Download Chapter 4: Solution Chemistry: The Hydrosphere

Chapter 4: Solution Chemistry: The Hydrosphere
Problems: 4.1-4.34, 4.39-4.86, 4.91-134, 4.139-4.141, 4.143-4.144
4.1 SOLUTIONS ON EARTH AND OTHER PLACES
aqueous solution: a solution where water is the dissolving medium (the solvent)
– For example, when table salt (NaCl) is dissolved in water, it results in an aqueous solution of
sodium chloride, NaCl(aq), with Na+ and Cl- ions dissolved in water.
– Note: The physical state aqueous,(aq), indicates an element or compound dissolved in
water while the physical state liquid,(l), means a pure substance in the liquid state.
Example: If you had to drink one on the following, which would you choose?
Explain why.
NaCl(l)
NaCl(aq)
Thus, recognize that aqueous and liquid are not the same!
A solution consists of a solute dissolved in a solvent.
solute: component present in smaller amount
solvent: component present in greater amount
The formation of a solution
– As a solute crystal is dropped into a
solution, the water molecules begin
to pull apart the ionic compound ion
by ion
→ solvent molecules surround the
solute particles, forming a
solvent cage
→ The ions are now hydrated
(surrounded by polar
water molecules).
→ The solute is now dissolved
in the solvent and cannot be
seen because the ions are far
apart, like the particles in a
gas
•
•
View Pearson’s NaCl Dissolving Animation: https://www.youtube.com/watch?v=9aYLonML69w
View the Molecular Dynamics Animation: https://www.youtube.com/watch?v=aKGJm6OGJNs
CHEM 161: Chapter 4 Notes v0417 page 1 of 29 EVIDENCE OF A CHEMICAL REACTION
a) A gas is produced.
b) A precipitate forms.
c) Heat is released/absorbed
Types of Chemical Reactions
• Precipitation Reactions
• Acid-Base Neutralization Reaction
• Oxidation-Reduction (Redox) Reactions
– Also classified as combination, decomposition, combustion, and single-replacement
reactions
4.7 PRECIPITATION REACTIONS
Solubility Rules: Indicate if an ionic compound is soluble or insoluble in water.
– Keep in mind that these are just general guidelines, and in reality, some ionic compounds
may only be slightly soluble, and solubility may depend on temperature.
Solubility Rules for Ionic Compounds in Water
Soluble if the ionic compound contains:
1. Li+, Na+, K+, NH4+ (ALWAYS!)
2. C2H3O2–, NO3–, ClO3–, ClO4–
3. Halide ions (X–): Cl–, Br–, or I–, but AgX, PbX2,
HgX, and Hg2X2 are insoluble
4. sulfate ion (SO42-), but CaSO4, SrSO4, BaSO4,
Ag2SO4, `and PbSO4 are insoluble.
Insoluble if the ionic compound contains:
5. carbonate ion, CO326. chromate ion, CrO427. phosphate ion, PO438. sulfide ion (S2–), but CaS, SrS, and
BaS are all soluble.
9. hydroxide ion (OH–), but Ca(OH)2,
Sr(OH)2, and Ba(OH)2 are soluble.
soluble = compound dissolves in water → exists as individual ions in solution
→ physical state is aqueous, (aq)
insoluble = compound does not dissolve in water but remains a solid
→ physical state is shown as solid, (s)
CHEM 161: Chapter 4 v0916 page 2 of 29 Note that Section 4.7 (PRECIPITATION REACTIONS) and other topics are covered in more
detail in the Chapter 4 Review Lecture Notes.
4.5 ACID-BASE (NEUTRALIZATION) REACTIONS: PROTON TRANSFER
Properties of Acids and Bases
Acids
– produce hydrogen ions, H+
– taste sour
– turn blue litmus paper red
Bases
– produce hydroxide ions, OH–
– taste bitter; feel soapy, slippery
– turn red litmus paper blue
Arrhenius Definitions:
acid: A substance that releases H+ when dissolved in water—i.e., the molecule that loses a
H+ ion to water to form the hydronium ion (H3O+), a hydrated proton: H+ + H2O = H3O+
– Some acids are monoprotic (able to release only one H+ per molecule)
– e.g. HCl, HBr, HI, HNO3, HClO4
– Some acids are polyprotic (able to release more than one H+ per molecule)
– e.g. H2SO4 and H2CO3 are both diprotic; H3PO4 is triprotic.
base: A substance that releases OH– when dissolved in water (e.g. NaOH, Ca(OH)2, etc.)
In an acid-base reaction,
– H+ from acid reacts with the OH– from base → water, H2O
– The cation (M+) from base combines with anion from acid (X–) → the salt.
A general equation for an acid-base neutralization reaction is shown below:
HX(aq)
+
MOH(aq) → H2O(l) + MX
acid
base
water
salt
Because water is always produced, an acid always reacts with a base!
Ex. 1 Complete and balance each of the equations below:
a.
HCl(aq) +
NaOH(aq) →
b.
H2SO4(aq) +
KOH(aq) →
c.
H3PO4(aq) +
Ca(OH)2(aq) →
CHEM 161: Chapter 4 Notes v0417 page 3 of 29 Acid-Base Reactions with Gas Formation
– Some acid-base reactions produce carbon dioxide gas, CO2(g), along with water and salt.
– When the base contains carbonate ion (CO32–) or hydrogen carbonate ion (HCO3–), then
the products of the acid-base reaction are a salt and H2CO3(aq), which breaks down into
water and carbon dioxide gas.
– Thus, for acid-base reactions, always show H2CO3(aq) as H2O(l) + CO2(g).
The general equations for the unbalanced acid-base reactions are below:
HX(aq)
+
MCO3(s)
acid
HX(aq)
base
+
+
water
MHCO3(s)
acid
→ H2O(l)
base
→ H2O(l)
water
CO2(g)
+
MX
carbon dioxide
+
CO2(g)
carbon dioxide
salt
+
MX
salt
Because water is always produced, an acid always reacts with a base!
Ex. 1 Complete and balance each of the equations below:
a.
HBr(aq) +
CaCO3(s) →
b.
H2SO4(aq) +
KHCO3(s) →
c.
HClO4(aq) +
Sr(HCO3)2(s) →
A double-replacement reaction that produces NH4OH(aq) actually produces ammonia, NH3(g).
NH4OH(aq)
→ NH3(g) + H2O(l)
Ex. 2 Complete and balance the equation below:
(NH4)2SO4(aq) +
CHEM 161: Chapter 4 v0916 KOH(aq) →
page 4 of 29 Why is H+ called a proton?
Brønsted-Lowry Definitions of Acids and Bases
Brønsted-Lowry acid: A substance that donates a proton (H+)—i.e., a proton donor
Brønsted-Lowry base: A substance that accepts a proton (H+)—i.e., a proton acceptor
– Unlike an Arrhenius base, a B-L base need not contain OH–.
An acid-base reaction can also be a Brønsted-Lowry reaction where a proton (H+) is transferred
from one reactant to another:
NH3(aq)
+
NH4+(aq)
H2O(l)
+
OH–(aq)
In this reaction the H2O donates the H+ ion to NH3 to produce NH4+ and OH–.
→ In this reaction, H2O is the Brønsted-Lowry acid, and NH3 is the Brønsted-Lowry base.
conjugate acid-base pairs: An acid and base differing only by the presence of a H+ are
conjugate acid-base pairs (i.e., the conjugate acid has one more H+ than its base)
– For the reaction above, when H2O donates H+ to NH3 (i.e. acts as a Brønsted-Lowry acid),
it leaves behind OH–, which can act as a Brønsted-Lowry base for the reverse reaction.
– For the reaction above, when NH3 accepts H+ from H2O (i.e. acts as a Brønsted-Lowry
base), it forms NH4+, which can act as a Brønsted-Lowry acid for the reverse reaction.
→ In this reaction, the conjugate acid-base pairs are NH3 and NH4+ and H2O and OH–.
CHEM 161: Chapter 4 Notes v0417 page 5 of 29 The general reaction for the dissociation (or ionization) of an acid can be represented as,
HA(aq)
+
H2O(l)
H3O+(aq)
+
A–(aq)
where the double-arrow indicates both the forward and reverse reactions can occur:
– Note: The double arrow ( ) indicates the reaction is reversible (can go in both directions).
Brønsted-Lowry acid = ____________
Brønsted-Lowry base= ____________
Ex. 1: HA is the conjugate acid of ________, and A– is the conjugate base of ________.
Ex.2 : Determine the Brønsted-Lowry acid and base in each of the following reactions:
a.
NH3(aq)
+
b.
H2O(l)
+
c.
CH3COOH(aq)
NH4+(aq)
H2O(l)
H2SO4(aq)
+
NH3(aq)
+
OH–(aq)
H3O+(aq)
NH4+(aq)
HSO4–(aq)
+
+
CH3COO–(aq)
Example: List the two conjugate acid-base pairs in the examples above:
a. __________________________ and __________________________
b. __________________________ and __________________________
c. __________________________ and __________________________
Note: H2O is amphoteric since it can behave as an acid or a base.
CHEM 161: Chapter 4 v0916 page 6 of 29 4.9 OXIDATION-REDUCTION (REDOX) REACTIONS
In oxidation-reduction (redox) reactions, electrons are transferred from one substance to another.
Types of Redox Reactions
•
•
•
•
Combination Reaction
Decomposition Reaction
Single-Replacement (or Displacement) Reaction
Combustion Reaction
COMBINATION REACTIONS:
A + Z → AZ
– There are many types of combination reactions, but you only need to predict products for a
metal reacting with a nonmetal to form a solid ionic compound:
metal + nonmetal
Δ
##→
ionic compound(s)
Δ
where ##
→ indicates the reactants are also heated.
Ex. 1 Complete and balance each of the equations below:
Δ
a.
Na(s) +
Cl2(g)
##→
b.
Al(s) +
O2(g)
##→
DECOMPOSITION REACTIONS:
Δ
AZ → A + Z
Be able to classify and balance decomposition reactions. (You won’t need to predict products.)
Δ
a. _____ KHCO3(s) ##
→ _____ K2CO3(s) + _____ H2O(l) + _____ CO2(g)
Δ
b. _____ Al2(CO3)3(s) ##
→ _____ Al2O3(s) + _____ CO2(g)
COMBUSTION REACTIONS: CxHy + O2(g)
Δ
##→ CO2(g) + H2O(g)
CxHyOz + O2(g)
Δ
a.
C2H5OH(l) +
O2(g)
b.
C2H2(g) +
Δ
O2(g) ##
→
CHEM 161: Chapter 4 Notes v0417 Δ
##→ CO2(g) + H2O(g)
##→
page 7 of 29 THE ACTIVITY SERIES OF METALS (SINGLE-REPLACEMENT REACTIONS)
Activity Series:
Relative order of elements arranged by their ability to replace cations in
aqueous solution
Li > K > Ba > Sr > Ca > Na > Mg > Al > Mn > Zn >
Fe > Cd > Co > Ni > Sn > Pb > (H) > Cu > Ag > Au
Note: The Activity Series will be given to you on quizzes and exams.
Single Replacement (or Displacement) Reactions: A + BZ → AZ + B
metal A + aqueous solution B → aqueous solution A + metal B
Zn(s) + Sn2+(aq) → Sn(s) + Zn2+(aq)
Cu(s) + 2 Ag+(aq) → 2 Ag(s) + Cu2+(aq)
To balance and complete the following rxns:
– Check the Activity Series to see which metal is more active.
→ The more active metal replaces the less active by going into solution as an ion and the
less active metal ion comes out as a solid.
1.
Mg(s) +
2.
Al(s) +
3.
Ag(s) +
CHEM 161: Chapter 4 v0916 AlCl3(aq)
→
CdSO4(aq)
→
Mg(NO3)2(aq)
→
page 8 of 29 To balance and complete the following reactions:
– Check the Activity Series to see which metal is more active,
the metal or H.
→
The more active metal replaces the less active by
going into solution as an ion and the H comes out as
hydrogen gas, H2(g).
metal A + acid solution → aqueous solution A + H2(g)
1.
Zn(s) +
HCl(aq)
2.
Al(s)
HNO3(aq)
3.
Cu(s) +
+
→
HI(aq)
→
→
ACTIVE METALS: Li > K > Ba > Sr > Ca > Na
– React directly with water
→
The active metal replaces the less active by going into
solution as an ion with hydroxide, OH–, and the H comes
out as hydrogen gas, H2(g).
active metal
+
H2O(l)
→ metal hydroxide + H2(g)
a.
Ca(s)
+
H2O(l)
b.
Na(s)
+
H2O(l) →
c.
Fe(s)
+
H2O(l)
CHEM 161: Chapter 4 Notes v0417 →
→
page 9 of 29 Ex. 1.
For each of the following,
i. Identify the type of reaction using the letters designated below:
– Combination (C)
– Precipitation (P)
– Decomposition (D)
– Acid-Base Neutralization (N)
– Combustion (B)
– Single Replacement/Displacement (SR)
TYPE ii. Balance the equation
_____ a. _____ Mg(NO3)2(aq) + _____ K3PO4(aq) → _____ Mg3(PO4)2(s) + _____ KNO3(aq)
_____ b. _____ Ni(OH)3(s) + _____ HCl(aq) → _____ H2O(l) + _____ NiCl3(aq)
Δ
_____ c. _____ Al(HCO3)3(aq) ##
→ _____ CO2(g) + _____ H2O(g) + _____ Al2(CO3)3(s)
_____ d. _____ Fe(s) + _____ Pb(NO3)2(aq) → _____ Pb(s) + _____ Fe(NO3)3(aq)
_____ e. _____ SO2(g) + _____ O2(g) + _____ H2O(l)
→ H2SO4(aq)
acid rain
Ex. 2. For each of the following sets of reactants, write the formulas for the products (including
physical states) and balance the equation if the reaction occurs, or write “NR” for no reaction.
a.
HNO3(aq)
b.
KOH(aq)
c.
Al(s) +
d.
HI(aq)
e.
Sr(s) +
CHEM 161: Chapter 4 v0916 +
Zn(s)
+
Ni2(SO4)3(aq)
S8(s)
+
→
→
Δ
##→
MgCO3(s)
→
H2O(l)
→
page 10 of 29 Note that Section 4.4 (ELECTROLYTES ), Net Ionic Equations, and other topics are covered
in more detail in the Chapter 4 Review Lecture Notes.
4.9 OXIDATION-REDUCTION (REDOX) REACTIONS
In most oxidation-reduction (abbreviated as “redox”) reactions, electrons are transferred
from one reactant to another.
Ex. 1: Indicate the number of protons and electrons for the following:
Na+
Na
Balance the equation:
Cl
_____ Na(s) + _____ Cl2(g)
Cl–
Δ
##→
_____ NaCl(s)
a. Which reactant lost electrons (i.e., was oxidized)? ______________
b. Which reactant gained electrons (i.e., was reduced)? ______________
c. How many electrons were transferred? ______
Ex. 2: Indicate the number of protons and electrons for the following:
Al+3
O
O2–
Al
Balance the equation:
€
_____ Al(s) + _____ O2(g)
Δ
##→
_____ Al2O3(s)
a. Which reactant lost electrons (i.e., was oxidized)? ______________
b. Which reactant gained electrons (i.e., was reduced)? ______________
c. How many electrons were transferred? ______
CHEM 161: Chapter 4 Notes v0417 page 11 of 29 OXIDATION NUMBER: the actual or hypothetical charge of an atom in a compound if it
(or OXIDATION STATE) existed as a monatomic ion
– used to track changes in electron distribution in compounds and to determine electron transfer
Guidelines for Assigning Oxidation Numbers
1. The oxidation number of an element in its natural form is 0.
– e.g. the oxidation number is zero for each element in H2, O2, Cl2, P4, Na, etc.
2. The oxidation number of a monatomic ion is the charge on the ion.
– e.g. In Na3N, the ions are Na+ and N3–, so oxidation #’s: Na = +1 and N = -3
In Al2O3, the ions are Al+3 and O2–, so oxidation #’s: Al = +3 and O = -2
3. In a compound or polyatomic ion,
– Group I elements are always +1.
– Group II elements are always +2.
– Fluorine is always -1.
– Oxygen is usually -2 (except in the peroxide ion, O22–, when O is -1)
– Hydrogen is usually +1
(except when it is with a metal, like NaH or CaH2, then it is -1)
4. In a compound, the sum of all oxidation numbers must equal 0.
In a polyatomic ion, the sum of all oxidation numbers must equal charge.
Example: Determine the oxidation number for each element in the following:
a. CrO42–: Cr: ____, O: ____
d. H2SO4: H: ____, S: ____, O: ____
b. NO3 : N: ____, O: ____
e. CaCr2O7: Ca: ____, Cr: ____, O: ____
c. C2O42–: C: ____, O: ____
f. C3H8: C: ____, H: ____
−
Oxidation: process of losing electrons (oxidation number ↑)
Reduction: process of gaining electrons (oxidation number ↓)
Electrons are negatively charged, so charge becomes confusing because more electrons results in
a lower charge. Think of electrons the way you think of debt—more debt means lower net worth.
– Thus, gaining electrons results in a lower oxidation number.
In a redox reaction
– One reactant Loses Electrons/is Oxidized (LEO)
– Another reactant Gains Electrons/is Reduced (GER)
An easy way to remember is “LEO the lion goes GER!”
– The element or reactant that is oxidized is the reducing agent.
– The element or reactant that is reduced is the oxidizing agent.
CHEM 161: Chapter 4 v0916 page 12 of 29 For each of the following reactions,
1. Balance the equation.
2. Identify the ion or element with its oxidation state in a compound that is oxidized and reduced.
3. Identify the oxidizing agent and the reducing agent.
4. Indicate the number of electrons transferred in the reaction.
a.
Zn(s) +
AgNO3(aq)
→
Zn(NO3)2(aq) +
Ag(s)
The reactant oxidized is ___________________, and the oxidizing agent is _______________.
The reactant reduced is ___________________, and the reducing agent is _______________.
The number of electrons transferred is ______,
b.
Al(s) +
HCl(aq)
→
AlCl3(aq) +
H2(g)
The reactant oxidized is ___________________, and the oxidizing agent is _______________.
The reactant reduced is ___________________, and the reducing agent is _______________.
The number of electrons transferred is ______,
c.
C2H2(g) +
O2(g)
→
CO2(g) +
H2O(g)
The reactant oxidized is ___________________, and the oxidizing agent is _______________.
The reactant reduced is ___________________, and the reducing agent is _______________.
The number of electrons transferred is ______,
CHEM 161: Chapter 4 Notes v0417 page 13 of 29 d.
Ca(s) +
H2O(l)
→
Ca(OH)2(aq) +
H2(g)
The reactant oxidized is ___________________, and the oxidizing agent is _______________.
The reactant reduced is ___________________, and the reducing agent is _______________.
The number of electrons transferred is ______,
e.
H2O2(aq) +
Mn(OH)2(aq)
→
Mn(OH)3(aq)
The reactant oxidized is ___________________, and the oxidizing agent is _______________.
The reactant reduced is ___________________, and the reducing agent is _______________.
The number of electrons transferred is ______,
f.
NO(g) + O3(g)
→ NO2(g) + O2(g)
The reactant oxidized is ___________________, and the oxidizing agent is _______________.
The reactant reduced is ___________________, and the reducing agent is _______________.
The number of electrons transferred is ______,
CHEM 161: Chapter 4 v0916 page 14 of 29 Balancing Redox Reactions Using Half Reactions (in Acidic Solution)
i. If given both the oxidation and reduction reactions, separate them into two half-reactions.
ii. Balance atoms and use oxidation numbers to determine the numbers of electrons
transferred.
iii. Balance the oxygen atoms by adding H2O molecules to the other side of the equation.
iv. Balance the number of hydrogen atoms by adding H+ ions.
v. Confirm that the overall sum of charges on the left equals the overall sum on the right.
Ex. 1: Use the steps above to balance each equation below.
a. One of the half-reactions that occur in a lead-acid storage battery (e.g. a car battery) is
shown in the following (unbalanced) half-reaction:
_____ PbO2(s) → _____ Pb2+(aq)
Use the steps above to balance the equation.
Is the reaction above an oxidation or reduction half-reaction?
oxidation
reduction
b. The other half-reaction is as follows:
_____ Pb(s) → _____ Pb2+(aq)
Balance the reaction using electrons.
Is the reaction above an oxidation or reduction half-reaction?
oxidation
reduction
c. Now, combine the two half-reactions, cancelling the electrons, and write the overall reaction
below:
Ex. 2. Consider the following unbalanced reaction: Cr2O72-(aq) + I–(aq) → Cr3+(aq) + I2(aq)
a. Indicate the oxidation number for each atom above, then separate it into half-reactions
below, and finish balancing using steps i to iv above.
b. Balance the electrons transferred, then write the overall reaction below.
CHEM 161: Chapter 4 Notes v0417 page 15 of 29 Balancing Redox Reactions Using Half Reactions (in Basic Solution)
i. Indicate the electrons transferred using the oxidation numbers. (If given both the oxidation
and reduction reactions, also balance the numbers of electrons transferred.)
ii. Balance the oxygen atoms by adding H2O molecules to the other side of the equation.
iii. Balance the number of hydrogen atoms by adding H+ ions.
iv. Since the reaction occurs in basic solution, add OH- ions equal to the number of H+ ions to
both sides of the equation. (This shows the half reactions in basic solution.)
v. Combine the two half-reactions into an overall reaction by balancing the number of electrons
transferred then cancel any ions or compounds that appear on both sides of the equation.
vi. Confirm that the overall sum of charges on the left equals the overall sum on the right.
Ex. 3. Consider the following unbalanced reaction: MnO4–(aq) + I–(aq) → MnO2(aq) + I2(aq)
a. Indicate the oxidation number for each atom above then separate it into half-reactions below,
and finish balancing using steps i to vi above.
b. Balance the electrons transferred, then write the overall reaction below.
Ex. 4. Consider the unbalanced reaction: Zn(s) + NO3–(aq) → NH3(aq) + Zn(OH)42-(aq)
a. Indicate the oxidation number for each atom above then separate it into half-reactions below,
and finish balancing using steps i to vi above.
b. Balance the electrons transferred, then write the overall reaction below.
CHEM 161: Chapter 4 v0916 page 16 of 29 4.2 CONCENTRATION UNITS
solution: homogeneous mixture of substances present as atoms, ions, and/or molecules
solute: component present in smaller amount
solvent: component present in greater amount
Note: Unless otherwise stated, the solvent for most solutions considered in this class will
almost always be water!
→ Aqueous solutions are solutions in which water is the solvent.
A concentrated solution has a large quantity of solute present for a given amount of solution.
A dilute solution has a small quantity of solute present for a given amount of solution.
SOLUTION CONCENTRATION:
concentration =
amount of solute
amount of solution
The more solute in a given amount of solution → the more concentrated the solution
Molar Concentration = Molarity =
moles of solute
liters (L) of solution
(reported in units of M=molar)
Ex. 1 Find the molarity of a solution prepared by dissolving 50.00 g of NaOH in 150.0 mL of
solution.
Ex. 2 Find the molarity of a solution prepared by dissolving 25.00 g of copper(II) sulfate in
250.0 mL of solution.
Ex. 3: Write a unit conversion factor for each of the following:
a. 2.50M KOH solution
CHEM 161: Chapter 4 Notes v0417 b. 0.125M MgSO4 solution
page 17 of 29 Ex. 4 Indicate the concentration of calcium and chloride ions in a 1.00M calcium chloride
solution.
Ex. 5 Indicate the molarity of each ion in the solutions indicated below:
a. In a 1.25M Na2SO4(aq) solution, [Na+]=___________ and [SO4-2]=___________.
b. In a 0.500M zinc nitrate solution, [Zn2+]=___________ and [NO3–]=__________.
c. In a 0.150M aluminum sulfate solution, [Al+3] =__________ and [SO4-2]=__________.
Ex. 6 Circle the solution in each set with the highest [H+] (or [H3O+]):
a.
0.100 M HF(aq)
b.
0.100 M H2CO3(aq)
0.100 M HCl(aq)
0.100 M HNO2(aq)
0.100 M H2SO3(aq)
0.100 M H2SO4(aq)
SOLVING MOLAR CONCENTRATION PROBLEMS:
Note that if molarity and volume are both given, you can calculate # of moles since
volume × molarity = volume (in L) ×
moles of solute
liters (L) of solution
so volume units will cancel → # of moles!
If you are given volume and molarity for a solution, multiply them together to get # of moles!
Ex. 1 Calculate the mass of NaCl required to make a 1.00 L of a 1.00M NaCl solution.
Ex. 2 Calculate the mass of acetic acid required to make 250.0 mL of a 0.500M acetic acid
solution.
CHEM 161: Chapter 4 v0916 page 18 of 29 We do these calculations, so we can prepare solutions in the lab as follows:
Ex. 3 What volume (in mL) of a 0.125M silver nitrate solution contains 5.00 g of silver nitrate?
Ex. 4 Calculate the mass of barium hydroxide required to make 500.0 mL of solution with a
1.500M hydroxide concentration.
CHEM 161: Chapter 4 Notes v0417 page 19 of 29 4.6 (ACID-BASE) TITRATIONS
standard solution: an acid or base solution where the concentration is known, generally
to at least 3 or more sig figs
– used to analyze properties of substances, such as the neutralizing power of commercial
antacids, the tartness of wine, etc.
acid-base indicators:
– Solutions that are pH sensitive and change color
– Generally have color changes occurring for pH close to 7 since reactions monitored are
neutralization reactions which occur near pH=7
titration: The gradual addition of a solution from a
buret to another solution in a flask or beaker until the
reaction between the two is complete, as signaled by
the indicator changing color.
titrant: the solution in the buret
analyte: the solution for which a property (e.g.
molar concentration) is being determined
– In some titrations, the titrant is also the analyte, in
which case a known amount of acid or base is
present in the flask, and the amount of titrant
necessary to neutralize it will be used to
determine the concentration of the titrant.
– In other titrations, the titrant’s concentration is
known, and the amount of titrant used to
neutralize it will determine the concentration of
the solution in the flask.
Example: A sample of vinegar (about 5% acetic acid by volume) is titrated in a flask with sodium
hydroxide using bromthymol blue indicator. Indicate the color of the solution in the
flask at the following times:
i. At the start: __________________
CHEM 161: Chapter 4 v0916 ii. At the endpoint: __________________
page 20 of 29 Let’s also consider what occurs at the molecular
level in an acid-base titration. Consider the
following reaction:
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
a. Draw the ions/molecules representing
hydrochloric acid in the circle at the right.
b. The solution’s color will be ___________
when phenolphthalein is added to the
hydrochloric acid.
c. Draw the ions/formula units representing
sodium hydroxide being added to the
circle and the products of the resulting
neutralization reactions.
equivalence point: Theoretical point in the titration when the amount of base added is exactly
equal to the acid present, so the base completely neutralizes the acid.
endpoint: The moment when the acid-base indicator changes color.
– Note in most acid-base titrations, the phenolphthalein indicator does not change
color until the solution is basic, so in reality, there is a slight excess of hydroxide
ion present in the solution when it turns pink.
– Ideally, the endpoint is reached with one drop or a fraction of a drop of titrant, so
the endpoint is very close to the equivalence point.
The figure below shows an acid-base titration.
Figure 1a
CHEM 161: Chapter 4 Notes v0417 Figure 1b
Figure 1c
page 21 of 29 Ex. 1: If phenolphthalein was added to the solution in Figure 1a, and
it remained colorless, then the solution in the flask must be _______.
Ex. 2: Given that the solution turns pink (Fig. 1b) upon addition of
the titrant, the titrant in the titration above must be _______.
Ex. 3: In Fig. 1c above, the solution is ________.
acidic
an acid
basic
acidic
basic
a base
neutral
Ex. 4: If the endpoint for the titration is close enough to the equivalence point, then one
assumes that the amount of acid and base are equal at the endpoint. For this
assumption to be valid, a student must catch the endpoint within how many drops? _____
Ex 5. Oxalic acid, H2C2O4(aq), is often sold as wood bleach and can be used to remove rust.
A student wanted to remove rust from an old bike chain and prepared the solution by
dissolving 1/8th of a cup of solid oxalic acid in 0.750 gal of hot tap water. The student then
analyzed a 10.00 mL sample of the oxalic acid solution, which required 28.39 mL of a
0.1550M NaOH solution for complete neutralization. Predict the products and balance the
equation below, then calculate the molarity of the oxalic acid solution:
H2C2O4(aq) +
NaOH(aq)
→
Ex 6. The active ingredient in Tums is often calcium carbonate, and Tums Extra Strength
often contains 750 mg of calcium carbonate. If stomach acid is primarily hydrochloric acid
with an approximate molarity of 1.0M, calculate the volume (in mL) of stomach acid that
can be neutralized by taking two Tums Extra Strength tablets. (Assume the amount of
calcium carbonate in the Tums is 750±0.05 mg of calcium carbonate.) Write the balanced
equation for the reaction below
CHEM 161: Chapter 4 v0916 page 22 of 29 MILLIMOLARITY, MICROMOLARITY, ETC.
Many environmental and biological systems have solution concentrations much lower than 1.0M.
→ Lower concentration units are required to describe these systems.
If 1 millimole (1 mmol) = 0.001 mole, so 1 mmol/L (mM) is 1000th of 1 mol/L (1 M), or a
thousand times less concentrated.
If 1 micromole (1 µmol) = 10-6 mole, so 1 µmol/L (µM) is one-millionth of 1 mol/L (1 M),
or a million times less concentrated.
For example, the concentration of K+ ions in a sample of seawater is about 10.5 mM while
the concentration of Sr2+ ions in the same sample of seawater is about 92.8 µM.
Example: Ethylenediaminetetraacetic acid (abbreviated as EDTA is a
chelating agent that can bind to and remove metal ions like Fe3+ and Ca2+.
Solutions of EDTA are prepared starting with a sodium salt of EDTA,
EDTA⋅Na2⋅2H2O, with a formula of [CH2N(CH2CO2Na)CH2CO2H]2⋅2H2O.
Calculate the mass of EDTA⋅Na2⋅2H2O necessary to produce 500.0 mL of
a 0.50mM EDTA solution.
Image from http://en.wikipedia.org/wiki/Ethylenediaminetetraacetic_acid
MASS PERCENT CONCENTRATION (M/M%)
M/M% =
mass of solute
mass of solution
× 100% =
mass of solute
mass of solute + mass of solvent
× 100%
Ex 1. What is the mass percent concentration of a solution made by dissolving 15.00 g of
HC2H3O2 in 250.0 g of water?
The solute is ______________, and the solvent is _______________.
CHEM 161: Chapter 4 Notes v0417 page 23 of 29 Ex 2. A person accused of a DUI violation submitted a 5.00 g sample of blood for alcohol
content analysis. The analysis determined the presence of 5.17 mg of alcohol in the
blood. If a person with a blood alcohol content (mass percent of alcohol in the blood) of
0.08% is considered legally impaired, was this person driving while impaired—i.e., was
the blood alcohol content greater than or equal to 0.08%?
Ex 3. Intravenous saline injections are sometimes administered to restore electrolyte balance
in trauma patients. What mass of NaCl is present in 1.00 L of a 0.90% saline solution
givent the density of the saline solution was 1.01 g/mL.
PARTS PER MILLION (ppm) and PARTS PER BILLION (ppb)
While the concentration of some solutions can be expressed as mass percentage, which is
defined as grams of solute per one hundred grams of solution, the concentrations of much more
dilute solutions can be expressed in mg of solute per kg of solution (parts per million or
ppm) or µg of solute per kg of solution (parts per billion or ppb).
ppm =
mass of solute (in mg)
mass of solution (in kg)
or
ppb =
mass of solute (in µg)
mass of solution (in kg)
Example: “In January 2011, the U.S. Department of Health and Human Services and the U.S.
Environmental Protection Agency proposed a recommended level of fluoride in drinking water of
0.7 parts per million.”1 If a 5.00 L sample of Seattle tap water was analyzed and found to contain
4.05 mg of fluoride ion, calculate the concentration of fluoride ion (in ppm) in the tap water given
that the tap water’s density is 1.005 g/mL.
1
http://www.seattle.gov/util/MyServices/Water/Water_Quality/Fluoride/index.htm
CHEM 161: Chapter 4 v0916 page 24 of 29 4.3 DILUTIONS (DILUTING CONCENTRATED SOLUTIONS)
More concentrated standard solutions (with accurately known concentrations) are often diluted
with deionized water to get a solution with a specific concentration, as shown below.
The graphic above shows how to prepare a solution with a specific concentration
(a) A determined volume of a more concentrated solution is measured out in the smaller flask.
(b) The more concentrated solution is then transferred to a larger empty volumetric flask.
(c) The solution is diluted with deionized water to obtain the calibrated volume of the flask.
Dilution Equation:
M1 V1 = M2 V2
The dilution equation above is used to determine the new molarity of the dilute solution (M2)
given the molarity of the more concentrated solution (M1) or the volume of the more
concentrated solution to use (V1) given the new total volume of the dilute solution (V2) required.
Ex. 1: What volume (in mL) of a 6.00M NaOH solution is required to prepare 250.0 mL of a
0.1200M NaOH solution?
Ex. 2: Calculate the molarity of a hydrochloric acid solution prepared by diluting 15.0 mL of
6.00M hydrochloric acid with 100.0 mL of deionized water.
CHEM 161: Chapter 4 Notes v0417 page 25 of 29 Ex. 3: Calculate the molarity of aluminum ion in a solution prepared by diluting 5.00 mL of
2.500M aluminum sulfate with 100.0 mL of deionized water.
Ex. 4: Calculate the molarity of hydroxide ion in a solution prepared by diluting 50.0 mL of
1.50M potassium hydroxide with 100.0 mL of 0.500M calcium hydroxide.
SOLUTION STOICHIOMETRY PROBLEMS
Many wood bleaching crystals are often close to pure oxalic acid, H2C2O4(aq), which can be
used to bleach wood as well as remove rust, Fe2O3, as shown in the following equation:
Fe2O3(s) + 6 H2C2O4(aq) →
2 Fe(C2O4)3–3(aq) + 3 H2O(l) + 6 H+(aq)
Ex. 1: Calculate the molarity of a 1.500% by mass oxalic acid solution if the density of the
solution is 1.015 g/mL.
Ex. 2: Calculate the volume of the 1.50% oxalic acid solution required to remove 50.0 g of rust
from an old bike chain.
CHEM 161: Chapter 4 v0916 page 26 of 29 CARBON DIOXIDE, SEAWATER, AND OCEAN ACIDIFICATION
Seawater is not only salty but basic (or alkaline) because of dissolved salts. In the pre-industrial
era, the average global surface ocean pH was about 8.2, but it has dropped and is currently
about 8.1, representing an increase in acidity of about 30%.
Carbon dioxide currently makes up about 0.039% of the atmosphere, and carbon dioxide in the
atmosphere dissolves in seawater to form carbonic acid, hydrogen carbonate, and carbonate
ions.
Ex. 1: Write the balanced chemical equations for the reactions described below:
a. Carbon dioxide reacts with water to form carbonic acid.
b. Carbonic acid reacts with water to form hydronium ion and hydrogen carbonate ion.
c. Hydrogen carbonate ion reacts with water to form hydronium ion and carbonate ion.
CHEM 161: Chapter 4 Notes v0417 page 27 of 29 The resulting carbonate ions allow the
production of corals. “Corals combine
calcium and carbonate ions in seawater
to make calcium carbonate skeletons.
Coral polyps produce a special kind of
calcium carbonate called aragonite,
fashioning aragonite crystals into
structures that create coral reefs. Lower
ocean pH means fewer carbonate ions
available for corals to build skeletons.”
http://www.whoi.edu/page.do?pid=130656&tid=3622&cid=177769 (Glenn Gaetani, Woods Hole Oceanographic Institution) In addition, other marine organisms like
pteropods (or “sea butterfly” as shown below) use calcium carbonate as the building blocks for
their skeletons and shells.
http://www.whoi.edu/oceanus/feature/tiny-­‐ubiquitous-­‐vital-­‐delicate-­‐vulnerable However, increasing levels of CO2 in the atmosphere result in a higher concentration of
dissolved CO2, resulting in more carbonic acid and H+ (or H3O+ ion) in ocean water.
Consequently, the additional H+ ions react with and reduce the concentration of CO32- ions.
Ex. 2: Write the balanced chemical equation for hydrogen ion reacting with carbonate ion to
produce its conjugate acid.
The images below show what happens to a pteropod’s shell placed in seawater with a pH and
carbonate ion levels projected for the year 2100 given the current increase in CO2 levels.
http://www.pmel.noaa.gov/co2/story/What+is+Ocean+Acidification%3F CHEM 161: Chapter 4 v0916 page 28 of 29 https://www.niwa.co.nz/news/investigating-­‐ocean-­‐acidification View the following videos about Ocean Acidification:
• Acid Test: Science of Ocean Acidification (3 min.): https://www.youtube.com/watch?v=WlwbqvNViSE
• OSU Ocean Acidification: Changing Waters on the Oregon Coast:
https://www.youtube.com/watch?v=7h08ok3hFSs
•
It’s Okay To Be Smart: Can Coral Reefs Survive Climate Change?
https://www.youtube.com/watch?v=P7ydNafXxJI
Example: Describe at least three problems with far reaching consequences associated with
ocean acidification.
CHEM 161: Chapter 4 Notes v0417 page 29 of 29