Worked Solutions
Chapter 19
Question 7
Calculate the pH of a solution with a H3O+ concentration of (a) 0.1 M
Answer:
pH = -log10 [H3O+]
= -log10 (0.1)
=1
(b) 0.15 M
Answer:
pH = -log10 [H3O+]
= -log10 (0.15)
= 0.82
(c) 0.0005 M
Answer:
pH = -log10 [H3O+]
= -log10 (0.0005)
= 3.30
(d) 0.002 M
Answer:
pH = -log10 [H3O+]
= -log10 (0.002)
= 2.70
(e) 0.04 M.
Answer:
pH = -log10 [H3O+]
= -log10 (0.04)
= 1.40
1
Question 8
Calculate the H3O+ concentration of a solution with a pH of
(a) 4
Answer:
pH = -log10 [H3O+] = 4
-pH = log10 [H3O+] = -4
[H3O+] = antilog (-4)
= 0.0001 M
(b) 6
Answer:
pH = -log10 [H3O+] = 6
-pH = log10 [H3O+] = -6
[H3O+] = antilog (-6)
= 0.000001 M
(c) 4.5
Answer:
pH = -log10 [H3O+] = 4.5
-pH = log10 [H3O+] = -4.5
[H3O+] = antilog (-4.5)
= 0.000032 M
(d) 2.8
Answer:
pH = -log10 [H3O+] = 2.8
-pH = log10 [H3O+] = -2.8
[H3O+] = antilog (-2.8)
= 0.0016 M
2
Question 11
Find the pH of each of the following solutions:
(a) 0.001 M HCl
Answer:
pH = -log10 [H3O+]
= -log10 (0.001)
=3
(b) 0.02 M HNO3
Answer:
pH = -log10 [H3O+]
= -log10 (0.02)
= 1.70
(c) 0.00004 M HCl
Answer:
pH = -log10 [H3O+]
= -log10 (0.00004)
= 4.40
(d) 1 M HCl
Answer:
pH = -log10 [H3O+]
= -log10 (1)
=0
Question 12
What is the pH of a solution containing 3.65 g of HCl in 1 litre of solution?
Answer:
The hydrochloric acid solution contains 3.65 g of hydrochloric acid in 1 litre
The molar mass of HCl is 36.5 g mol-1
Therefore [HCl] = 3.65 / 36.5 mol l-1
= 0.1 mol l-1
Since HCl is a strong acid, [H3O+] = [HCl] = 0.1 mol l-1
pH = -log10 [H3O+]
= -log10 (0.1)
=1
3
Question 13
What is the pH of a solution containing 3.15 g of HNO3 in 500 cm3 of solution?
Answer:
The nitric acid solution contains 3.15 g of nitric acid in 500 cm3
= 3.15 × 1000 g l-1
500
= 6.3 g l-1
The molar mass of HNO3 is 63 g mol-1
Therefore [HNO3] = 6.3 / 63 mol l-1
= 0.1 mol l-1
Since HNO3 is a strong acid, [H3O+] = [HNO3] = 0.1 mol l-1
pH = -log10 [H3O+]
= -log10 (0.1)
=1
Question 14
Find the pH of each of the following solutions:
(a) 0.001 M NaOH
Answer:
[NaOH] = 0.001 mol l-1
[OH-] = [NaOH]
pOH = – log10 [OH-]
= – log10 (0.001)
= 3.
pH = 14 –pOH
= 14 –3
= 11.
(b) 0.05 M KOH
Answer:
[KOH] = 0.05 mol l-1
[OH-] = [KOH]
pOH = – log10 [OH-]
= – log10 (0.05)
4
= 1.301
pH = 14 –pOH
= 14 – 1.301
= 12.70
(c) 0.06 M KOH
Answer:
[KOH] = 0.001 mol l-1
[OH-] = [KOH]
pOH = – log10 [OH-]
= – log10 (0.06)
= 1.222.
pH = 14 –pOH
= 14 –1.222
= 12.78
(d) 0.003 M NaOH
Answer:
[NaOH] = 0.003 mol l-1
[OH-] = [NaOH]
pOH = – log10 [OH-]
= – log10 (0.003)
= 2.523.
pH = 14 –pOH
= 14 –2.523
= 11.48
Question 15
What is the pH of a solution containing 0.8 g of NaOH per litre?
Answer:
The NaOH solution contains 0.8 g of NaOH in 1 litre
The molar mass of NaOH is 40 g mol-1
Therefore [NaOH] = 0.8 / 40 mol l-1
= 0.02 mol l-1
Since NaOH is a strong base, [OH-] = [NaOH] = 0.02 mol l-1
pOH = – log10 [OH-]
= – log10 (0.02)
5
= 1.6990
pH = 14 –pOH
= 14 – 1.6990
= 12.30
Question 16
Find the pH of a solution containing 1.6 g of NaOH in 250 cm3 of solution.
Answer:
The NaOH solution contains 1.6 g of NaOH in 250 cm3
= 1.6 × 1000 g l-1
250
= 6.4 g l-1
The molar mass of NaOH is 40 g mol-1
Therefore [NaOH] = 6.4 / 40 mol l-1
= 0.16 mol l-1
Since NaOH is a strong base, [OH-] = [NaOH] = 0.16 mol l-1
pOH = – log10 [OH-]
= – log10 (0.16)
= 0.796
pH = 14 –pOH
= 14 – 0.796
= 13.20
Question 18
A weak acid, HX, is 3.5% dissociated in a 0.1 M aqueous solution. Find the value of the acid
dissociation constant, Ka.
Answer:
The solution can be made up by adding 0.1 moles of the pure acid to water, and making up
the solution to 1 litre with water. The reaction is
HX + H2O
X- + H3O+
Ignoring the concentration of water, which is present in excess, the initial concentrations of
each of the other species are
[HX]
0.1
[X-]
[H3O+]
0
0
At equilibrium, 3.5% of the acid has dissociated, and so the equilibrium concentrations are
6
[HX]
[X-]
[H3O+]
0.1 – 0.0035
0.0035
0.0035
= 0.0965
0.0035
0.0035
Substituting these values into the equilibrium constant expression,
Ka = [X- ] [H3O+] / [HX] = (0.0035)2 / 0.0965 = 1.27 × 10-4
Question 19
A weak acid, HX, is 2.5% dissociated in a 0.2 M aqueous solution. Find the value of the acid
dissociation constant, Ka.
Answer:
The solution can be made up by adding 0.2 moles of the pure acid to water, and making up
the solution to 1 litre with water. The reaction is
HX + H2O
X- + H3O+
Ignoring the concentration of water, which is present in excess, the initial concentrations of
each of the other species are
[HX]
0.2
[X-]
[H3O+]
0
0
At equilibrium, 2.5% of the acid has dissociated, and so the equilibrium concentrations are
[X-]
[HX]
[H3O+]
0.2 – 0.005
0.005
0.005
= 0.195
0.005
0.005
Substituting these values into the equilibrium constant expression,
Ka = [X- ] [H3O+] / [HX] = (0.005)2 / 0.195= 1.28 × 10-4
Question 20
Calculate the pH of each of the following solutions, using the Ka values given in Table 19.3:
(a) 0.01 M CH3COOH
Answer:
A 0.01 M solution of CH3COOH can be made up by adding 0.01 moles of the pure acid to
water, and making up the solution to 1 litre with water.
The reaction is
CH3COOH + H2O
CH3COO- + H3O+
7
Ignoring the concentration of water, which is present in excess, the initial concentrations of
each of the other species are
[CH3COOH]
0.01
[CH3COO-]
0
[H3O+]
0
At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the
equilibrium concentrations are
[CH3COOH]
[CH3COO-]
[H3O+]
x
x
0.01 - x
Substituting these values into the dissociation constant expression,
Ka = [CH3COO-] [H3O+] / [CH3COOH] = 1.8 × 10-5
= (x2) / (0.01 - x )
Since x is very small compared to 0.01, (0.01 - x ) can be taken to be equal to 0.01, and so
Ka = [CH3COO-] [H3O+] / [CH3COOH] = 1.8 × 10-5
= (x2) / (0.01)
Therefore, x2 = 0.01 × 1.8 × 10-5 = 0.00000018
Therefore, x =  (0.00000018) = 0.000424
[H3O+] = x = 0.000424
pH = -log10 [H3O+]
= -log10 (0.000424)
= 3.37
(b) 1 M HCOOH
Answer:
A 1 M solution of HCOOH can be made up by adding 1 mole of the pure acid to water, and
making up the solution to 1 litre with water.
The reaction is
HCOOH + H2O
HCOO- + H3O+
Ignoring the concentration of water, which is present in excess, the initial concentrations of
each of the other species are
[HCOOH]
[HCOO-]
1
[H3O+]
0
0
At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the
equilibrium concentrations are
[HCOOH]
1-x
[HCOO-]
[H3O+]
x
x
8
Substituting these values into the dissociation constant expression,
Ka = [HCOO-] [H3O+] / [HCOOH] = 1.6 × 10-4
= (x2) / (1 - x )
Since x is very small compared to 1, (1 - x ) can be taken to be equal to 1, and so
Ka = [HCOO- ] [H3O+] / [HCOOH] = 1.6 × 10-4
= (x2) / (1)
Therefore, x2 = 1 × 1.6 × 10-4 = 0.00016
Therefore, x =  (0.00016) = 0.0126
[H3O+] = x = 0.0126
pH = -log10 [H3O+]
= -log10 (0.0126)
= 1.90
(c) 0.005 M CH3CH2COOH
Answer:
A 0.005 M solution of CH3CH2COOH can be made up by adding 0.005 moles of the pure
acid to water, and making up the solution to 1 litre with water.
The reaction is
CH3CH2COO- + H3O+
CH3CH2COOH + H2O
Ignoring the concentration of water, which is present in excess, the initial concentrations of
each of the other species are
[CH3CH2COOH]
[CH3CH2COO -]
0.005
0
[H3O+]
0
At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the
equilibrium concentrations are
[CH3CH2COOH]
[CH3CH2COO -]
[H3O+]
0.005 - x
x
x
Substituting these values into the dissociation constant expression,
Ka = [CH3CH2COOH -] [H3O+] / [CH3CH2COOH] = 1.3 × 10-5
= (x2) / (0.005 - x )
Since x is very small compared to 0.005, (0.005 - x ) can be taken to be equal to 0.005, and so
Ka = [CH3CH2COO- ] [H3O+] / [CH3CH2COOH] = 1.3 × 10-5
= (x2) / (0.005)
9
Therefore, x2 = 0.005 × 1.3 × 10-5 = 0.000000065
Therefore, x =  (0.000000065) = 0.000255
[H3O+] = x = 0.000255
pH = -log10 [H3O+]
= -log10 (0.000255)
= 3.59
(d) 0.1 M HNO2
Answer:
A 0.1 M solution of HNO2 can be made up by adding 0.1 moles of the pure acid to water, and
making up the solution to 1 litre with water.
The reaction is
HNO2 + H2O
NO2- + H3O+
Ignoring the concentration of water, which is present in excess, the initial concentrations of
each of the other species are
[HNO2]
[NO2-]
[H3O+]
0.1
0
0
At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the
equilibrium concentrations are
[HNO2]
0.1 - x
[NO2-]
x
[H3O+]
x
Substituting these values into the dissociation constant expression,
Ka = [NO2-] [H3O+] / [HNO2] = 4.7 × 10-4
= (x2) / (0.1 - x )
Since x is very small compared to 0.1, (0.1 - x) can be taken to be equal to 0.1, and so
Ka = [NO2- ] [H3O+] / [HNO2] = 4.7 × 10-4
= (x2) / (0.1)
Therefore, x2 = 0.1 × 4.7 × 10-4 = 0.000047
Therefore, x =  (0.000047) = 0.00686
[H3O+] = x = 0.00686
pH = -log10 [H3O+]
= -log10 (0.00686)
= 2.16
10
(e) 0.02M HF
Answer:
A 0.02 M solution of HF can be made up by adding 0.02 moles of the pure acid to water, and
making up the solution to 1 litre with water.
The reaction is
HF + H2O
F- + H3O+
Ignoring the concentration of water, which is present in excess, the initial concentrations of
each of the other species are
[HF]
[F-]
[H3O+]
0.02
0
0
At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the
equilibrium concentrations are
[F-]
[HF]
0.02 - x
x
[H3O+]
x
Substituting these values into the dissociation constant expression,
Ka = [F-] [H3O+] / [HF] = 5.6 × 10-4
= (x2) / (0.02 - x )
Since x is very small compared to 0.02, (0.02 - x) can be taken to be equal to 0.02, and so
Ka = [F- ] [H3O+] / [HF] = 5.6 × 10-4
= (x2) / (0.02)
Therefore, x2 = 0.02 × 5.6 × 10-4 = 0.0000112
Therefore, x =  (0.0000112) = 0.00335
[H3O+] = x = 0.00335
pH = -log10 [H3O+]
= -log10 (0.00335)
= 2.47
Question 21
Find the pH of a solution containing 30 g of CH3COOH per litre of solution.
Answer:
The ethanoic acid solution contains 30 g of ethanoic acid in 1 l
The molar mass of CH3COOH is 60 g mol-1
Therefore [CH3COOH] = 30 / 60 mol l-1
= 0.5 mol l-1
11
The reaction is
CH3COO- + H3O+
CH3COOH + H2O
Ignoring the concentration of water, which is present in excess, the initial concentrations of
each of the other species are
[CH3COOH]
[CH3COO-]
[H3O+]
0.5
0
0
At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the
equilibrium concentrations are
[CH3COOH]
[CH3COO-]
[H3O+]
x
x
0.5 - x
Substituting these values into the dissociation constant expression,
Ka = [CH3COO-] [H3O+] / [CH3COOH] = 1.8 × 10-5
= (x2) / (0.5 - x )
Since x is very small compared to 0.5, (0.5 - x) can be taken to be equal to 0.5.
Ka = [CH3COO- ] [H3O+] / [CH3COOH] = 1.8 × 10-5
= (x2) / (0.5)
Therefore, x2 = 0.5 x 1.8 × 10-5 = 0.000009
Therefore, x =  (0.000009) = 0.003
[H3O+] = x = 0.003
pH
= -log10 [H3O+]
= -log10 (0.003)
= 2.52
Question 22
Find the pH of a solution containing 0.23 g of HCOOH per 100 cm3 of solution.
Answer:
The methanoic acid solution contains 0.23 g of ethanoic acid in 100 cm3
= 0.23 × 1000 g l-1
100
= 2.3 g l-1
The molar mass of HCOOH is 46 g mol-1
Therefore [CH3COOH] = 2.3 / 46 mol l-1
= 0.05 mol l-1
12
The reaction is
HCOOH + H2O
HCOO- + H3O+
Ignoring the concentration of water, which is present in excess, the initial concentrations of
each of the other species are
[HCOO-]
[HCOOH]
0.05
0
[H3O+]
0
At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the
equilibrium concentrations are
[HCOO-]
[HCOOH]
0.05 - x
x
[H3O+]
x
Substituting these values into the dissociation constant expression,
Ka = [HCOO-] [H3O+] / [HCOOH] = 1.6 × 10-4
= (x2) / (0.05 - x )
Since x is very small compared to 0.05, (0.05 - x) can be taken to be equal to 0.05.
Ka = [HCOO-] [H3O+] / [HCOOH] = 1.6 × 10-4
= (x2) / (0.05)
Therefore, x2 = 0.05 x 1.6 × 10-4 = 0.000008
Therefore, x =  (0.000008) = 0.00283
[H3O+] = x = 0.00283
pH = -log10 [H3O+]
= -log10 (0.00283)
= 2.55
Question 24
A weak base, XOH, is 2.1% dissociated in a 0.1 M aqueous solution. Find the value of the
dissociation constant, Kb.
Answer:
The solution can be made up by adding 0.1 moles of the pure base to water, and making up
the solution to 1 litre with water. The reaction is
XOH
X+ + OH-
Ignoring the concentration of water, which is present in excess, the initial concentrations of
each of the other species are
13
[XOH]
0.1
[X+]
0
[OH-]
0
At equilibrium, 2.1% of the base has dissociated, and so the equilibrium concentrations are
[X+]
[OH-]
0.1 – 0.0021
0.0021
0.0021
= 0.0979
0.0021
0.0021
[XOH]
Substituting these values into the equilibrium constant expression,
Kb = [X+] [OH-] / [XOH] = (0.0021)2 / 0.0979 = 4.5 × 10-5
Question 25
Calculate the pH of each of the following solutions, using the Kb values given in Table 19.4:
(a) 0.001 M NH3
Answer:
pOH = -log10[ (Kb × Mb)]
= -log10[ (1.8 × 10-5 × 0.001)]
= -log10[ (0.000000018)]
= -log10 (0.000134)
= 3.872
pH = 14 –pOH
= 14 – 3.872
= 10.13
(b) 1 M NH3
Answer:
pOH = -log10[ (Kb × Mb)]
= -log10[ (1.8 × 10-5 × 1)]
= -log10[ (0.000018)]
= -log10 (0.00424)
= 2.373
pH = 14 –pOH
= 14 – 2.373
= 11.63
(c) 0.02 M CH3NH2
Answer:
pOH = -log10[ (Kb × Mb)]
14
= -log10[ (4.4 × 10-4 × 0.02)]
= -log10[ (0.0000088)]
= -log10 (0.00297)
= 2.527
pH = 14 –pOH
= 14 – 2.527
= 11.47
(d) 0.1 M Be(OH)2
Answer:
pOH = -log10[ (Kb × Mb)]
= -log10[ (5.0 × 10-11 × 0.1)]
= -log10[ (5 × 10-12)]
= -log10 ( 2.236 × 10-6)
= 5.65
pH = 14 –pOH
= 14 – 5.65
= 8.35
(e) 0.02 M C2H5NH2
Answer:
pOH = -log10[ (Kb × Mb)]
= -log10[ (5.6 × 10-4 × 0.02)]
= -log10[ (0.0000112)]
= -log10 (0.00335)
= 2.475
pH = 14 –pOH
= 14 – 2.475
= 11.53
Question 26
Find the pH of a solution containing 6.8 g of NH3 per litre of solution.
Answer:
The ammonia solution contains 6.8 g of ammonia in 1 l.
The molar mass of NH3 is 17 g mol-1
Therefore [NH3] = 6.8 / 17 mol l-1
15
= 0.4 mol l-1
pOH = -log10[ (Kb × Mb)]
= -log10[ (1.8 × 10-5 × 0.4)]
= -log10[ (0.0000072)]
= -log10 (0.00268)
= 2.572
pH = 14 –pOH
= 14 – 2.572
= 11.43
Question 27
Find the pH of a solution containing 1.36 g of NH3 per 100 cm3 of solution.
Answer:
The ammonia solution contains 1.36 g of ammonia in 100 cm3.
= 1.36 × 1000 g l-1
100
= 13.6 g l-1
The molar mass of NH3 is 17 g mol-1
Therefore [NH3] = 13.6 / 17 mol l-1
= 0.8 mol l-1
pOH = -log10[ (Kb × Mb)]
= -log10[ (1.8 × 10-5 × 0.8)]
= -log10[ (0.0000144)]
= -log10 (0.00379)
= 2.421
pH = 14 –pOH
= 14 – 2.421
= 11.58
Question 31
Calculate the pH of
(a) a 0.001 M solution of sodium hydroxide
Answer:
[NaOH] = 0.001 mol l-1
16
[OH-] = [NaOH]
pOH = – log10 [OH-]
= – log10 (0.001)
= 3.
pH = 14 –pOH
= 14 –3
= 11.
(b) a solution of hydrochloric acid containing 3.65 g HCl in 500 cm3 of solution.
Answer:
The hydrochloric acid solution contains 3.65 g of hydrochloric acid in 500 cm3
= 3.65 × 1000 g l-1
500
= 7.3 g l-1
The molar mass of HCl is 36.5 g mol-1
Therefore [HCl] = 7.3 / 36.5 mol l-1
= 0.2 mol l-1
Since HCl is a strong acid, [H3O+] = [HCl] = 0.2 mol l-1
pH = -log10 [H3O+]
= -log10 (0.2)
= 0.70
Question 32
Calculate (a) the pH of a 0.03 M solution of NaOH
Answer:
Since NaOH is a strong base, [OH-] = [NaOH] = 0.03 mol l-1
pOH = – log10 [OH-]
= – log10 (0.03)
= 1.523
pH = 14 –pOH
= 14 – 1.523
= 12.48
(a) a solution containing 0.014 g of potassium hydroxide in 250 cm3 of solution.
Answer:
The NaOH solution contains 0.014 g of KOH in 250 cm3
17
= 0.014 × 1000 g l-1
250
= 0.056 g l-1
The molar mass of KOH is 56 g mol-1
Therefore [KOH] = 0.056 / 56 mol l-1
= 0.001 mol l-1
Since KOH is a strong base, [OH-] = [KOH] = 0.001 mol l-1
pOH = – log10 [OH-]
= – log10 (0.001)
=3
pH = 14 –pOH
= 14 – 3
= 11
Question 34
Propanoic acid is a weak monobasic acid that dissolves in water, dissociating slightly
according to the equation
C2H5COOH + H2O
C2H5COO- + H3O+
(b) Find the pH of a 0.01 M solution of the acid, given that the value of Ka for this acid is
1.3 × 10-5.
Answer:
A 0.01 M solution of C2H5COOH can be made up by adding 0.01 moles of the pure acid to
water, and making up the solution to 1 litre with water.
The reaction is
C2H5COOH + H2O
C2H5COO- + H3O+
Ignoring the concentration of water, which is present in excess, the initial concentrations of
each of the other species are
[C2H5COOH]
[C2H5COO -]
0.01
[H3O+]
0
0
At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the
equilibrium concentrations are
[C2H5COOH]
0.01 - x
[C2H5COO -]
x
[H3O+]
x
Substituting these values into the dissociation constant expression,
18
Ka = [C2H5COOH -] [H3O+] / [C2H5COOH] = 1.3 × 10-5
= (x2) / (0.01 - x )
Since x is very small compared to 0.01, (0.01 - x) can be taken to be equal to 0.01, and so
Ka = [C2H5COO- ] [H3O+] / [C2H5COOH] = 1.3 × 10-5
= (x2) / (0.01)
Therefore, x2 = 0.01 × 1.3 × 10-5 = 0.00000013
Therefore, x =  (0.00000013) = 0.000361
[H3O+] = x = 0.000361
pH = -log10 [H3O+]
= -log10 (0.000361)
= 3.44
Question 35
The dissociation of the weak base ammonia may be represented as
NH3 + H2O
NH4+ + OH-
(b) Find the pH of a 0.002 M solution of ammonia, given that its dissociation constant is
1.8× 10-5.
Answer:
pOH = -log10[ (Kb × Mb)]
= -log10[ (1.8 × 10-5 × 0.002)]
= -log10[ (0.000000036)]
= -log10 (0.000190)
= 3.721
pH = 14 –pOH
= 14 – 3.721
= 10.28
Question 37
(d) Calculate the pH of 0.1 M chloroethanoic acid.
Answer:
A 0.1 M solution of CH2ClCOOH can be made up by adding 0.1 moles of the pure acid to
water, and making up the solution to 1 litre with water.
The reaction is
CH2ClCOOH + H2O
CH2ClCOO- + H3O+
19
Ignoring the concentration of water, which is present in excess, the initial concentrations of
each of the other species are
[CH2ClCOO -]
[CH2ClCOOH]
0.1
[H3O+]
0
0
At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the
equilibrium concentrations are
[CH2ClCOO -]
[CH2ClCOOH]
0.1 - x
x
[H3O+]
x
Substituting these values into the dissociation constant expression,
Ka = [CH2ClCOO -] [H3O+] / [CH2ClCOOH] = 1.3 × 10-3
= (x2) / (0.1 - x )
Since x is very small compared to 0.1, (0.1 - x) can be taken to be equal to 0.1, and so
Ka = [CH2ClCOO- ] [H3O+] / [CH2ClCOOH] = 1.3 × 10-3
= (x2) / (0.1)
Therefore, x2 = 0.1 × 1.3 × 10-3 = 0.00013
Therefore, x =  (0.00013) = 0.0114
[H3O+] = x = 0.0114
pH = -log10 [H3O+]
= -log10 (0.0114)
= 1.94
Question 38
H3PO4 dissociates in water at first as follows:
H3PO4 + H2O
H2PO4- + H3O+
(c) Given that the acidity of the H3PO4 is almost entirely due to its first dissociation, calculate
the pH of 0.01 M H3PO4, given that Ka for this dissociation is 8 × 10-3.
Answer:
A 0.01 M solution of H3PO4 can be made up by adding 0.01 moles of the pure acid to water,
and making up the solution to 1 litre with water.
The reaction is
H3PO4 + H2O
H2PO4- + H3O+
Ignoring the concentration of water, which is present in excess, the initial concentrations of
each of the other species are
20
[H3PO4]
0.01
[H2PO4-]
[H3O+]
0
0
At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the
equilibrium concentrations are
[H3PO4]
[H2PO4-]
[H3O+]
0.1 - x
x
x
Substituting these values into the dissociation constant expression,
Ka = [H2PO4-] [H3O+] / [H3PO4] = 8 × 10-3
= (x2) / (0.01 - x )
Since x is very small compared to 0.01, (0.01 - x ) can be taken to be equal to 0.01, and so
Ka = [H2PO4- ] [H3O+] / [H3PO4] = 8 × 10-3
= (x2) / (0.01)
Therefore, x2 = 0.01 × 8 × 10-3 = 0.00008
Therefore, x =  (0.00008) = 0.00894
[H3O+] = x = 0.00894
pH = -log10 [H3O+]
= -log10 (0.00894)
= 2.05
21