Download Aqueous Reactions and Solution Stoichiometry (Chapter 4)

Aqueous Reactions and Solution Stoichiometry
(Chapter 4)
Water has many unique chemical and physical properties. Possibly one of the most important
is its ability to dissolve other substances to form solutions. Solutions are homogeneous
mixtures of two or more substances. The solvent (usually the substance present in the
greatest quantity) causes the other substance(s), the solute(s), to dissolve and enter into the
solution.
In an aqueous solution, water is the solvent, or dissolving medium. The symbol “(aq)” in a
chemical equation indicates that the substance is dissolved in water.
CaCO3 (s) + H2O (l) + CO2 (aq) Ca(HCO3)2 (aq)
Some substances dissociate (split apart) in solution to form ions. Since such solutions can
conduct an electric current, the dissolved substance is classified as an electrolyte. These
substances consist of ions that are held together by the electrostatic forces of attraction
between their opposite charges -- ionic bonding (see Chapter 8).
H 2O
NaCl (s) Na+(aq) + Cl–(aq)
Nonelectrolytes do not dissociate in aqueous solution and thus their solutions do not conduct
electricity. These substances consist of molecules where the atoms are covalently bonded
(see Chapter 8).
H 2O
C12H22O11 (s) C12H22O11 (aq)
Due to the strength of their ionic bonds, some ionic substances dissociate almost completely in
aqueous solution. These substances, such as hydrobromic acid, are classified as strong
electrolytes.
HBr (aq) H+(aq) + Br–(aq)
Some molecular substances, such as acetic acid, partially dissociate in aqueous solution and
are classified as weak electrolytes.
H+(aq) + CH3COO–(aq)
CH3COOH (aq) The double arrows indicate that the reaction occurs in both directions. The longer arrow
(pointing to the left) indicates that at chemical equilibrium, the concentration of the
undissociated acetic acid is greater than that of its dissociated ions.
NOTE:
Please do not confuse concentration (mole per liter) with strength (degree of
dissociation).
Page 1 of 11
Water Solubility
A substance’s solubility is usually expressed in terms of the grams of the solute that can
dissolve in 100.00 cm3 of water. Sometimes is it more meaningful to discuss the solubility in
terms of moles of solute per liter of solution. A substance’s solubility changes as the solution’s
temperature varies.
As a general rule, substances whose solubility is less than 0.01 mol/L at 25 oC are considered
to be insoluble (a.k.a. slightly soluble). Some substances that are insoluble at 25 oC dissolve
at slightly higher temperatures and are referred to as moderately soluble.
The following solubility guidelines are taken from a variety of textbooks, but the primary source
was pp. 134-135 of General Chemistry (6th Ed.), Whitten, Davis, and Peck, Saunders College
Publishing.
1.
2.
3.
4.
5.
6.
7.
8.
9.
The common inorganic acids (HCl, HNO3, etc.) are water soluble. Low-molecular-mass
organic acids (HCOOH, CH3COOH, etc.) are also water soluble.
All common compounds of the Group 1 metal ions (Li+, Na+, K+, Rb+, Cs+) and the
ammonium ion, NH4+, are water soluble.
The common nitrites, NO2–; nitrates, NO3–; acetates, CH3COO–; chlorates, ClO3–; and
perchlorates, ClO4–, are water soluble. AgCH3COO, AgNO2 and KClO4 are moderately
soluble.
(a)
The common chlorides, Cl–, are water soluble except for AgCl, Hg2Cl2, and PbCl2
which is soluble in hot water.
(b)
The common bromides, Br–, and iodides, I–, show approximately the same
solubility behavior as chlorides, but there are some exceptions. As these halide
ions (Cl–, Br–, I–) increase in size, the solubilities of their slightly soluble
compounds decrease. HgBr2 is moderately soluble.
(c)
The common fluorides, F–, are water soluble except MgF2, CaF2, SrF2, BaF2, and
PbF2.
The common sulfates, SO42–, and sulfites, SO32–, are water soluble except PbSO4,
BaSO4, HgSO4; CaSO4, SrSO4, and Ag2SO4 are moderately soluble.
The common metal hydroxides, OH–, are insoluble in water except those of Group 1
metals and the heavier Group 2 metals: Ca(OH)2, Sr(OH)2, Ba(OH)2, and Ra(OH)2 are
moderately soluble.
The common carbonates, CO32–; chromates, CrO42-; oxalate, C2O42–; phosphates, PO43–
; and arsenates, AsO43–, are insoluble in water except those of Group 1 metals and
NH4+. MgCO3 is moderately soluble.
The common sulfides, S2–, are insoluble in water except for NH4+ and metal ions in
Groups 1 and 2.
Metal oxides, O2–, are generally insoluble. The following exceptions are soluble Li2O,
Na2O, K2O, and BaO while CaO and SrO are moderately soluble. These exceptions
dissolve to form hydroxides in an exothermic manner.
Page 2 of 11
Ionic Equations
During aqueous chemical reactions involving ionic substances, some ions are not involved in
the actual reaction. These uninvolved ions are known as spectator ions. When writing the
chemical equation to represent a reaction involving ions, we can choose to represent the
reaction in one of three manners.
The molecular equation shows all of the involved species and their phases:
AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)
The complete ionic equation shows all of the involved species with the strong electrolytes
dissociated into aqueous ions. Solids, liquids, gases, and aqueous molecular species retain
their molecular formulas:
Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) AgCl (s) + Na+(aq) ) + NO3–(aq)
In the net ionic equation, the spectator ions are eliminated to show only the species that are
actually involved in the chemical reaction:
Ag+(aq) + Cl–(aq) AgCl (s)
REACTION TYPES
EXCHANGE (metathesis) reactions involve the cation from one solute exchanging anions with
the cation from the other solute.
Precipitation reactions occur when solute ions from two, or more, aqueous solutions combine
to form a solid.
Aqueous solutions of barium chloride and sodium sulfate react as follows:
BaCl2 (aq) + Na2SO4 (aq) BaSO4 (s) + 2 NaCl (aq)
The barium sulfate forms a solid precipitate that settles to the bottom of the solution.
Acid-Base reactions: An acid dissociates in water to form H+(aq). A base dissociates in water
to form OH–(aq). The neutralization reaction between the acid’s H+(aq) and the base’s OH–(aq)
forms water. A salt is formed when the cation from the base combines with the anion from the
acid. Soluble salts can be recovered by evaporating the water from the solution.
Page 3 of 11
The reaction of nitric acid with potassium hydroxide (the base) produces water and potassium
nitrate (the salt):
HNO3 (aq) + KOH (aq) H2O (l) + KNO3 (aq)
Some acid-base reactions produce salts that decompose to form water and a gas:
2 HNO3 (aq) + Na2CO3 (aq) 2 NaNO3 (aq) + H2O (l) + CO2 (g)
Most acids and bases are weak… they are only partially dissociated in water. Strong acids
and bases dissociate almost completely in water. Since there are so few strong acids and
bases, it is easier to remember which common acids and base are strong rather than trying to
learn rules to determine whether or not an acid or base is weak or strong.
Common Strong Acids
Hydrochloric
Hydrobromic
Hydroiodic
Chloric
Perchloric
Nitric
Sulfuric
**
HCl
HBr
HI
HClO3
HClO4
HNO3
H2SO4**
Common Strong Bases
Lithium hydroxide
Sodium hydroxide
Potassium hydroxide
Rubidium hydroxide
Cesium hydroxide
Calcium hydroxide
Strontium hydroxide
Barium hydroxide
LiOH
NaOH
KOH
RbOH
CsOH
Ca(OH)2
Sr(OH)2
Ba(OH)2
Note: H2SO4’s first ionization is strong while its second ionization is weak
H2SO4 (aq) H+(aq) + HSO4–(aq)
H+(aq) + SO42–(aq)
HSO4–(aq) OXIDATION-REDUCTION reactions involve the transfer of electrons between atoms.
Oxidation involves the loss of electrons which makes the oxidized substance more positive,
thus increasing its ability to combine with oxygen.
Reduction involves the gain of electrons which makes the reduced substance more negative
thus reducing its ability to combine with oxygen.
The substance that is oxidized gives up electrons – allowing another substance to gain
electrons (reducing it). Thus the oxidized substance is classified as the reducing agent.
The substance that is reduced gains electrons – allowing another substance to lose electrons
(oxidizing it). Thus the reduced substance is classified as the oxidizing agent.
In a chemical reaction, oxidation cannot occur without an accompanying reduction.
Page 4 of 11
Oxidation numbers (a.k.a. oxidation states) are a “bookkeeping” tool to help determine whether
or not electrons have been transferred during a chemical reaction.
Oxidation numbers are assigned on a per atom basis. The following general “rules” are
adapted mainly from pp. 138-139 of General Chemistry (6th Ed.), Whitten, Davis, and Peck,
Saunders College Publishing will help with the assignment of oxidation numbers. In order of
precedence the rules are:
1.
2.
3.
4.
5.
6.
7.
8.
9.
The atoms of free, uncombined elements have oxidation numbers of zero. This
includes the diatomic elements: H2, N2, O2, F2, Cl2, Br2, and I2 as well as O3, P4, S8, and
C60.
The oxidation number of an element in a monatomic ion is equal to the charge on the
ion.
The sum of the oxidation numbers of all the atoms in a compound is zero.
The sum of the oxidation numbers of all the atoms in a polyatomic ion equals the charge
on the ion.
Fluorine’s oxidation number is 1- in all of its compounds.
Hydrogen’s oxidation number is 1+ in most of its compounds, except its oxidation
number is 1- in metallic hydrides.
Oxygen’s oxidation number is 2- in its compounds, except for:
(a)
peroxides, O22–, where each O has a 1-. Examples: H2O2, Na2O2, and CaO2.
(b)
superoxides, O2–, where each O has a 1/2–. Examples: KO2 and RbO2.
(c)
In OF2, O has a 2+.
The element’s position on the periodic table often helps with the assignment of oxidation
numbers:
(a)
Group 1 metals have an oxidation number of 1+ in all of their compounds.
(b)
Group 2 metals have an oxidation number of 2+ in all of their compounds.
(c)
Group 3 metals have an oxidation number of 3+ in all of their compounds with a
few rare exceptions.
(d)
Group 14 can an oxidation number of 2+ or 4+, except that C and Si often have
other charges.
(e)
Group 15 elements have an oxidation number of 3– in binary compounds with
metals, hydrogen, or in ammonium. They can also have an oxidation number of
3+ or 5+ in compounds with more electronegative elements.
(f)
Group 16 (except O) elements have 2– in binary compounds with metals or
hydrogen, or with ammonium. They can also have an oxidation number of 4+ or
6+ with O and the lighter halogens.
(g)
Group 17 elements have 1– in binary compounds with metals or hydrogen, or
with ammonium. Cl, Br, and I can have an oxidation number of 1+, 3+, 5+, or 7+
with O or a lighter halogen.
The oxidation number for the transition (Groups 3-12) and the inner-transition metals
must be determined based on what they are combined with – or from the Roman
numeral in their compound name. Exceptions: Ag forms 1+, Cd and Zn form 2+,
Page 5 of 11
Example:
Zn(s) + 2 HBr(aq) ZnBr2(aq) + H2(g)
Zn(s) and H2(g) have oxidation numbers of zero
H+(aq) has a 1+ oxidation number
Br–(aq) is a spectator ion and has an oxidation number of 1–.
Zn2+(aq) has an oxidation number of 2+.
Zn(s) Zn2+(aq) + 2 e–
2 H+(aq) + 2 e– H2(g)
is the oxidation, Zn becomes more +
is the reduction, H+ becomes less +
Zn(s) is the reducing agent and H+(aq) is the oxidizing agent
REPLACEMENT
The Activity Series (Table 4.5, p. 124) can be used to predict whether or not a metal will
“replace” another in an aqueous compound. Metals listed higher on the activity series list are
more easily oxidized and thus reduce the lower metal:
Ni(s) + Cu2+(aq) Ni2+(aq) + Cu(s)
A similar series can be created for the halogens:
F would replace Cl–, Br–, or I–
Cl would replace Br–, or I–
Br would replace I–
I would not replace F–, Cl–, or Br–
Cl2 (aq) + 2 NaI (aq) 2 NaCl (aq) + I2 (aq)
I2 (aq) + 2 NaF (aq) No Reaction
These “activities” are directly related to the elements’ standard reduction potentials (Chapter
20: Electrochemistry).
Since most of the chemical reaction they we study are in aqueous solution, we need to be able
to determine the amount of reactants and products present in a given volume of solution.
There are several methods of expressing the concentration of the solutes; we will mainly be
using molarity (M).
Molarity =
moles of solute
liters of solution
mol = M x L
L =
Page 6 of 11
mol
M
Example Problem #1:
What is the molarity of 27.5 g of potassium nitrate in 235 mL of solution?
mol KNO3
=
L solution
M =
1 mol KNO3
101.10324 g KNO3
= 1.16 M KNO3
1 L
235 mL x
1000 mL
27.5 g KNO3 x
Example Problem #2
What is molar concentration for the ions formed when 125 g Ca(NO3)2 is dissolved in 437 mL
of solution?
M Ca(NO3)2 =
M Ca
2+
mol Ca(NO3) 2
=
L solution
1 mol Ca(NO3)2
164.08788 g Ca(NO3) 2
= 1.74 M Ca(NO3)2
1 L
437 mL x
1000 mL
125 g Ca(NO3) 2 x
1 mol Ca 2+
= 1.74 M Ca(NO3)2 x
= 1.74 M Ca2+
1 mol Ca(NO3) 2
Note: Round-off error accounts for the fact that the NO3– concentration exceeds double that
for Ca2+
In the laboratory, reagent solutions are typically prepared by dissolving solid, liquid, or gaseous
substances with deionized water. Some solutions are prepared by diluting “stock” solutions
with deionized water.
NOTE:
The addition of water during the dilution process has an insignificant impact on
the number of moles of solute present in the solution.
Example Problem #3
What mass of solid barium chloride must be dissolved to prepare 1250 mL of a 0.245 M
0.245 mol
solution? Note: 0.245 M is equivalent to
1 L
0.245 M BaCl2 x 1250 mL x
1 L
208.2324 g BaCl2
x
= 63.8 g BaCl2
1000 mL
1 mol BaCl 2
Page 7 of 11
Example Problem #4
What volume of concentrated hydrochloric acid (11.7 M) must be diluted with water to prepare
2250.0 mL of 0.347 M solution?
Method 1:
2250.0 mL solution x 0.347 M HCl x
1 L solution
1 L HCl
x
= 0.0667 L HCl
1000 mL solution
11.7 M HCl
We would probably convert the volume to 66.7 mL.
Method 2:
Since the moles of solute is unchanged by the dilution process, and moles = molarity x liters,
we can use the relationship: M1V1 = M2V2.
Since we are solving for a volume, V1, we do not necessarily need to change the 2230.0 mL to
liters.
M 2V 2
M1
(0.347 M HCl)(2250.0 mL solution)
V1 =
= 66.7 mL solution
11.7 M HCl
M1V 1 = M 2V 2 V 1 =
SOLUTION STOICHIOMETRY
When we are performing stoichiometric calculations involving aqueous solutions, we must
follow pretty much the same steps as we did in Chapter 3:
1.
Write the balanced chemical equation for the reaction.
2.
Convert the mass, or volume, of each reactant to moles.
3.
Use the coefficient ratios from the balanced chemical equation to determine the moles
of product(s) formed.
4.
Convert the moles of products to grams to obtain the theoretical yield.
5.
If an actual yield is given, determine the reaction’s percentage yield.
Page 8 of 11
Example Problem #5:
What is the percent yield for a reaction that produces 17.9 g of lead (II) sulfate when 125 mL of
0.500 M lead (II) nitrate is combined with 75.5 mL of 0.295 M aluminum sulfate?
3 Pb(NO3)2 (aq) + Al2(SO4)3 (aq) 3 PbSO4 (s) + 2 Al(NO3)3 (aq)
125 mL Pb(NO 3)2 x
1L
1000 mL
75.5 mL Al 2( SO 4)3 x
x 0.500 M Pb(NO3)2 x
3 mol PbSO4
3 mol Pb(NO3)2
1 L
x
303.2636 g PbSO 4
1 mol PbSO 4
3 mol PbSO 4
303.2636 g PbSO 4
x 0.295M Al 2(SO 4) 3 x
x
1 mol Al 2(SO 4) 3
1 mol PbSO 4
1000 mL
= 19.0 g PbSO 4
= 20.3 g PbSO 4
The Pb(NO3)2 is the limiting reactant and the theoretical yield for the lead (II) sulfate is 19.0 g.
% yield=
actual yield
x 100
theoretical yield
% yield =
17.9 g
x 100 = 94.2 %
19.0 g
TITRATIONS
Titration is an analytical procedure where the amount of an unknown substance is determined
by the addition of a standard solution (a solution with a known concentration) until the
equivalence point is reached. The equivalence point is reached when stoichiometrically equal
amounts of each reactant have been added to the solution.
In most titrations, if a suitable indicator (a substance that changes color at a specific color) is
used, the indicator will change color (the end-point) at approximately the equivalence point.
There are several indicators to choose from for a particular titration. The choice of indicator is
determined by the pH at which the equivalence point occurs (see Chapter 17 for additional
details).
Indicators such as phenolphthalein, a weak organic acid, is often used in acid-base titrations
involving a strong base (such as NaOH) and a weak acid (such as HC2H3O2). Phenolphthalein
is colorless in acidic solution and turns to pink/red in base (around pH = 8.2).
Methyl red or methyl orange might be used for titrations involving strong acids, such as
HCl (aq), and weak bases, such as NH3 (aq). The equivalence point for this titration would occur
at a pH of about 4 … well below where phenolphthalein changes color.
Indicators such as bromthymol blue or litmus change color at a pH of about 7, which would be
appropriate for use during a strong acid-strong base titration.
Page 9 of 11
Example Problem #6
What is the molarity of formic acid, HCOOH, a monoprotic acid with one ionizible H+, if 45.73
mL of 0.1025 M strontium hydroxide was required to titrate a 20.0 mL sample of the formic acid
solution?
2 HCOOH (aq) + Sr(OH)2 (aq) 2 H2O (l) + Sr(COOH)2 (aq)
Method 1:
M HCOOH =
mol HCOOH
L HCOOH
45.73 mL Sr (OH ) 2 x
=
1 L Sr (OH) 2
1000 mL Sr (OH) 2
20.0 mL HCOOH x
x 0.1025 M Sr (OH ) 2 x
2mol HCOOH
1 mol Sr (OH ) 2
1 L HCOOH
= 0.469 M HCOOH
1000 mL HCOOH
Method 2:
At the equivalence point, the mol H+(aq) = mol OH–(aq), thus:
mol ionized H + ions
mol ionized OH ions
x Macid x Vacid =
x M base x Vbasse
mol acid molecules
mol base molecules
HCOOH donates one H+ per molecule, and Sr(OH)2 donates two OH– for each of its molecules
Thus we have:
1 x M acid x Vacid = 2 x M base x Vbase
M acid =
M acid =
2 x Mbase x Vbase
1 x Vacid
1 L
1000 mL = 0.469 M HCOOH
1 L
1 x 20.0 mL x
1000 mL
2 x 0.1025 M x 45.73 mL x
Page 10 of 11
Example Problem #7
What is the mass percent of HCOOH in the solution if the density is 1.03 g/cm3?
Assume we have 1.0000 L of the solution.
The mass of HCOOH in one liter of the solution would be found by multiplying the molarity by
the volume to get moles and then multiplying by HCOOH’s molar mass.
The mass of the solution is determined by multiplying its volume by its density.
mass % =
g HCOOH
g solution
0.469 mol HCOOH x
x 100 =
46.02538 g HCOOH
1 mol HCOOH
1.03 g
1000 mL
x
1.0000 L x
1 mL
1 L
Page 11 of 11
x 100 = 2.10 % HCOOH