Chapter 16
16.1
HBr is a strong acid and will therefore give its hydrogen atom to a water molecule,
making a hydronium cation and a bromine anion.
16.2
HNO3 is a strong acid and will therefore give its hydrogen atom to a water molecule,
making a hydronium cation and a nitrate anion.
16.3
HClO4 is a strong acid and dissolves in water to generate H3O+cations and ClO −4 anions.
Since there are no bases present for the hydronium ion to react with, the only equilibrium
occurring is proton transfer reaction between water molecules:
OH- + H3O+
Kw = 1.00 x 10-14
H2O + H2O
To determine the concentrations of hydroxide and hydronium ions, set up a concentration
table, write the equilibrium expression, and solve for the final concentrations.
The initial concentration of hydronium ions is the same as the concentration of perchloric
acid, [H3O+] = 1.25 x 10-3M. The concentration table is:
Reaction:
H2O +
H3O+ +
H2O
1.25 x 10-3
Start (M)
----
----
Change (M)
----
----
solvent
solvent
Final (M)
+x
1.25 x 10-3 + x
The equilibrium expression is:
Kw = [H3O+][OH-]
1.00 x 10-14 = [0.00125 + x][x],
We can make the assumption that x << 0.00125M
1.00 x 10-14 = [0.00125][x],
x = 8.00 x 10-12 M = [OH-];
[H3O+] = 1.25 x 10-3 M + 8.00 x 10-12 M = 1.25 x 10-3 M
Thus, the assumption that x<<0.00125 is valid.
506
OH–
0
+x
x
Chapter 16
16.4
NaOH is a strong base and dissolves in water to generate Na+ cations and OH- anions.
Since there are no acids present to react with the hydroxide, the only equilibrium
occurring is proton transfer reaction between water molecules:
H2O + H2O
OH- + H3O+
Kw = 1.00 x 10-14
To determine the concentrations of hydroxide and hydronium ions set up a concentration
table, write the equilibrium expression, and solve for the final concentrations. The initial
concentration of hydroxide ions will be the concentration of NaOH,
[OH-] = 3.65 x 10-2M. The concentration table is:
H3O+ +
Reaction:
H2O +
Start (M)
----
----
0
Change (M)
----
----
+x
solvent
solvent
x
Final (M)
H2O
OH–
0.0365
+x
0.0365 + x
The equilibrium expression is:
Kw = [H3O+][OH-]
1.00 x 10-14 = [x][0.0365 + x]
We can make the assumption that x << 0.0365
1.00 x 10-14 = [x][0.0365],
x = 2.74 x 10-13 M = [H3O+]; our assumption is valid
[OH-] = 0.0365M + 2.74 x 10-13 M = 0.0365 M
16.5
We are asked to determine the final concentrations of all the ions in a final solution.
Begin by analyzing the chemistry. HCl is strong acid and dissolves in water to generate
H3O+cations and Cl- anions. Any water solution always has OH- and H3O+ ions with the
equilibrium:
OH- + H3O+
Kw = 1.00 x 10-14
H2O + H2O
Therefore, the ions present in this solution are: H3O+, OH-, and ClThis is a dilution type problem. The first step is to determine the concentration of HCl in
the flask after the dilution.
MiVi = MfVf
Mf =
M iVi (12.1 M)(1.00 mL)
=
= 0.121 M HCl in final solution.
Vf
100. mL
Since Cl- is a spectator ion, its concentration is the same as that of HCl, [Cl-] = 0.121 M.
507
Chapter 16
The rest of the ion concentrations are determined by the equilibrium. Set up a
concentration table, write the equilibrium expression, and solve for the final
concentrations.
H3O+ +
H2O +
Start (M)
----
----
0.121
Change (M)
----
----
+x
+x
solvent
solvent
0.121 + x
x
Final (M)
H2O
OH–
Reaction:
0
The equilibrium expression is:
Kw = [H3O+][OH-]
1.00 x 10-14 = x(0.121 +x );
assume x << 0.121
1.00 x 10-14 = x(0.121),
x = [OH-] = 8.26 x 10-14M
[H3O+] = 0.121 M + 8.26 x 10-14M = 0.121 M
Here are the final concentrations:
[Cl-] = [H3O+] = 0.121 M
[OH-] = 8.26 x 10-14 M
16.6
We are asked to determine the final concentrations of all the ions in a final solution.
Begin by analyzing the chemistry. HClO4 is a strong acid and dissolves in water to
generate H3O+cations and ClO −4 anions. Any water solution always has OH- and H3O+
ions with the equilibrium:
H2O + H2O
OH- + H3O+
Kw = 1.00 x 10-14
Therefore, the ions present in this solution are: H3O+, OH-, and ClO4This is a dilution type problem. The first step is to determine the concentration of HClO4
in the flask after the dilution.
MiVi = MfVf
Mf =
M iVi (14.8 M)(5.00 mL)
=
= 0.0740 M HClO4 in final solution.
Vf
1000. mL
Since ClO −4 is a spectator ion, its concentration is the same as that of HClO4,
[ ClO −4 ]= 0.0740M.
The rest of the ion concentrations are determined by the equilibrium. Set up a
concentration table, write the equilibrium expression, and solve for the final
concentrations.
508
Chapter 16
H3O+ +
Reaction:
H2O +
Start (M)
----
----
Change (M)
----
----
solvent
solvent
Final (M)
H2O
0.0740
+x
0.0740 + x
OH–
0
+x
x
The equilibrium expression is:
Kw = [H3O+][OH-]
1.00 x 10-14 = x(0.0740 +x );
assume x<<0.0740
1.00 x 10-14 = x(0.0740),
x = [OH-] = 1.35 x 10-13 M
[H3O+] = 0.0740 M + 1.35 x 10-13 M = 0.0740 M
Here are the final concentrations:
[ClO4-] = [H3O+] = 0.0740 M
[OH-] = 1.35 x 10-13 M
16.7
We are asked to determine the concentrations of hydroxide and hydronium ions in the
solution. Begin by analyzing the chemistry. HCl is a strong acid and dissolves in water to
generate H3O+cations and Cl- anions. Any water solution always has OH- and H3O+ ions
with the equilibrium:
H2O + H2O
OH- + H3O+
Kw = 1.00 x 10-14
The first step is to determine the initial number of moles of HCl gas and convert that to
concentration of HCl dissolved in the solution:
MMHCl = 1.008 g/mol + 35.453 g/mol = 36.461 g/mol
 1 mol 
0.488 g 
36.461 g 
=0.01338 mol HCl dissolved.


0.01338 mol
[HCl] =
= 0.0412 M
0.325 L
To determine the concentrations of the ions, construct a concentration table, write the
equilibrium expression, and solve for the final concentrations. The initial concentration
of hydronium ions will be the same as the concentration of HCl, [H3O+] = 0.0412 M.
509
Chapter 16
H3O+ +
H2O
OH–
Reaction:
H2O +
Start (M)
----
----
0.0412
Change (M)
----
----
+x
+x
solvent
solvent
0.0412 + x
x
Final (M)
0
The equilibrium expression is:
Kw = [H3O+][OH-]
1.00 x 10-14 = (0.0412 +x)x;
Assume x << 0.0412
1.00 x 10-14 = (0.0412)x;
x = 2.43 x 10-13 M = [OH-]; the assumption is vaild
[H3O+] = 0.0412 M + 2.43 x 10-13 M = 0.0412 M
16.8
We are asked to determine the concentrations of hydroxide and hydronium ions in the
solution. Begin by analyzing the chemistry. NaOH is a strong base and dissolves in
water to generate Na+ cations and OH- anions. Any water solution always has OH- and
H3O+ ions with the equilibrium:
H2O + H2O
OH- + H3O+
Kw = 1.00 x 10-14
The first step is to determine the concentration of NaOH:
MMNaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
 1 mol 
0.345 g 
 40.00 g 
= 0.008625 mol NaOH dissolved.


0.008625 mol
[NaOH] =
= 0.03833 M
0.225 L
H3O+ +
Reaction:
H2O +
Start (M)
----
----
0
Change (M)
----
----
+x
Solvent
solvent
x
Final (M)
H2O
Kw = [H3O+][OH-];
1.00 x 10-14 = x(0.03833 +x )
assume x << 0.03833
510
OH–
0.03833
+x
0.03833 + x
Chapter 16
1.00 x 10-14 = x(0.03833);
x = 2.61 x 10-13 M = [H3O+]; assumption is valid
[OH-] = 0.03833 M + 2.61 x 10-13 M = 0.03833 M
16.9
Conversion from hydronium ion molarity to pH is accomplished by taking logarithm to
base ten and changing sign, pH = – (log[H3O+]):
(a) – 0.60; (b) 5.426; (c) 2.32; and (d) 3.593.
16.10 Conversion from hydronium ion molarity to pH is accomplished by taking logarithm to
base ten and changing sign, pH = – (log[H3O+]):
(a) -0.097; (b) 5.530; (c) 1.437; and (d) 3.128 .
16.11 Because pH + pOH = 14.00, pH = 14.00 – pOH. Convert from hydroxide ion molarity to
pOH by taking logarithm to base ten and changing sign, pOH = – (log[OH-]).
Thus, pH = 14 + log[OH-]:
(a) 14.60 ; (b) 8.574; (c) 11.68; and (d) 10.407.
16.12 Because pH + pOH = 14.00, pH = 14.00 – pOH. Convert from hydroxide ion molarity to
pOH by taking logarithm to base ten and changing sign, pOH = – (log[OH-]).
Thus, pH = 14 + log[OH-]:
(a) 14.097 ; (b) 8.470; (c) 12.563; and (d) 10.872.
16.13 Take 10-pH to convert pH into hydronium ion concentration:
(a) 0.22 M; (b) 1.4 x 10-8 M; (c) 2.1 x 10-4 M; (d) 4.7 x 10-15 M.
16.14 Take 10-pH to convert pH into hydronium ion concentration:
(a) 0.028 M; (b) 1.4 x 10-4 M; (c) 1.8 x 10-10 M; (d) 6.0 x 10-12 M.
16.15 Use pH + pOH = 14.00 to convert pH to pOH. Then take 10-pOH to convert pOH into
hydroxide ion concentration:
(a) pOH = 13.34, [OH-] = 4.6 x 10-14 M;
(b) pOH = 6.15, [OH-] = 7.1 x 10-7 M;
(c) pOH = 10.32, [OH-] = 4.8 x 10-11 M;
(d) pOH = – 0.33, [OH-] =2.1 M.
16.16 Use pH + pOH = 14.00 to convert pH to pOH. Then take 10-pOH to convert pOH into
hydroxide ion concentration:
(a) pOH = 12.44, [OH-] = 3.6 x 10-13 M;
(b) pOH = 10.15, [OH-] = 7.1 x 10-11 M;
(c) pOH = 4.25, [OH-] = 5.6 x 10-5 M;
(d) pOH = 2.78, [OH-] = 1.7 x 10-3 M.
16.17 To calculate the pH of a solution, it is necessary to determine either the hydronium ion
concentration or the hydroxide ion concentration.
511
Chapter 16
For each species determine the initial concentrations, construct a concentration table,
write the equilibrium expression, and solve for the concentrations.
(a) Strong base, carry out the water equilibrium to determine H3O+
OH- +
Reaction:
H2O + H2O
Start (M)
----
----
Change (M)
----
----
+x
solvent
1.5 + x
Final (M)
solvent
1.5
H3O+
0
+x
x
The equilbrium expression is:
Kw = 1.0 x 10-14 = [H3O+][OH-] = x(1.5 + x); assume x << 1.5
1.0 x 10-14 = x(1.5); from which x = 6.7 x 10-15 M = [H3O+]; assumption valid
pH = -log(6.7 x 10-15) = 14.18.
(b) weak base, carry out equilibrium calculation to determine [OH-]:
C5H5NH+ +
OH–
Reaction:
H2O +
Start (M)
----
1.5
0
Change (M)
----
-x
+x
+x
1.5 – x
x
x
Final (M)
Solvent
C5H5N
0
The equilibrium expression is:
[C 5 H 5 NH + ]eq [OH - ]eq
x2
Kb = 1.7 x 10-9 =
=
; Assume x << 1.5:
[C 5 H 5 N]eq
1.5 − x
1.7 x 10-9 =
x2
, from which x2 = 2.55 x 10-9 and x = 5.0 x 10-5; assumption is valid
1.5
[OH-] = 5.0 x 10-5 M,
pOH = -log (5.0 x10-5) = 4.30, and
pH = 14.00 – 4.30 = 9.70;
512
Chapter 16
(c) weak base, carry out equilibrium calculation to determine [OH-]:
NH3OH + +
OH–
Reaction:
H2O + NH2OH
Start (M)
----
1.5
0
Change (M)
----
-x
+x
+x
1.5 – x
x
x
Final (M)
solvent
0
The equilibrium expression is:
[NH 3 OH + ]eq [OH - ]eq
x2
Kb = 8.7 x 10-9 =
=
; Assume x << 1.5:
[NH 2 OH]eq
1.5 − x
8.7 x 10-9 =
x2
, from which x2 = 1.31 x 10-8 and x = 1.1 x 10-4; assumption valid
1.5
[OH-] = 1.1 x 10-4 M,
pOH = -log(1.1 x 10-4) = 3.96, and
pH = 14.00 – 3.96 = 10.04;
(d) Weak acid, carry out equilibrium calculation to determine [H3O+]:
HCO2- +
H3O+
Reaction:
H2O +
Start (M)
----
1.5
0
Change (M)
----
-x
+x
+x
1.5 - x
x
x
Final (M)
solvent
HCO2H
0
The equilibrium expression is:
[HCO 2 ]eq [H 3 O + ]eq
x2
-4
Ka= 1.8 x 10 =
=
; Assume x << 1.5:
[HCO 2 H]eq
1.5 − x
1.8 x 10-4 =
x2
, from which x2 = 2.7 x 10-4 and x = 1.6 x 10-2; assumption is valid
1.5
[H3O+] = 1.6 x 10-2 M, and
pH = -log(1.6 x 10-2) = 1.80
16.18 To calculate the pH of a solution, it is necessary to determine either the hydronium ion
concentration or the hydroxide ion concentration.
513
Chapter 16
For each species determine the initial concentrations, construct a concentration table,
write the equilibrium expression, and solve for the concentrations.
(a) weak base, carry out equilibrium calculation to determine [OH-]:
OH - +
NH4+
Reaction:
H2O +
Start (M)
----
2.5 x 10-2
0
0
Change (M)
----
-x
+x
+x
x
x
Final (M)
solvent
NH3
2.5 x 10-2- x
The equilibrium expression is:
+
[NH 4 ]eq [OH - ]eq
x2
-5
Kb = 1.8 x 10 =
=
; Assume that x << 2.5 x 10-2
[NH 3 ]eq
2.5 x 10 -2 − x
x2
2.5 x 10-2
-4
x = 6.7 x 10 M
[OH-] = 6.7 x 10-4 M,
pOH = -log(6.7 x 10-4) = 3.17, and pH =10.83;
1.8 x 10-5 =
(b) weak acid, carry out an equilibrium calculation to determine [H3O+]:
ClO - +
H3O+
Reaction:
H2O +
Start (M)
----
2.5 x 10-2
0
Change (M)
----
-x
+x
+x
x
x
Final (M)
solvent
HClO
2.5 x 10-2- x
0
The equilibrium expression is:
[ClO - ]eq [H 3 O + ]eq
x2
Ka= 4.0 x 10-8 =
; Assume x << 2.5 x 10-2:
=
[HClO] eq
2.5 x 10 -2 − x
x2
, from which
(2.5 x 10-2 )
x2 = 1.0 x 10-9 and x = 3.2 x 10-5; assumption is valid
4.0 x 10-8 =
[H3O+] = 3.2 x 10-5 M, and
pH = -log(3.2 x 10-5) = 4.49;
(c) weak acid, carry out an equilibrium calculation to determine [H3O+]:
514
Chapter 16
CN- +
H3O+
Reaction:
H2O +
Start (M)
----
2.5 x 10-2
0
Change (M)
----
-x
+x
+x
x
x
Final (M)
solvent
HCN
2.5 x 10-2- x
0
The equilibrium expression is:
[CN - ]eq [H 3 O + ]eq
x2
Ka= 6.2 x 10-10 =
=
; Assume x << 2.5 x 10-2:
-2
[HCN]eq
2.5 x 10 − x
x2
, from which
2.5 x 10-2
x2 = 1.55 x 10-11 and x = 3.9 x 10-6; assumption is valid
6.2 x 10-10 =
[H3O+] = 3.9 x 10-6 M, and
pH = -log(3.9 x 10-6) = 5.41;
(d) Strong base. The initial hydroxide concentration will be twice the concentration of
Ba(OH)2 since there are two hydroxide ions per molecule.
[OH-] = 2(2.5 x 10-2 M) = 5.0 x 10-2 M,
OH- +
Reaction:
H2O +
Start (M)
----
----
Change (M)
----
----
Final (M)
solvent
H2O
Solvent
H3O+
5.0 x 10-2
+x
0
+x
5.0 x 10-2 + x
x
The equilbrium expression is:
Kw = 1.0 x 10-14 = [H3O+][OH-] = x(5.0 x 10-2 + x); assume x<<5.0 x 10-2
1.0 x 10-14 = x(5.0 x 10-2); from which x = 2.0 x 10-13 M = [H3O+]; assumption valid
pH = -log(2.0 x 10-13) = 12.70.
16.19 HONH2 is a weak base and will therefore take a proton from water to form hydroxide
ions.
515
Chapter 16
HCO2H is a weak acid and will give a proton to water to form hydronium ions.
16.20 NH3 is a weak base and will take a proton from water to form hydroxide ions:
HCN is a weak acid and will give a proton to water to form hydronium ions.
16.21 Follow standard procedures for dealing with equilibrium problems:
(a) HN3 is a weak acid. Major species: HN3 and H2O, Minor species: N3- and H3O+, and
OH(b) Construct the concentration table, write the equilibrium expression, and solve for the
concentrations.
N3- +
Reaction:
H2O +
Start (M)
----
1.50
Change (M)
----
-x
Final (M)
Ka= 2.5 x 10-5 =
HN3
solvent
[N 3- ]eq [H 3 O + ]eq
[HN 3 ]eq
1.50 - x
=
0
H3O+
0
+x
+x
x
x
x2
; assume that x << 1.50
1.50 - x
x2
;
1.50
x2 =3.75 x 10-5 and x = 6.1 x 10-3
2.5 x 10-5 =
[H3O+] = [N3-] = 6.1 x 10-3 M, and [HN3] = 1.50 M – 6.1 x 10-3 M = 1.50 M
1.0 x 10 -14
[OH-] =
= 1.6 x 10-12 M
-3
6.1 x 10
(c) pH = – log (6.1 x 10-3) = 2.21;
516
Chapter 16
(d) The dominant equilibrium is proton transfer from water to HN3:
16.22 Follow standard procedures for dealing with equilibrium problems:
(a) HCNO is a weak acid. Major species: HCNO and H2O, Minor species: CNO- and
H3O+ and OH-;
(b) Construct the concentration table, write the equilibrium expression, and solve for the
concentrations.
Reaction:
H2O +
Start (M)
----
0.0275
Change (M)
----
-x
Final (M)
HCNO
solvent
pKa = 3.46, Ka= 3.5 x 10-4 =
0.0275 - x
[CNO - ]eq [H 3O + ]eq
[HCNO]eq
CNO - +
H3O+
0
0
+x
+x
x
x
; Substitute and solve for x:
x2
, so x2 = (3.5 x 10-4)(0.0275 – x) = 9.6 x 10-6 – 3.5 x 10-4 x;
0.0275 − x
0 = x2 + 3.5 x 10-4 x – 9.6 x 10-6;
3.5 x 10-4 =
− b ± b 2 − 4ac − (3.5 x 10 − 4 ) ± (3.5 x 10 − 4 ) 2 − 4(9.6 x 10 − 6 )
=
= 2.9 x 10 − 3 ;
2a
2
[H3O+] = [CNO-]= 2.9 x 10-3 M, and [HCNO] = 0.0275 – 0.0029 = 0.0246 M
-14
[OH-] = 1.00 x 10 = 3.4 x 10-12 M
2.9 x 10-3
x=
(c) pH = – log( 2.9 x 10-3) = 2.54;
(d)
16.23 Follow standard procedures for dealing with equilibrium problems:
517
Chapter 16
(a) N(CH3)3 is a weak base. Major species: N(CH3)3, H2O, Minor species:
HN(CH3)3+,OH-, and H3O+;
(b) Construct the concentration table, write the equilibrium expression, and solve for the
concentrations.
Reaction:
H2O +
Start (M)
----
0.350
Change (M)
----
-x
Final (M)
solvent
N(CH3)3
OH - +
0.350 - x
HN(CH3)3+
0
0
+x
+x
x
x
The equilibrium expression is:
[HN(CH 3 ) 3+ ]eq [OH - ]eq
x2
-5
=
; assume x << 0.350
Kb = 6.5 x 10 =
[N(CH 3 ) 3 ]eq
0.350 − x
x2 ;
0.350
2
x = 2.28 x 10-5 and x = 4.8 x 10-3; assumption is valid
6.5 x 10-5 =
[OH-] = [HN(CH3)3+] = 4.8 x 10-3 M,
[N(CH3)3] = 0.350 – 4.8 x 10-3 = 0.345 M;
-14
[H3O+] = 1.0 x 10 = 2.1 x 10-12 M
4.8 x 10 -3
(c) pH =-log(2.1 x 10-12) = 11.68;
(d) The dominant equilibrium is proton transfer from water to trimethylamine:
16.24 Follow standard procedures for dealing with equilibrium problems:
(a) C6H5NH2 is a weak base. Major species: C6H5NH2, H2O, Minor species:
C6H5NH+,OH- , and H3O+;
(b) Construct the concentration table, write the equilibrium expression, and solve for the
concentrations.
518
Chapter 16
Reaction:
H2O +
Start (M)
----
1.85x10-3
Change (M)
----
-x
Final (M)
Kb = 7.4 x 10-10 =
solvent
C6H5NH2
OH - +
C6H5NH3+
0
0
+x
+x
x
x
1.85x10-3– x
[OH − ][C6 H 5 NH 3+ ]
x2
; assume x << 1.85 x 10-3
=
-3
[C 6 H 5 NH 2 ]
1.85 x 10 − x
x2
, from which x2 = 1.37 x 10-12;
-3
1.85 x 10
x = 1.2 x 10-6; assumption correct
7.4 x 10-10 =
[OH-] = [C6H5NH3+] = 1.2 x 10-6 M,
[C6H5NH2] = 1.85 x 10-3– 1.2 x 10-6 = 1.85 x 10-3 M;
-14
[H3O+] = 1.00 x 10 = 8.3 x 10-9 M
1.2 x 10 -6
(c) pH =-log(8.3 x 10-9) = 8.08;
(d)
16.25 Examine the chemical formula to determine the nature of a compound: (a) weak base;
(b) weak acid, (c) weak acid; (d) strong base.
16.26 Examine the chemical formula to determine the nature of a compound: (a) strong acid;
(b) weak acid; (c) amphiprotic, weak acid and weak base; (d) weak acid; (e) weak base.
16.27 The conjugate base will have one less H and one less charge, a conjugate acid will have
one more H and one higher charge:
Problem 16.17
C5H5N conjugate acid is C5H5NH+;
HONH2 conjugate acid is HONH3+;
HCO2H conjugate base is HCO2-;
519
Chapter 16
16.28 The conjugate base will have one less H and one less charge, a conjugate acid will have
one more H and one higher charge:
Problem 16.18
Problem 16.26
+
NH3 conjugate acid is NH4 ;
CH3CO2H conjugate base is CH3CO2-;
HClO conjugate base is ClO-;
HOH conjugate base is OH-,
HCN conjugate base is CN-;
HOH conjugate acid is H3O+;
HOCl conjugate base is OCl-;
NH2OH conjugate acid is NH3OH+.
16.29 Conjugate pairs are connected. Any aqueous solution always has OH- and H3O+ ions
with the equilibrium:
(a) NH3 is a weak base:
(b) HCNO is a weak acid:
(c) HClO is a weak acid:
(d) Ba(OH)2 is a strong base:
520
Chapter 16
16.30 Any aqueous solution always has OH- and H3O+ ions with the equilibrium:
(a) HNO3 is a strong acid:
(b) CH3CO2H is a weak acid:
(c) HOH is water:
(d) HOCl is a weak acid:
(e) NH2OH is a weak base:
16.31 Follow standard procedures for dealing with equilibrium problems:
(a) The compound is a salt, so major species are Na+, SO32-, and H2O;
(b) The species with acid-base properties are SO32- (a weak base) and H2O, so the
dominant equilibrium is: H2O(l) + SO32-(aq)
HSO3-(aq) + OH-(aq)
(c) Construct the concentration table, write the equilibrium expression, and carry out an
equilibrium calculation to determine [OH-]:
521
Chapter 16
SO32-
Reaction:
H2O
Start (M)
---
0.45
Change (M)
---
–x
Final (M)
+
solvent
HSO3- +
OH-
0
0
+x
+x
x
x
0.45 – x
The equilibrium reaction is a weak base proton transfer. HSO3- is the species resulting
from the gain of a proton from SO32-, so use Ka2 to determine pKb = 14.00 – pKa:
pKa = 7.20, pKb= 6.80;
The equilibrium expression is:
[HSO -3 ]eq [OH - ]eq
x 2 ; Assume x << 0.45:
Kb = 1.6 x 10-7 =
=
0.45 − x
[SO 32- ]eq
1.6 x 10-7 =
x2
, from which x2 = 7.2 x 10-8 and x = 2.7 x 10-4; assumption valid
0.45
[OH-] = 2.7 x 10-4 M,
pOH = -log(2.7 x 10-4) = 3.57, and
pH = 14.00 – 3.57 = 10.43.
16.32 Follow standard procedures for dealing with equilibrium problems:
(a) The compound is a salt, so major species are Na+, C6H5CO2-, and H2O;
(b) The species with acid-base properties are C6H5CO2- (a weak base) and H2O, so the
C6H5CO2H(aq) + OH-(aq)
dominant equilibrium is: H2O(aq) + C6H5CO2- (aq)
(c) Construct the concentration table, write the equilibrium expression, and carry out an
equilibrium calculation to determine [OH-]:
Reaction:
H2O + C6H5CO2-
Start (M)
---
6.75 x 10-3
Change (M)
---
–x
Final (M)
solvent
6.75 x 10-3 – x
C6H5CO2H +
OH-
0
0
+x
+x
x
x
The equilibrium reaction is a weak base proton transfer. C6H5CO2H is the species
resulting from the gain of a proton from C6H5CO2-, so use Ka to determine
pKb = 14.00 – pKa:
pKa = 4.20, pKb= 9.80;
522
Chapter 16
The equilibrium expression is:
K = 1.6 x 10-10 = [C6 H 5CO 2 H]eq [OH ]eq
b
[C 6 H 5CO -2 ]eq
=
x2
; Assume x << 0.00675:
0.00675 − x
x2
, from which
0.00675
x2 = 1.1 x 10-12 and x = 1.0 x 10-6; assumption valid
1.6 x 10-10 =
[OH-] = 1.0 x 10-6 M,
pOH = -log(1.0 x 10-6) = 6.00, and
pH = 14.00 - 6.00 = 8.00.
16.33 Follow standard procedures for dealing with equilibrium problems:
(a) The compound is a salt, so major species are NH4+, NO3-, and H2O;
(b) The species with acid-base properties are NH4+ (a weak acid) and H2O, so the
dominant equilibrium is: H2O(l) + NH4+(aq)
NH3(aq) + H3O+(aq)
(c) carry out equilibrium calculation to determine [H3O+]:
H2O
Start (M)
---
0.0100
Change (M)
---
–x
Final (M)
solvent
pKa = 9.25, Ka= 5.6 x 10-10 =
+
NH4+
Reaction:
NH3 +
0.0100 – x
[NH +4 ]eq [H 3 O + ]eq
[NH 3 ]eq
=
H3O+
0
0
+x
+x
x
x
x2
; Assume x << 0.0100:
0.0100 − x
x2
, from which
0.0100
x2 = 5.6 x 10-12 and x = 2.37 x 10-6; assumption valid
[H3O+] = 2.37 x 10-6 M, and
pH = -log(2.37 x 10-6) = 5.63.
5.6 x 10-10 =
16.34 Follow standard procedures for dealing with equilibrium problems:
(a) The compound is a salt, so major species are NH+4 , Br , and H2O;
(b) The species with acid-base properties are NH+4 (a weak acid) and H2O, so the
dominant equilibrium is: H2O(l) + NH4+(aq)
NH3(aq) + H3O+(aq)
+
(c) carry out an equilibrium calculation to determine [H3O ]:
523
Chapter 16
Reaction:
H2O
Start (M)
---
Change (M)
---
Final (M)
+
solvent
NH4+
NH3 +
4.75 x 10-2
–x
H3O+
0
0
+x
+x
x
x
4.75 x 10-2 – x
The equilibrium reaction is a weak acid proton transfer, NH3 is the species resulting from
the loss of a proton from NH+4 , so use Kb to determine pKa =14.00 - pKb:
pKb = 4.75, pKa= 9.25,
Ka = 5.6 x 10-10 =
[NH3 ]eq [H3O+ ]eq
[NH +4 ]eq
=
x2
; Assume x << 0.0475:
0.0475-x
x2
, from which x2 = 2.7 x 10-11 and x = 5.2 x 10-6; assumption valid
0.0475
+
[H3O ] = 5.2 x 10-6 M,
pH = -log(5.2 x 10-6) = 5.28.
5.6 x 10-10 =
16.35 H2O + SO2
H2SO3 + H2O
HSO3- + H2O
H2SO3
HSO3- + H3O+ (determines pH)
SO32- + H3O+
16.36 H2O + CO2
H2CO3 + H2O
HCO3- + H2O
H2CO3
HCO3- + H3O+ (determines pH)
CO32- + H3O+
16.37
FeO(s) + 2 H3O+(aq)
16.38
Fe2O3 (s) + 6 H3O +(aq)
16.39 (a)
(b)
Fe2+ (aq) + 3 H2O(l)
2 Fe3+(aq) + 9 H2O(l)
H2SO4 is stronger because anions are poorer proton donors than neutral species.
HClO is stronger because Cl is a more electronegative atom than I. A higher
electronegativity means that Cl attracts more of the electron density around it than
I, weakening the H-X bond and making it easier to break (hence a better proton
donor/acid).
524
Chapter 16
(c)
HClO2 is stronger. O atoms are highly electronegative and attract electron density
around them. Having two O atoms, HClO2 will have less electron density in the
H-X bond than HClO, thus, making the bond easier to break.
16.40 (a)
HBrO3 is stronger. O atoms are highly electronegative and attract electron
density around them. Having three O atoms, HBrO3 will have less electron
density in the H-X bond than HBrO2, thus making the bond easier to break and
making it a stronger acid.
H2S is stronger because the larger S atom provides less bond overlap with the H
making it easier to remove the H.
H2S is a stronger acid because the negative charge on HS- makes it harder to
remove the proton.
(b)
(c)
16.41 Use arrows of different sizes to show differences in electron density shifts.
(b)
O
O
Cl
I
H
H
(c)
O
Cl
O
H
H
Cl
O
16.42 Use arrows of different sizes to show differences in electron density shifts.
(b)
O
H
H
S
H
H
16.43 This problem asks for the concentration of all ionic species present in the solution. Begin
by analyzing the chemistry. NaC2H3O2 is a salt that dissolves in solution to form Na+ and
C2H3O2-. Acetate is a weak base with equilibrium:
C2H3O2H + OHKb= 5.6 x 10-10
H2O + C2H3O2Every aqueous solution has the water equilibrium:
H2O + H2O
H3O+ + OH-
Kw = 1.0 x 10-14
Thus, the ionic species present in the solution are: Na+, C2H3O2-, H3O+, and OH-
525
Chapter 16
Na+ is a spectator ion and will have the same concentration as the initial salt,
[Na+] = 0.250 M
For the remaining ions we will need to set up concentration tables, write the equilibrium
expressions, and solve for the ionic concentrations.
C2H3O2-
C2H3O2H +
OH-
0.250
0
0
----
-x
+x
+x
----
0.250-x
x
x
Reaction:
H2O +
Start (M)
----
Change (M)
Final (M)
The acid-base equilibrium expression is:
x2
-10
Kb = 5.6 x 10 =
; assume that x << 0.250
0.250 − x
x2
5.6 x 10-10 =
; for which x2 = 1.4 x 10-10 and x = 1.2 x 10-5; assumption valid
0.250
[C2H3O2H] = [OH-] = 1.2 x 10-5 M
[C2H3O2-] = 0.250 M – 1.2 x 10-5 M = 0.250 M
Now use a second concentration table to determine the concentrations of hydronium and
hydroxide ions:
Reaction:
H2O +
H2O
H3O+ +
OHStart (M)
----
----
0
1.2 x 10-5
Change (M)
----
----
+x
+x
Final (M)
solvent
solvent
x
1.2 x 10-5 + x
Kw = 1.00 x 10-14 = (x)(1.2 x 10-5 + x); assume x << 1.2 x 10-5
1.00 x 10-14 = (x)(1.2 x 10-5);
x = 8.3 x 10-10 M = [H3O+]; assumption valid
[OH-] = 1.2 x 10-5 M
The ionic concentrations are:
[Na+] = [C2H3O2-] = 0.250 M
[OH-] = 1.2 x 10-5 M
[H3O+] = 8.3 x 10-10 M
16.44 This problem asks for the concentration of all ionic species present in the solution. Begin
by analyzing the chemistry. KBrO is a salt that dissolves in solution to form K+ and
BrO-. BrO- is a weak base with equilibrium:
HBrO + OHKb= 3.6 x 10-6
H2O + BrO-
526
Chapter 16
Every aqueous solution has the water equilibrium:
H2O + H2O
H3O+ + OH-
Kw = 1.00 x 10-14
Thus, the ionic species present in the solution are: K+, BrO-, H3O+, and OHK+ is a spectator ion and will have the same concentration as the initial salt,
[K+] = 3.45 x 10-2 M
For the remaining ions we will need to set up concentration tables, write the equilibrium
expressions, and solve for the ionic concentrations.
BrO-
HBrO +
OH-
3.45 x 10-2
0
0
----
-x
+x
+x
----
3.45 x 10-2 -x
x
x
Reaction:
H2O +
Start(M)
----
Change (M)
Final (M)
The equilibrium expression is:
x2
; assume that x << 3.5 x 10-2
Kb = 3.5 x 10-6 =
-2
3.45 x 10 − x
x2
3.5 x 10 =
;
3.45 x 10-2
x = 3.5 x 10-4 M = [HBrO] = [OH-]
-6
[BrO-] = 3.45 x 10-2 M – 3.5 x 10-4 M = 3.42 x 10-2 M
Now use a second concentration table to determine the concentrations of hydronium and
hydroxide ions:
H3O+ +
OH-
----
0
3.5 x 10-4
----
----
+x
+x
solvent
solvent
x
3.5 x 10-4 + x
Reaction:
H2O +
Start(M)
----
Change (M)
Final (M)
H2O
Kw = 1.00 x 10-14 = (x)(3.5 x 10-4 + x); assume x << 3.5 x 10-4
1.00 x 10-14 = (x)(3.5 x 10-4)
x = 2.9 x 10-11 M = [H3O+]
527
Chapter 16
The concentrations are:
[K+] = 3.45 x 10-2 M
[HBrO] = [OH-] = 3.5 x 10-4 M
[BrO-] = 3.42 x 10-2 M
[H3O+] = 2.9 x 10-11 M
16.45 This problem asks for the concentration of all ionic species present in the solution. Begin
by analyzing the chemistry. H2CO3 is a diprotic acid with equilibria:
H2O + H2CO3
HCO3- + H3O+
Ka1 = 4.5 x 10-7
-2
+
H2O + HCO3
CO3 + H3O
Ka2 = 4.7 x 10-11
Every aqueous solution has the water equilibrium:
H2O + H2O
H3O+ + OH-
Kw = 1.00 x 10-14
Thus, the ionic species present in the solution are: HCO3-, CO3-2, H3O+, and OHSet up concentration tables, write equilibrium expressions, and solve for the ionic
concentrations.
HCO3- +
H3O+
Reaction:
H2O + H2CO3
Start(10 –2 M)
----
1.55
0
0
Change(10 –2 M)
----
-x
+x
+x
Final(10 –2 M)
solvent
1.55-x
x
x
x2
-2
Ka1 = 4.5 x 10 =
; assume x << 1.55 x 10
−2
1.55 x 10 − x
x2
;
4.5 x 10 − 7 =
1.55 x 10 − 2
from which x2 = 6.98 x 10-9 and x = 8.4 x 10-5; assumption valid
−7
[H3O+] = [HCO3-] = 8.4 x 10-5 M
[H2CO3] = 1.55 x 10-2 M
Now set up a concentration table for the second equilibrium:
Reaction:
H2O +
Start(10 –5 M)
----
Change(10 –5 M)
----
HCO3-
CO3-2 +
H3O+
8.4
0
8.4
-x
+x
+x
528
Chapter 16
Final(10 –5 M)
solvent
Ka2 = 4.7 x 10 − 11 =
8.4-x
x
8.4 + x
x(8.4 x 10 − 5 + x )
; assume x << 8.4 x 10-5
−5
8.4x 10 − x
x = 4.7 x 10-11M = [CO32-];
Now use a third concentration table to determine the ionic concentrations of hydronium
and hydroxide ions:
H3O+ +
OH-
----
8.4 x 10-5
0
----
----
+x
+x
solvent
solvent
8.4 x 10-5 + x
x
Reaction:
H2O +
Start(M)
----
Change (M)
Final (M)
H2O
Kw = 1.00 x 10-14 = (x)(8.4 x 10-5 + x); assume x << 8.4 x 10-5
1.00 x 10-14 = (x)(8.4 x 10-5)
x = 1.2 x 10-10 M = [OH-]; assumption valid
ionic concentrations:
[H3O+] = [HCO3-] = 8.4 x 10-5M
[CO32-] = 4.7 x 10-11M
[OH-] = 1.2 x 10-10 M
16.46 This problem asks for the concentration of all ionic species present in the solution. Begin
by analyzing the chemistry. H2SO3 is a diprotic acid with equilibria:
HSO3- + H3O+
Ka1 = 1.4 x 10-2
H2O + H2SO3
-2
+
H2O + HSO3
SO3 + H3O
Ka2 = 6.3 x 10-8
Every aqueous solution has the water equilibrium:
H2O + H2O
H3O+ + OHKw = 1.00 x 10-14
Thus, the ionic species present in the solution are: HSO3-, SO3-2, H3O+, and OHSet up concentration tables, write equilibrium expressions, and solve for the ionic
concentrations.
Reaction:
H2O +
H2SO3
HSO3- +
H3O+
Start(M)
----
0.355
0
0
Change(M)
----
-x
+x
+x
529
Chapter 16
Final(M)
solvent
0.355 - x
x
x
x2
;
0.355 − x
This must be solved using the quadratic equation:
x = 0.064
x = 0.064 M = [H3O+] = [HSO3-]
[H2SO3] = 0.355 M – 0.064 M = 0.291 M
Ka1 = 1.4 x 10 − 2 =
Use these values as initial concentrations in the concentration table for the second
equilibrium reaction:
HSO3-
SO3-2 +
H3O+
0.064
0
0.064
----
-x
+x
+x
solvent
0.064 - x
x
0.064 - x
Reaction:
H2O +
Start(M)
----
Change (M)
Final (M)
The equilibrium expression is:
Ka2 = 6.3 x 10 − 8 =
x(0.064 + x )
; assume x << 0.064
0.064 − x
x = 6.3 x 10-8 M = [SO32-]; assumption valid
Because the hydronium concentration is large the water equilibrium can be considered
negligible:
Kw
1.00 x 10-14
[OH-] =
=
= 1.6 x 10-13 M
+
[H 3O ]
0.064
The ionic concentrations are:
[H3O+] = [HSO3-]= 0.064 M
[SO32-]= 6.3 x 10-8M
[OH-] = 1.6 x 10-13 M
16.47 There are two amine groups, one at either end of the molecule, each of which can accept
a proton from a water molecule. Convert the line structure to a Lewis structure using the
standard procedures, then show the transfer of one proton to each N atom:
530
Chapter 16
16.48 First determine the Lewis structures following standard procedures, then show proton
transfer from the O–H group of acetic acid to the N atom of ammonia:
-
+
16.49 To determine the pH of a solution, follow the standard procedure for working equilibrium
problems:
1.) Major species are H2O, Na+, and F-;
2.) The dominant acid-base equilibrium is H2O + FHF + OH-;
K
3.) Keq = Kb = w ; from Table 16-3, Ka = 6.3 x 10-4, from which
Ka
Kb =
1.0 x 10 − 14
= 1.6 x 10-11;
6.3 x 10 − 4
531
Chapter 16
4.) Concentration table is:
F-
Reaction:
H2O
Start (M)
---
0.250
Change (M)
---
–x
Final (M)
Kb = 1.6 x 10-11 =
+
solvent
HF +
OH-
0
0
+x
+x
x
x
0.250 – x
[HF]eq [OH - ]eq
[F - ]eq
=
x2
; Assume x << 0.250:
0.250 − x
x2
, from which x2 = 4.0 x 10-12 and x = 2.0 x 10-6; assumption valid
0.250
[OH-] = 2.0 x 10-6 M,
pOH = -log(2.0 x 10-6) = 5.70, and
pH = 14.00 – 5.70 = 8.30.
1.6 x 10-11 =
16.50 To determine the pH of a solution, follow the standard procedure for working equilibrium
problems:
1.) Major species are H2O, NH4+, and Cl-;
2.) The dominant acid-base equilibrium is H2O + NH4+
NH3 + H3O+;
-14
K
1.0 x 10
= 5.6 x 10-10;
3.) Keq = Ka; from Appendix E, Ka = w =
K b 1.8 x 10 -5
4.) The concentration table is:
H2O
Start (M)
---
0.025
Change (M)
---
–x
Final (M)
Ka = 5.6 x 10-10 =
+
NH4+
Reaction:
solvent
[NH 3 ]eq [H 3 O + ]eq
[NH +4 ]eq
0.025 – x
NH3 +
H3O+
0
0
+x
+x
x
x
x2
; Assume x << 0.025:
=
0.025 − x
x2
, from which x2 = 1.4 x 10-11 and x = 3.7 x 10-6; assumption valid
0.025
[H3O+] = 3.7 x 10-6 M, and
pH = -log(3.7 x 10-6) = 5.43.
5.6 x 10-10 =
16.51 To determine concentrations of species in a solution, follow the standard procedure:
532
Chapter 16
1.) This is a strong acid. Major species are H2O, H3O+, and HSO4-;
2.) The dominant acid-base equilibrium is H2O + HSO4SO42- + H3O+;
3.) Keq = Ka2; from Table 16-2, Ka2 = 1.0 x 10-2;
4.) In this solution, there is a hydronium ion from the strong acid present initially:
SO42- +
H3O+
H2O
Start (M)
----
2.00
0
2.00
Change (M)
----
-x
+x
+x
Final (M)
solvent
x
2.00 + x
Keq = 1.0 x 10-2 =
+
HSO4-
Reaction:
2.00 – x
[SO 24- ]eq [H 3 O + ]eq
4 eq
[HSO ]
=
x(2.00 + x) ; Assume x << 2.00:
2.00 − x
(2.00)(x )
, from which x = 1.0 x 10-2; assumption valid
(2.00)
[H3O+] = 2.00 + 0.010 = 2.01 M;
[SO42-] = 1.0 x 10-2 M;
[HSO4-] = 2.00 – x = 1.99 M
1.0 x 10-2 =
16.52 To determine concentrations of species in a solution, follow the standard procedure:
1.) This is a salt. Major species are H2O, Na+, and NO2-;
2.) The dominant acid-base equilibrium is H2O + NO2HNO2 + OH-;
K
3.) Keq = w = Kb; from Table 16-1, Ka = 5.6 x 10-4, from which
Ka
1.0 x 10 − 14
= 1.8 x 10-11
Kb =
5.6 x 10 − 4
4.)
OH-
H2O
Start (M)
----
0.200
0
0
Change (M)
----
-x
+x
+x
Final (M)
Solvent
x
x
Keq = 1.8 x 10-11 =
1.8 x 10-11 =
+
NO2-
Reaction:
[NO -2 ]eq [OH - ]eq
[HNO 2 ]eq
0.200 – x
=
HNO2 +
x2
; Assume x << 0.200:
0.200 − x
( x)2
, from which x2 = 3.6 x 10-12; x = 1.9 x 10-6; assumption valid
0.200
[OH-] = [HNO2] = 1.9 x 10-6 M;
533
Chapter 16
[NO2- ] = 0.200 M
Kw
1.0 x 10 − 14
[H3O+] =
=
= 5.3 x 10-9 M
[OH − ] 1.9 x 10 − 6
16.53 (a) H2SO4 is a strong acid, so the major species in solution are H2O, HSO4-, and H3O+.
The hydrogensulfate ion is a weak acid, so the equilibrium reaction that determines pH is:
H2O + HSO4SO42- + H3O+;
(b) Na2SO4 is a salt, so the major species in solution are H2O, SO42-, and Na+. The
sulfate ion is the conjugate base of a weak acid, so the equilibrium reaction that
determines pH is:
SO42- + H2O
HSO4- + OH-;
(c) CO2 associates with H2O when it dissolves in water, so the major species in solution
are CO2, H2CO3, and H2O. Carbonic acid is a weak acid, so the equilibrium reaction that
determines pH is:
HCO3- + H3O+;
H2O + H2CO3
(d) NH4Cl is a salt, so the major species in solution are H2O, Cl-, and NH4+. The
ammonium ion is the conjugate acid of a weak base, so the equilibrium reaction that
determines pH is:
H2O + NH4+
NH3 + H3O+.
16.54 (a) NH4NO3 is a salt, so the major species in solution are H2O, NO3-, and NH4+. The
ammonium ion is the conjugate acid of a weak base, so the equilibrium reaction that
NH3 + H3O+;
determines pH is:
H2O + NH4+
(b) KH2PO4 is a salt, so the major species in solution are H2O, H2PO4-, and K+. H2PO4is the conjugate base of a weak acid and is also a weak acid itself, so there are two
equilibrium reactions that might determines pH:
H2O + H2PO4HPO42- + H3O+, pKa2 = 7.21, and
H2O + H2PO4H3PO4 + OH-,
pKeq = pKw – pKa1 = 14.00 – 2.16 = 11.84;
The equilibrium with smaller pK dominates: H2O + H2PO4HPO42- + H3O+;
(c) Na2O is a salt that dissolves to form O2-, an extremely strong base that reacts to
completion with water: O2- + H2O → 2 OH-, so the major species in solution are H2O,
Na+, and OH-. There is no acid-base equilibrium to consider, so pH is determined by
[OH-]initial;
(d) HCO2H is a weak acid, so the major species in solution are H2O and HCO2H. The
equilibrium reaction that determines pH is: H2O + HCO2H
HCO2- + H3O+.
16.55 Tabulated equilibrium constants for acid-base reactions always refer to reactions in which
H2O is one of the reactants. The reaction in this problem is the reverse of a base reaction:
HPO42- (aq) + OH- (aq)
PO43- (aq) + H2O (l)
Table 16-2 lists Ka values for phosphoric acid:
534
Chapter 16
HPO42- (aq) + H2O (l)
PO43- (aq) + H3O+(aq)
Ka3 = 4.8 x 10-13
Ka and Kb for a conjugate acid-base pair are related through Ka Kb = Kw:
1.0 x 10 − 14
Kb =
= 2.1 x 10-2;
4.8 x 10 − 13
1
= 48
Thus, Keq =
Kb
16.56 Tabulated equilibrium constants for acid-base reactions always refer to reactions in which
H2O is one of the reactants. The reaction in this problem is the reverse of an acid
reaction:
HPO42- (aq) + H3O+ (aq)
H2PO4- (aq) + H2O (l)
Table 16-2 lists Ka values for phosphoric acid:
H2PO4- (aq) + H2O (l)
HPO42- (aq) + H3O+ (aq)
Ka2 = 6.2 x 10-8
1
Keq =
= 1.6 x 107
Ka
16.57 (a) An acid-base equilibrium reaction involves proton transfer, in this case from boric
acid to water:
(b) To calculate the pH of a solution, follow the standard procedure for equilibrium
calculations:
+
H3BO3
H2BO3- +
H3O+
Reaction:
H2O
Start (M)
----
0.050
0
0
Change (M)
----
-x
+x
+x
Final (M)
solvent
x
x
0.050 - x
-
Ka= 5.4 x 10-10 =
[H 3 BO 2 ]eq [H 3 O + ]eq
[H 3 BO 3 ]eq
=
x2
; Assume x << 0.050:
0.050 − x
x2
,
0.050
from which x2 = 2.7 x 10-11 and x = 5.2 x 10-6; assumption valid
[H3O+] = 5.2 x 10-6 M, and
pH = -log(5.2 x 10-6) = 5.28
5.4 x 10-10 =
535
Chapter 16
16.58 (a) An acid-base equilibrium reaction involves proton transfer, in this case from water to
hydrazine:
(b) To calculate the pH of a solution, follow the standard procedure for equilibrium
calculations:
+
N2H4
HN2H4+ +
OH-
Reaction:
H2O
Start (M)
----
0.200
0
0
Change (M)
----
-x
+x
+x
Final (M)
solvent
x
x
0.200 – x
+
Kb = 1.3 x 10-6 =
[HN 2 H 4 ]eq [OH - ]eq
[N 2 H 4 ]eq
x2
; Assume x << 0.200:
0.200 − x
=
2
x
,
0.200
from which x2 = 2.6 x 10-7 and x = 5.1 x 10-4; assumption valid
[OH-] = 5.1 x 10-4 M,
pOH = -log(5.1 x 10-4) = 3.29, and pH = 14.00 – 3.29 = 10.71
1.3 x 10-6 =
16.59 There is much interesting chemical information provided in the statement of this
problem, but the calculation is a straightforward equilibrium determination for a solution
of a weak base. Follow the standard procedures to determine the pH. Begin by
constructing an equilibrium table, write the equilibrium expression, and solve for the
concentration of hydroxide ions.
+
LSDH + +
OH-
Reaction:
H2O
Start (M)
----
0.55
0
0
Change (M)
----
-x
+x
+x
Final (M)
Solvent
0.55 - x
x
x
LSD
The equilibrium expression is:
[LSDH + ]eq [OH - ]eq
x2
Kb = 7.6 x 10-7 =
=
; Assume x << 0.55:
[LSD]eq
0.55 − x
536
Chapter 16
x2
, from which x2 = 4.2 x 10-7 and x = 6.5 x 10-4; assumption valid
0.55
[OH-] = 6.5 x 10-4 M,
pOH = -log(6.5 x 10-4) = 3.19, and pH = 14.00 – 3.19 = 10.81.
7.6 x 10-7 =
16.60 There is much interesting chemical information provided in the statement of this
problem, but the calculation is a straightforward equilibrium determination for a solution
of a weak base. Follow the standard procedures to determine the pH. Begin by
constructing an equilibrium table, write the equilibrium expression, and solve for the
concentration of hydroxide ions.
MorphineH + +
OH-
0.015
0
0
----
-x
+x
+x
solvent
0.015 - x
x
x
Reaction:
H2O +
Start (M)
----
Change (M)
Final (M)
Kb = 7.9 x 10-7 =
7.9 x 10-7 =
Morphine
[MorphineH + ]eq [OH - ]eq
[Morphine] eq
=
x2
; Assume x << 0.015
0.015 − x
x2
, from which x2 = 1.2 x 10-8 and x = 1.1 x 10-4; assumption valid
0.015
[OH-] = 1.1 x 10-4 M,
pOH = -log(1.1 x 10-4) = 3.96, and pH = 14.00 – 3.96 = 10.04.
16.61 To identify an acid from the pH of its solution, use the pH to calculate the equilibrium
constant of the acid. For pH = 2.71, [H3O+]eq = 1.95 x 10-3 M = [A-];
[HA]eq = (0.060 M – 0.00195 M) = 0.058 M;
Ka =
[A - ]eq [H3O+ ]eq
(1.95 x 10-3 ) 2
=
= 6.6 x 10-5; pKa = 4.18;
(0.058)
[HA]eq
The acid is benzoic acid, listed in Appendix E , pKa = 4.19.
16.62 Follow the standard procedure for dealing with equilibrium calculations:
The major species are Na+, A-, and H2O, and the acid-base equilibrium is:
HA (aq) + OH- (aq)
A-(aq) + H2O (l)
Set up a concentration table and use it to determine Keq:
pOH = 14.00 – pH = 3.00;
[OH-] = 10-3.00 = 1.00 x 10-3 M;
537
Chapter 16
A-
OH-
Reaction:
H2O +
Start (M)
----
0.0100
0
Change (M)
----
-0.00100
+0.00100
+0.00100
solvent
0.0090
0.00100
0.00100
Final (M)
HA +
0
The equilibrium expression is:
Keq = Kb =
[HA]eq [OH - ]eq
[A - ]eq
(1.00 x 10 -3 ) 2
=
= 1.1 x 10-4;
(0.0090)
K w 1.00 x 10-14
=
Ka =
= 9.1 x 10-11
Kb
1.1 x 10-4
16.63 Molecular pictures must show the correct relative numbers of the various species in the
solution. From the starting condition (six molecules of oxalic acid), make appropriate
changes and then draw new pictures:
(a) Hydroxide ions react with oxalic acid to form water and hydrogen oxalate ions:
H2O + HC2O4- The picture shows 2 molecules of oxalic acid and
H2C2O4 + OH4 each of water and hydrogenoxalate:
(b) When all oxalic acid has reacted, hydroxide ions react with hydrogen oxalate ions to
form water and oxalate ions:
HC2O4- + OHH2O + C2O42-. The picture shows 4 hydrogen oxalate ions, 8
water molecules, and 2 oxalate ions:
538
Chapter 16
(c) NH3, a weak base, accepts a proton from oxalic acid, a weak acid:
H2C2O4 + NH3
NH4+ + HC2O4The picture shows 2 oxalic acid molecules and four each ammonium and
hydrogenoxalate ions:
16.64 The acidic or basic nature of a solution depends on the dominant equilibrium, which is
determined by the major species present:
(a) CH3CO2H is a weak acid, major species are H2O and CH3CO2H; dominant
equilibrium is CH3CO2H + H2O
CH3CO2- + H3O+, and the solution is acidic;
(b) NH3 is a weak base, major species are H2O and NH3; dominant equilibrium is
NH4+ + OH-, and the solution is basic;
NH3 + H2O
(c) NH4Cl is a salt, major species are H2O, NH4+, and Cl-; dominant equilibrium is
NH4+ + H2O
NH3 + H3O+, and the solution is acidic;
(d) NaCH3CO2 is a salt, major species are H2O, CH3CO2-, and Na+; dominant
equilibrium is CH3CO2- + H2O
CH3CO2H + OH-, and the solution is basic;
539
Chapter 16
(e) NH4CH3CO2 is a salt, major species are H2O, CH3CO2-, and NH4+; dominant
equilibrium is CH3CO2- + NH4+
CH3CO2H + NH3, for which Ka = Kb; thus
this solution is neutral.
16.65 The chemical reaction that occurs is:
P4O10 + 6 H2O → 4 H3PO4
(a) The major species present are H2O, and H3PO4
(b) The minor species are (in order of highest concentration to lowest):
H2PO4-, H3O+, HPO42-, OH- , and PO43(c) The dominant equilibrium that determines the pH is:
H3PO4 + H2O
H2PO4- + H3O+
Set up a concentration table, solve for hydronium ion concentration, then calculate the
pH. Determine the initial concentration using standard stoichiometric procedures:
 1 mol  4 mol H 3PO 4 

n(H3PO4) = 3.5 g P4O10 
 283.88 g 


 = 0.0493 mol;

 1 mol P4O10 
[H3PO4] =
0.0493 mol
= 0.033 M
1.50 L
Reaction:
H2O +
Start (M)
----
0.033
Change (M)
----
-x
Final (M)
solvent
H3PO4
0.033 - x
H2PO4- +
H3O+
0
0
+x
+x
x
x
Now substitute into the equilibrium constant expression and solve for x:
+
x2
; solve by the quadratic equation
Keq = 0.0069 = [H 2 PO 4 ]eq [H3O ]eq =
0.033-x
[H 3PO 4 ]eq
[H3O+] = 0.012 M,
pH = -log(0.012) = 1.92
16.66 (a) The salt generates Na+ and SO32-, CH3CO2H is a weak acid, and H2O is always a
major species in aqueous solution;
(b and c) CH3CO2H is the acid, SO32- is the base, HSO3- is the conjugate acid, and
CH3CO2- is the conjugate base:
540
Chapter 16
16.67 Proton transfer occurs from the carboxylic acid O–H (shown screened below) and the
amino nitrogen atom:
16.68 To determine the mass percent of vinegar we must first determine how much of each
species is present in the solution using the standard procedure:
1.) This is a weak acid. Major species are H2O and CH3CO2H
2.) The dominant acid-base equilibrium is CH3CO2H + H2O
-
CH3CO2 + H3O+;
3.) Keq = Ka = 1.8 x 10-5
4.) In this solution, the pH corresponds to the equilibrium concentrations of hydronium
ions and acetate ions:
[H3O+] = [CH3CO2 ] = 10-2.39 = 0.00407 M
Assuming that the concentration of CH3CO2H is much larger than 0.00407 M:
[H 3O + ][CH 3CO −2 ] (0.00407) 2
Ka =1.8 x 10-5 =
=
[CH 3CO 2 H]
x
x = 0.92 M = [HCH3CO2]; assumption valid
0.92 mol 60.05 g 
m(acetic acid) = 

 =55 g/L or 0.055 g/mL acetic acid
 1 L  1 mol 
mass % =
density of acetic acid 0.055
=
×100% = 5.1%
density of sample
1.07
16.69 Net ionic equations show only the reacting species. Remember that strong acids generate
H3O+ in solution and react to completion with weak bases, and strong bases generate
OH- in solution and react to completion with weak acids:
(a) strong base reacting with weak acid: OH- + C6H5CO2H
H2O + C6H5CO2- ;
(b) strong acid reacting with weak base: H3O+ + (CH3)3N
H2O + (CH3)3NH+;
541
Chapter 16
(c) weak base reacting with weak acid: SO42- + CH3CO2H
HSO4- + CH3CO2HSO4-, pKa = 1.99; CH3CO2H, pKa = 4.75; HSO4- is stronger, so this reaction proceeds
to a small extent;
(d) strong base reacting with weak acid: OH- + NH4+
H2O + NH3;
PO43- + NH4+ ;
(e) weak base reacting with weak acid: HPO42- + NH3
HPO42-, pKa = 12.32; NH4+, pKa = 9.25; NH4+ is stronger, so this reaction proceeds to
a small extent.
16.70 (a)
H2CO3 is acidic.
H2CO3 + H2O
(b)
KHCO3 is both.
HCO3 + H2O
−
−
HCO3 + H2O
(c)
(d)
(e)
(f)
NH3 is basic.
NaCl is neither.
Na2SO4 is basic.
SO2 is acidic.
(g)
Li2O is basic.
−
HCO3 + H3O+
2−
CO 3 + H3O+
(acting as acid)
H 2CO3 + OH- (acting as base)
+
NH4 + OH-
NH3 + H2O
2−
SO 4 + H2O
SO2 + H2O
H2SO 3 + H2O
SO2 + 2 H2O
Li2O + H2O
−
HSO4 + OHH2SO 3
–
HSO 3 + H3O+
–
HSO 3 + H3O+
2 Li+(aq) + 2 OH-(aq)
16.71 Follow the standard procedure for dealing with equilibrium calculations:
(a) The major species are Na+, HCO3-, and H2O;
There are two equilibria involving major species:
HCO3- (aq) + H2O (l)
CO32- (aq) + H3O+ (aq)
pKeq = pKa2 = 10.33
HCO3 (aq) + H2O (l)
H2CO3 (aq) + OH (aq)
pKeq = pKw – pKa1 =14.00 – 6.35 = 7.65;
The equilibrium with the larger Keq (smaller pKeq) dominates, making this solution basic;
(b) Set up a concentration table, solve for hydroxide ion concentration, then calculate the
pH. Determine the initial concentration using standard stoichiometric procedures:
n 0.0228 mol
[HCO3-] = =
= 0.152 M;
V
0.150 L
HCO3-
Reaction:
H2O +
Start (M)
----
0.152
Change (M)
----
-x
Final (M)
solvent
0.152 - x
H2CO3 +
OH-
0
0
+x
+x
x
x
Now substitute into the equilibrium constant expression and solve for x:
542
Chapter 16
Keq = 10-7.65 = 2.2 x 10-8;
2.2 x 10-8 =
[H2CO 3]eq [OH- ]eq
[HCO-3]eq
=
x2
; assume x << 0.152
0.152-x
x2 = 3.3 x 10-9 , from which x = 5.8 x 10-5; assumption valid
[OH-] = 5.8 x 10-5 M,
pOH = -log(5.8 x 10-5) = 4.24 and
pH = 14.00 – 4.24 = 9.76
16.72 (a)
CH3CO2- + H2O
NH4+ + H2O
H3O+ + OHNH4+ + CH3CO2Keq =
CH3CO2H + OHNH3 + H3O+
2 H2 O
Kw
Ka
Kw
Kb
1
Kw
NH3 + CH3CO2H
Kw
Ka Kb
(b) to determine all proton transfer reactions, first determine all major acid/base species
in the solution. From part (a) we can see that the major species are:
Bases: CH3CO2-, NH3, H2O, and OH-.
Acids: CH3CO2H, NH4+, H3O+, and H2O.
Now make a list of the reactions between each acid/base pair
CH3CO2H + OHCH3CO2- + H2O
+
NH4 + H2O
NH3 + H3O+
2 H2O
H3O+ + OHNH3 + H2O
NH4+ + OHCH3CO2H + H2O
CH3CO2- + H3O+
+
NH3 + CH3CO2H
NH4 + CH3CO2
+
CH3CO2 + H3O
CH3CO2H + H2O
NH3 + H3O+
NH4+ + H2O
CH3CO2H + OHCH3CO2- + H2O
NH4+ + OHNH3 + H2O
(c) Using the equations from part a, set up an equilibrium table, write the Keq expression
and solve for the desired concentrations:
543
Chapter 16
Reaction:
CH3CO2- +
Start (M)
0.25
----
0
0
Change (M)
-x
----
+x
+x
solvent
x
x
Final (M)
Kb =
0.25 - x
OH-
CH3CO2H +
H2O
K w 1.00 x 10 − 14 [OH - ][CH 3CO 2 H]
( x)2
; assume x << 0.25
=
=
=
Ka
1.8 x 10 − 10
[CH 3CO 2− ]
0.25 − x
x2
0.25
-5
x = 1.2 x 10 ; assumption correct
[OH-] = 1.2 x 10-5 M
5.6 x 10-5 =
Repeat with ammonium to determine the concentration of hydronium ions:
Reaction:
NH4+ +
Start (M)
0.25
----
0
0
Change (M)
-x
----
+x
+x
solvent
x
x
Final (M)
Kb =
0.25 - x
NH3 +
H2O
H3O+
( x)2
K w 1.00 x 10 − 14 [H 3O + ][NH 3 ]
; assume x << 0.25
=
=
=
1.8 x 10 − 10
[NH 4+ ]
0.25 − x
Kb
x2
0.25
-5
x = 1.2 x 10 ; assumption correct
[H3O+] = 1.2 x 10-5 M
5.6 x 10-5 =
(d) Since the concentrations of hydroxide and hydronium ions are equal, we can expect
both to react completely with each other to form water. The resulting equilibrium is
that of water which has a pH of 7.0
2 H2O
H3O+ + OH16.73 Compare Ka values to determine the extent of the reaction.
(a) H2S is the stronger acid : H2S + NH3
HS- + NH4+;
(b) HSO4 is the stronger acid, therefore the reaction is significant:
C2H5NH2 + HSO4C2H5NH3+ +SO42-;
(c) HCN is a weaker acid than C5H5NH+, therefore the reaction proceeds only to a small
extent : C5H5N + HCN
C5H5NH+ + CN-;
(d) ammonium is the stronger acid: NH4+ + PO43NH3 + HPO42(e) HClO is the slightly weaker acid:
544
Chapter 16
HClO(aq) + HONH2(aq)
ClO-(aq) + HONH3+(aq)
16.74 (a) HBr is a strong acid, so when this gas bubbles through water it generates hydronium
ions. The major species are H2O, H3O+, Br-, Ca2+, and OH-, and the reaction that goes
to completion is H3O+ + OH- → 2 H2O;
(b) The major species are H2O, Na+, HSO4-, and OH-, and the reaction that goes to
completion is HSO4- + OH- → H2O + SO42-;
(c) The major species are H2O, NH4+, I-, Pb2+, and NO3-, and the reaction that goes to
completion is formation of PbI2 precipitate, Pb2+ + 2 I- → PbI2(s).
16.75 (a) H2SO4(l) + H2SO4(l)
H3SO4+(solv) + HSO4-(solv);
Either doubly bonded oxygen atom can accept a proton.
(c) H2SO4(l) + HClO4(l)
H3SO4+(solv) + ClO4-(solv);
545