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