Worked Solutions - EDCO



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

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

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

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)

= 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)

= 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 [pic] X- + H3O+

Ignoring the concentration of water, which is present in excess, the initial concentrations of each of the other species are

[HX] [X-] [H3O+]

0.1 0 0

At equilibrium, 3.5% of the acid has dissociated, and so the equilibrium concentrations are

[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 [pic] X- + H3O+

Ignoring the concentration of water, which is present in excess, the initial concentrations of each of the other species are

[HX] [X-] [H3O+]

0.2 0 0

At equilibrium, 2.5% of the acid has dissociated, and so the equilibrium concentrations are

[HX] [X-] [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 [pic] CH3COO- + H3O+

Ignoring the concentration of water, which is present in excess, the initial concentrations of each of the other species are

[CH3COOH] [CH3COO-] [H3O+]

0.01 0 0

At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the equilibrium concentrations are

[CH3COOH] [CH3COO-] [H3O+]

0.01 - x x 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 [pic] HCOO- + H3O+

Ignoring the concentration of water, which is present in excess, the initial concentrations of each of the other species are

[HCOOH] [HCOO-] [H3O+]

1 0 0

At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the equilibrium concentrations are

[HCOOH] [HCOO-] [H3O+]

1 - x x x

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

CH3CH2COOH + H2O [pic] CH3CH2COO- + H3O+

Ignoring the concentration of water, which is present in excess, the initial concentrations of each of the other species are

[CH3CH2COOH] [CH3CH2COO -] [H3O+]

0.005 0 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)

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 [pic] 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] [NO2-] [H3O+]

0.1 - x x 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

(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 [pic] 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

[HF] [F-] [H3O+]

0.02 - x x 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

The reaction is

CH3COOH + H2O [pic] CH3COO- + H3O+

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+]

0.5 - x x 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

The reaction is

HCOOH + H2O [pic] HCOO- + H3O+

Ignoring the concentration of water, which is present in excess, the initial concentrations of each of the other species are

[HCOOH] [HCOO-] [H3O+]

0.05 0 0

At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the equilibrium concentrations are

[HCOOH] [HCOO-] [H3O+]

0.05 - x x 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 [pic] X+ + OH-

Ignoring the concentration of water, which is present in excess, the initial concentrations of each of the other species are

[XOH] [X+] [OH-]

0.1 0 0

At equilibrium, 2.1% of the base has dissociated, and so the equilibrium concentrations are

[XOH] [X+] [OH-]

0.1 – 0.0021 0.0021 0.0021

= 0.0979 0.0021 0.0021

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)]

= -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

= 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

[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

= 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 [pic] 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 [pic] C2H5COO- + H3O+

Ignoring the concentration of water, which is present in excess, the initial concentrations of each of the other species are

[C2H5COOH] [C2H5COO -] [H3O+]

0.01 0 0

At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the equilibrium concentrations are

[C2H5COOH] [C2H5COO -] [H3O+]

0.01 - x x x

Substituting these values into the dissociation constant expression,

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 [pic] 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 [pic] CH2ClCOO- + H3O+

Ignoring the concentration of water, which is present in excess, the initial concentrations of each of the other species are

[CH2ClCOOH] [CH2ClCOO -] [H3O+]

0.1 0 0

At equilibrium, a certain amount (say x moles) of the acid has dissociated, and so the equilibrium concentrations are

[CH2ClCOOH] [CH2ClCOO -] [H3O+]

0.1 - x x 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 [pic] 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 [pic] H2PO4- + H3O+

Ignoring the concentration of water, which is present in excess, the initial concentrations of each of the other species are

[H3PO4] [H2PO4-] [H3O+]

0.01 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

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