mL KOH in the sample is known. By
inspection, moving the decimal 3 times,
Bottom unit = ? L KOH = 19.29 mL KOH = 0.01929 L KOH
Find the final WANTED ratio.
* * * * *
If you want a ratio, you must convert from a ratio. The answer unit divides the moles
KOH in the sample by the L KOH in the sample. Both of those amounts are solved above.
WANT:
? mol KOH =
L KOH
8.669 x 10─4 mol KOH =
0.01929 L KOH
0.0449 mol KOH
L KOH
In the original DATA, 0.338 g has 3 sf. This limits the final answer to 3 sf.
Practice.
Do Problem 1 today. Save Problem 2 for your next study session.
1. A 10.00 mL sample of NaOH solution completely neutralizes 55.1 mg H2SO4.
a. How many moles of H2SO4 were neutralized?
b. What was the original concentration of the base solution?
2. If 10.91 mL of 0.110 M HCl is required to neutralize 25.00 mL of Ba(OH)2 solution,
a. how many moles of hydroxide ions were in the original base solution ?
b. What was the original [base]?
ANSWERS
1. (For reaction calculations with parts, do stoichiometry steps 2, 3, 4, then 1)
2. DATA: 10.00 mL NaOH = 55.1 mg H2SO4
98.1 g H2SO4 = 1 mol H2SO4
3. Balance:
4. Bridge.
2 NaOH + 1 H2SO4 Æ 2 H2O + …..
2 mol NaOH = 1 mol H2SO4
(equivalent at endpoint)
( mg = g prompt)
(partial balancing provides the bridge)
WANT: ? mol H2SO4
Part A:
* * * * *
(Using the rules in Lesson 11F: if you want a single unit and all of the DATA is in equalities,
• Label the equality that has two amounts that are equivalent;
• Choose one side of the equivalency as the given;
• As the given, pick the side that most easily converts to the WANTED unit. In stoichiometry,
that will usually be the side of the equivalency that has the WANTED formula.)
? mol H2SO4 = 55.1 mg H2SO4 • 1 g
• 1 mol H2SO4 = 5.617 x 10─4 mol H2SO4
103 mg 98.1 g H2SO4
* * * * *
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Module 14 — Acid-Base Neutralization
Part B:
WANT: ? mol NaOH
L NaOH soln
(non-moles):
(solve for non-moles, then moles)
? L NaOH soln. = 10.00 mL NaOH = 0.01000 L NaOH
(Solve for moles NaOH:
You can use stoichiometry steps, or use the Part A answer to solve more quickly. The moles of acid from
Part A are known. Mol/mol ratios for reactions are always simple. Here, the moles of base that reacts is
2x the moles of acid that reacts.)
? mol NaOH = 5.617 x 10─4 mol H2SO4
• 2 = 11.23 x 10─4 mol NaOH in sample
* * * * *
? mol NaOH = 11.23 x 10─4 mol NaOH in sample =
L NaOH
0.01000 L NaOH in sample
0.112 mol NaOH
L NaOH
(Original DATA has 3 sf or higher; round the final answer to 3 sf)
2.
(For reaction calculations with parts, do stoichiometry steps 2, 3, and 4 first.)
2. DATA:
25.00 mL of Ba(OH)2 = 10.91 mL HCl
(Equivalent at endpoint)
0.110 mol HCl = 1 L HCl soln.
3. Balance: (Since the problem includes quantities of both Ba(OH)2 and OH─ , balance in 3 parts.)
2 HCl + 1 Ba(OH)2 Æ 2 H+ + 2 Cl ─ + Ba2+ + 2 OH─ Æ 2 H2O + 1 BaCl2
4. Bridge:
2 mol HCl = 1 mol Ba(OH)2 = 2 mol OH─
Part A. WANT: ? moles OH─
(There are several ways to solve, but whenever moles of H+ or OH─ are involved in a
problem, consider using the rule that 1 mol H+ neutralizes 1 mol OH─.
Take a look at the DATA. Can you find moles of OH─ or base directly? No.
But you know a lot about the HCl. If you can find moles of HCl that reacted, that equals the
moles of H+ that reacted (1 HCl Æ 1 H+), which equals the moles of OH─ that reacts.)
* * * * *
? mol HCl = 10.91 mL HCl •
1 L • 0.110 mol HCl = 1.20 x 10─3 mol HCl = mol H+ = mol OH─
103 mL
1 L HCl soln.
Part B: WANT: ? mol Ba(OH)2
L Ba(OH)2
(non-moles)
(solve stoichiometry ratios for non-moles, then moles)
? L Ba(OH)2 soln. = 25.00 mL Ba(OH)2 = 0.02500 L Ba(OH)2
(Find moles Ba(OH)2: You can use stoichiometry or you can solve by inspection.
Since 1 Ba(OH)2 Æ 2 OH─ , the mol Ba(OH)2 is 1/2 the mol OH─ from Part A.)
? mol Ba(OH)2 = 1/2 times
1.20 x 10─3 mol OH─ = 0.600 x 10─3 mol Ba(OH)2
(Finish from here.)
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Module 14 — Acid-Base Neutralization
* * * * *
? mol Ba(OH)2
L Ba(OH)2
* * * * *
=
6.00 x 10─4 mol Ba(OH)2 =
0.02500 L Ba(OH)2 soln.
0.0240 mol Ba(OH)2
L Ba(OH)2
Lesson 14E: Carbonate Neutralization
Timing: Do this lesson if you are assigned problems involving the acid-base neutralization
of carbonates and/or hydrogen carbonates.
Prerequisites: Lessons 7C, 12C, and 14A.
Pretest: If you can do all steps of the following problem, you may skip this lesson. Check
answers at the end of this lesson.
Q. Assuming complete neutralization of all protons and basic ions in an open system,
write the final product formulas and balance these.
a.
CaCO3 + HCl Æ
b.
c.
H2SO4 + Al2(CO3)3 Æ
HCO3─ + OH─ Æ
d.
HCO3─ +
H+ Æ
* * * * *
Carbonic Acid
At room temperature and pressure, CO2 gas dissolves to a slight extent in water. If
pressure is applied to a CO2 gas and water mixture, more gas dissolves. This pressurized
mixing is used to make carbonated beverages.
Some CO2 molecules, when dissolved in water, react with water form carbonic acid:
H2CO3(aq). Carbonic acid is a weak acid. In water, it ionizes to form H+ plus HCO3─,
but it does so only slightly; less than 1% of the carbonic acid molecules ionize at room
temperature. The dissolved CO2 gas gives carbonated beverages their effervescence, and
the H+ ions, though low in concentration, give carbonated beverages their tart (acidic)
taste. Together, these reactions can be represented as
H2O(l) + CO2(g) Å Æ
H2CO3(aq)
Å Æ H+(aq)
+ HCO3─(aq)
The two-way arrows indicate that these reactions are reversible: the steps can go forward
and backward. Kept under pressure at constant temperature, a carbonated solution is a
stable mixture of all of the particles above. However, if the solution is left open at room
pressure, the CO2 gas will slowly leak out. This “drives the reaction to the left,” using up
the H+, HCO3─, and H2CO3. As the CO2 gas leaves, the solution loses its fizz and acid
tartness.
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Module 14 — Acid-Base Neutralization
Hydrogen Carbonate Ion
A second way to make a carbonic acid solution is to react an acid with hydrogen carbonate
ion (HCO3─, also known as bicarbonate ion). The result is the same two-way reaction
written above, but it is usually written in the reverse direction to emphasize the addition of
acid to the HCO3─.
H+ + HCO3─
ÅÆ
[ H2CO3 ] Å Æ
H2O + CO2↑
When CO2 is mixed with water, carbonic acid forms slowly and a low concentration, but
when typical lab concentrations of acid and HCO3─ are mixed, the carbonic acid forms
quickly and at relatively high concentration compared to its formation by mixing carbon
dioxide and water. At room pressure, most of the carbonic acid formed will quickly break
down into water and carbon dioxide gas. The mixing of acid and hydrogen carbonate ion
thus causes the solution to form bubbles of CO2 gas.
In the reaction above, the brackets [ ] are a way to indicate a substance is formed
temporarily. When acid and hydrogen carbonate are mixed, most of the carbonic acid
formed is an intermediate particle that quickly breaks apart. If the container is open, the
up-arrow ↑ after the CO2 is one way to indicate that a gas formed is allowed to escape.
In the reaction above, HCO3─ reacts with an acid as a base. However, HCO3─ is
amphoteric, meaning that it can react with both acids and bases. How it behaves will
depend on whether it is mixed with an acid or a base.
•
Mixed with an acid:
H+ + HCO3─ Æ
•
Mixed with a base:
OH─ + HCO3─ Æ H─OH + CO32─
[ H2CO3 ] Æ
H2O + CO2↑
When HCO3─ is mixed with acids, it forms H2CO3 and then CO2 gas. When HCO3─ is
mixed with a strong base such as OH─, it loses a proton to form carbonate ion, and the
solution does not bubble.
Practice A
Check answers as you go.
1. Predict intermediate (if any) and final products of these reactions in open containers,
then balance the equations. Ion combinations in final products may be written in solid
or separated formulas.
a.
HCl + NaHCO3
b.
HCO3─ +
c.
Sodium hydroxide plus sodium hydrogen carbonate Æ
d.
LiHCO3 +
©2008 ChemReview.net v. f9
Æ
H2SO4 Æ
OH─ Æ
Page 312
Module 14 — Acid-Base Neutralization
Carbonate Neutralization
A third way to make a carbonic acid solution is to react excess acid with carbonate ion
(CO32─), a base. In the complete neutralization of CO32─, one product is always H2CO3.
As in the reaction of acid and hydrogen carbonate, if carbonate ion is reacted with excess
acid in an open container, the carbonic acid formed it will quickly break down to form
water and bubbles of carbon dioxide.
2 H+
+ CO32─
[ H2CO3 ] Å Æ H2O + CO2↑
If acid is slowly mixed with CO32─ ions, HCO3─ ions form initially. Once equivalent
moles of carbonate and protons are mixed, all of the carbonate ion is converted to hydrogen
carbonate ion, and further addition of acid forms H2CO3, most of which breaks down
quickly into water and CO2 gas.
ÅÆ
Complete Balancing of Acid-Carbonate Neutralization
To predict the complete products for the reaction of acids with CO32─ or HCO3─, use these
rules.
•
If an acid reacts with HCO3─ or completely neutralizes CO32─, one product is
H2CO3.
•
H2CO3 breaks apart to form liquid H2O and CO2 gas.
When representing all of the products carbonate neutralization in equations, it is helpful to
balance the equation in five steps.
For example: Calcium carbonate is found in minerals including limestone, calcite,
chalk, and marble. When excess hydrochloric acid solution is added to calcium
carbonate, the solution bubbles. Why?
Step 1. Write molecular formulas:
2 HCl + CaCO3 Æ
Step 2. Separate reactants into ions:
Æ 2 H+ + 2 Cl─ + Ca2+ + CO32─ Æ
Step 3. Write the initial products:
Æ [ H2CO3 ] + Ca2+ + 2 Cl─
Æ
When excess acid mixes with carbonate ion, one of the products is H2CO3.
Step 4. Write final products:
Æ H2O + CO2↑ + CaCl2(aq)
Step 5. Check that the atoms and total charge are balanced in all parts.
The fundamental rule: to predict the reactions of ionic compounds, re-write the solid and
molecular formulas for the reactants as separated ions, then look for ways that the new
possible combinations of ions will react.
In the final step, the spectator ions in the products can be represented by either an ionic
solid formula labeled aqueous or as the separated ions actually present.
Ca2+ + 2 Cl─
=
CaCl2(aq)
* * * * *
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Module 14 — Acid-Base Neutralization
Practice B:
Assume that these acids and bases ionize and react completely. Write
balanced equations for the steps of the reaction using the 5 step method above. Ions in final
products may be written in solid or separated formulas. Check answers as you go.
1. Write formulas for reactants (if needed) and products, then balance the equation.
a.
HCl +
K2CO3
b.
CO32─ +
c.
Sulfuric acid plus sodium hydrogen carbonate Æ
Æ
H3PO4 Æ
Partial Balancing of Carbonate Neutralization
To solve most neutralization stoichiometry, we only need the coefficients of the acid and
the base. This partial balancing can be completed if know the formula for one product. In
the complete neutralization of HCO3─ or CO32─ by acid, one product is always H2CO3.
For this problem, write H2CO3 as one product, balance the left side, and then check below.
Q. What is the ratio of reaction for
H2SO4 +
Al2(CO3)3 Æ
* * * * *
1. Below the two reactants, write the number of protons and basic ions in the reactants.
2. On the right, add one product of the reaction.
H2SO4
+
Al2(CO3)3
Æ
H2CO3 + …
( 2 H+ )
+
(3 CO32─ )
Æ
H2CO3 + …
3. Add coefficients to balance the atoms and charge.
H2SO4
+
Al2(CO3)3
3 ( 2 H+ ) +
Æ
1 (3 CO32─ )
H2CO3 + …
Æ 3 H2CO3 + …
4. Transfer the coefficients in front of the ( ) to the reactants.
3 H2SO4
+
1 Al2(CO3)3
Æ 3 H2CO3 + …
3 ( 2 H+ ) +
1 (3 CO32─ )
Æ 3 H2CO3 + …
The top reactant ratio will be all that is needed in most neutralization stoichiometry.
Knowing one product, and with practice at separating molecular formulas into ions, you
can often balance the reactants by inspection. Try that on this one.
Q. What is the ratio of reaction for
HNO3 +
BaCO3 Æ
* * * * *
One product for acid-carbonate neutralization is H2CO3. Add the coefficients by inspection.
2 HNO3 +
1 BaCO3 Æ
1 H2CO3 + …
* * * * *
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Module 14 — Acid-Base Neutralization
Practice C
First review the rules in the summary at the end of this module. Then try these problems.
Assume that the following acids and bases react completely. Balance the acid and base
reactants using the partial balancing methods above. Try every other part, and more if you
need more practice. Check answers as you go.
1. Write formulas for reactants (if needed) and products, then balance the equation.
a.
HNO3 + Al2(CO3)3 Æ
b.
Calcium carbonate plus nitric acid Æ
c.
Potassium hydroxide plus potassium hydrogen carbonate Æ
2. Supply the coefficients for the two reactants.
a.
H2Unk +
NaHCO3 Æ
b.
H2SO4 +
Unk(CO3)3 Æ
c.
HCl +
Unk(CO3)2 Æ
Carbonate and Hydrogen Carbonate Neutralization Calculations
In carbonate and hydrogen carbonate neutralization, the ratio of reaction for the acid and
the base can be determined by partial balancing using the steps above.
Once those acid and base coefficients are known, neutralization calculations can be solved
using the same solution stoichiometry steps that were practiced in Lesson 12C and 14D. If
needed, refer to those steps when solving the problems below.
Practice D:
Answers are at the end of this lesson. If you get stuck, read a portion of the
answer until you are unstuck, then try again.
1. If 38.58 mL of 0.150 M HCl is needed to react completely with a sample of solid CaCO3,
how many grams of calcium carbonate were in the sample?
2. If 872 mg of a dry solid acid is neutralized by 24.0 mL of 0.150 M NaHCO3, assuming
that each acid particle contains two acidic hydrogens, what is the molar mass of the
acid?
3. Review Lessons 14D-E and write additional flashcards that cover the fundamentals.
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Module 14 — Acid-Base Neutralization
ANSWERS
Pretest: Coefficients of 1 may be omitted as understood.
a. 1 CaCO3 + 2 HCl Æ 1 H2O + 1 CO2↑ + 1 CaCl2
b. 3 H2SO4 + 1 Al2(CO3)3 Æ 3 H2O + 3 CO2↑ + 1 Al2(SO4)3
c. HCO3─ + OH─ Æ H2O + CO32─
d. HCO3─ + H+ Æ 1 H2O + 1 CO2↑
Practice A: Balance atoms and charge in all parts of each equation. Coefficients of 1 may be omitted.
1. a.
b.
1 HCl + 1 NaHCO3 Æ 1 [ H2CO3 ] + 1 NaCl Æ H2O + CO2↑ + NaCl
2 HCO3─ + 1 H2SO4 Æ 2 [ H2CO3 ] + SO42─ Æ 2 H2O + 2 CO2↑ + SO42─
c.
1 NaOH + 1 NaHCO3 Æ 1 H─OH + 1 Na2CO3
d.
1 LiHCO3 + 1 OH─ Æ 1 H─OH + 1 Li+ + 1 CO32─
Practice B: Balance atoms and charge in all 4 parts of each equation.
2 H+ + 2 Cl─ + 2 K+ + CO32─ Æ
Æ 1 [ H2CO3 ] + 2 KCl Æ H2O + CO2↑ + 2 KCl
1. a. 2 HCl + 1 K2CO3
Æ
b. 3 CO32─ + 2 H3PO4 Æ 3 CO32─ + 6 H+ + 2 PO43─ Æ
Æ 3 [ H2CO3 ] + 2 PO43─ Æ 3 H2O + 3 CO2↑ + 2 PO43─
c. 1 H2SO4 + 2 NaHCO3 Æ 2 H+ + 1 SO42─ + 2 Na+ + 2 H+ + 1 CO32─ Æ
Æ 2 [ H2CO3 ] + 1 Na2SO4 Æ 2 H2O + 2 CO2↑ + 1 Na2SO4
Practice C
1. a. 6 HNO3 + 1 Al2(CO3)3 Æ 3 [ H2CO3 ] + …
b. 1 CaCO3 + 2 HNO3 Æ 1 [ H2CO3 ] + …
c. 1 KOH + 1 KHCO3 Æ H─OH + CO32─ … (mixed with OH─, HCO3─ acts as an acid)
2. a. 1 H2Unk + 2 NaHCO3 Æ 2 [ H2CO3 ] + …
b. 3 H2SO4 + 1 Unk(CO3)3 Æ
3 [ H2CO3 ] + …
c. 4 HCl + 1 Unk(CO3)2 Æ 2 [ H2CO3 ] + …
Practice D
1.
1. WANT: ? g CaCO3
2. DATA: 100.1 g CaCO3 = 1 mol CaCO3
38.58 mL HCl
0.150 mol HCl = 1 L HCl soln.
( g prompt)
(the single unit given)
(M prompt)
(For a reaction with DATA for two substances, use stoichiometry steps. Start with WDBB)
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Module 14 — Acid-Base Neutralization
3. Balance. (For acid-base neutralization, try partial balancing – reactants only.)
1 CaCO3 + 2 HCl Æ 1 [H2CO3] + …..
4. Bridge. 1 mol CaCO3 Æ 2 mol HCl
* * * * *
(Since a single unit is WANTED, use single-unit stoichiometry steps 5-7, chaining the conversions.)
? g CaCO3 = 38.58 mL HCl • 1 L • 0.150 mol HCl • 1 mol CaCO3 • 100.1 g CaCO3 = 0.290 g CaCO3
1 L HCl
2 mol HCl
1 mol CaCO3
103 mL
(Write the unit WANTED)
2.
1. WANTED: g acid
mol acid
2. DATA:
872 mg acid = 24.0 mL NaHCO3
(Equivalent at endpoint)
0.150 mol NaHCO3 = 1 L NaHCO3
(M prompt)
3. Balance. Let acid with two acidic hydrogens = H2Unk
1 H2Unk + 2 NaHCO3 Æ 2 [H2CO3] + ….
4. Bridge.
1 mol H2Unk (or 1 mol acid) = 2 mol NaHCO3
5. SOLVE.
(Not moles): ? g acid = 872 mg acid •
(Moles):
=
1g
103 mg
0.872 g acid in sample
? mol acid = 24.0 mL NaHCO3 • 1 L • 0.150 mol NaHCO3 • 1 mol acid =
103 mL
1 L NaHCO3
2 mol NaHCO3
= 1.800 x 10─3 mol acid in sample.
WANTED:
? g acid
mol acid
=
0.872 g acid
1.800 x 10─3 mol acid
=
484 g acid
mol acid
3. Your flashcards might include the following.
One-way cards (with notch)
Back Side -- Answers
Product of complete carbonate neutralization
H2CO3
[ H2CO3 ]
Breaks down to form H2O and CO2
Two-way cards (without notch):
If stoichiometry WANTED and DATA includes
both solid formulas and ions
Write solid formulas Æ separated-ions Æ
products
Define amphoteric
A particle that can act as an acid or a base
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Module 14 — Acid-Base Neutralization
SUMMARY: Acid-Base Neutralization
1. Acid-base neutralization is an ionic reaction. To understand ionic reactions, write the
separated-ion formulas, then look for new ion combinations that react.
2. Ions: Acids contain H+. The reacting particle in acids = H+ = a proton.
Bases include compounds with hydroxide OH─, carbonate CO32─, and hydrogen
carbonate HCO3─ ions.
3. Products: For acids + OH─, one product is water: H-OH.
H+ + OH─ Æ H-OH
For excess acid plus CO32─, one product is H2CO3.
2 H+ + CO32─ Æ H2CO3
4. Balancing: To predict the products and balance the equations,
•
Write the separated-ion formulas in ( ).
•
Write one product.
•
Finish by balancing atoms, familiar ions, and charge.
5. To balance if a formula is unknown,
•
If an acid has 2 acid hydrogens, use H2Unk.
•
If a base has 3 hydroxides, write Unk(OH)3.
•
If a base has 1 carbonate ion, use UnkCO3. If three, use Unk(CO3)3.
6. Stoichiometry: Do partial balancing at Step 3. Add one product; balance the left side,
and use the left side numbers to write the key bridge conversion.
7. To solve acid-base calculations, use stoichiometry steps plus these 3 rules.
1. When the WANTED and DATA terms include units and particle formulas that
include both an un-ionized particle and its ions,
a. balance the reaction equation in three parts:
Reactants in molecular formulas Æ Reactants in separated ions ÆProducts
b. In the bridge conversion, write all formulas in the WANTED and DATA that
have units attached.
2. In the case of stoichiometry problems that contain more than one question about a
common set of data,
•
first do stoichiometry steps 2, 3, and 4,
•
then do step 1 (list the WANTED unit) for each part.
3. Hydroxide neutralization steps can often be solved by inspection using this rule:
At the endpoint of acid-OH─ titration, mol H+ reacted = mol OH─ reacted
# # # #
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Module 16: Half-Reaction Balancing
* * * * *
NOTE on the Table of Elements.
The atomic masses in this Table of Elements use fewer significant figures than most similar
tables in college textbooks. By keeping the numbers simple, it is hoped that you will use
mental arithmetic to do easy numeric cancellations and simplifications before you use a
calculator for arithmetic.
Many calculations in these lessons have been set up so that you should not need a
calculator at all to solve, if you look for easy cancellations first.
After any use of a calculator, use mental arithmetic and simple cancellations to estimate the
answer, in order to catch errors in calculator use.
# # # # #
©2008 ChemReview.net v. f9
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The ELEMENTS –
The third column shows the atomic number:
The protons in the nucleus of the atom.
The fourth column is the molar mass, in
grams/mole. For radioactive atoms, ( ) is the
molar mass of most stable isotope.
Actinium
Aluminum
Americium
Antimony
Argon
Arsenic
Astatine
Barium
Berkelium
Beryllium
Bismuth
Boron
Bromine
Cadmium
Calcium
Californium
Carbon
Cerium
Cesium
Chlorine
Chromium
Cobalt
Copper
Curium
Dysprosium
Erbium
Europium
Fermium
Fluorine
Francium
Gadolinium
Gallium
Germanium
Gold
Hafnium
Helium
Holmium
Hydrogen
Indium
Iodine
Iridium
Iron
Krypton
Lanthanum
Lawrencium
Lead
Lithium
Lutetium
Magnesium
Ac
Al
Am
Sb
Ar
As
At
Ba
Bk
Be
Bi
B
Br
Cd
Ca
Cf
C
Ce
Cs
Cl
Cr
Co
Cu
Cm
Dy
Er
Eu
Fm
F
Fr
Gd
Ga
Ge
Au
Hf
He
Ho
H
In
I
Ir
Fe
Kr
La
Lr
Pb
Li
Lu
Mg
89
13
95
51
18
33
84
56
97
4
83
5
35
48
20
98
6
58
55
17
24
27
29
96
66
68
63
100
9
87
64
31
32
79
72
2
67
1
49
53
77
26
36
57
103
82
3
71
12
(227)
27.0
(243)
121.8
39.95
74.9
(210)
137.3
(247)
9.01
209.0
10.8
79.9
112.4
40.1
(249)
12.0
140.1
132.9
35.5
52.0
58.9
63.5
(247)
162.5
167.3
152.0
(253)
19.0
(223)
157.3
69.7
72.6
197.0
178.5
4.00
164.9
1.008
114.8
126.9
192.2
55.8
83.8
138.9
(257)
207.2
6.94
175.0
24.3
Manganese
Mendelevium
Mercury
Molybdenum
Neodymium
Neon
Neptunium
Nickel
Niobium
Nitrogen
Nobelium
Osmium
Oxygen
Palladium
Phosphorus
Platinum
Plutonium
Polonium
Potassium
Praseodymium
Promethium
Protactinium
Radium
Radon
Rhenium
Rhodium
Rubidium
Ruthenium
Samarium
Scandium
Selenium
Silicon
Silver
Sodium
Strontium
Sulfur
Tantalum
Technetium
Tellurium
Terbium
Thallium
Thorium
Thulium
Tin
Titanium
Tungsten
Uranium
Vanadium
Xenon
Ytterbium
Yttrium
Zinc
Zirconium
Mn
Md
Hg
Mo
Nd
Ne
Np
Ni
Nb
N
No
Os
O
Pd
P
Pt
Pu
Po
K
Pr
Pm
Pa
Ra
Rn
Re
Rh
Rb
Ru
Sm
Sc
Se
Si
Ag
Na
Sr
S
Ta
Tc
Te
Tb
Tl
Th
Tm
Sn
Ti
W
U
V
Xe
Yb
Y
Zn
Zr
25
101
80
42
60
10
93
28
41
7
102
76
8
46
15
78
94
84
19
59
61
91
88
86
75
45
37
44
62
21
34
14
47
11
38
16
73
43
52
65
81
90
69
50
22
74
92
23
54
70
39
30
40
54.9
(256)
200.6
95.9
144.2
20.2
(237)
58.7
92.9
14.0
(253)
190.2
16.0
106.4
31.0
195.1
(242)
(209)
39.1
140.9
(145)
(231)
(226)
(222)
186.2
102.9
85.5
101.1
150.4
45.0
79.0
28.1
107.9
23.0
87.6
32.1
180.9
(98)
127.6
158.9
204.4
232.0
168.9
118.7
47.9
183.8
238.0
50.9
131.3
173.0
88.9
65.4
91.2
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