Solutions for the problems about „Calculation of pH in the case of monoprotic acids and bases” 1. What is the pH of a 0.1 M acetic acid solution? Acetic acid is a weak acid with Ka = 1.86 × 10–5 and in this case cweak acid >>> Ka, that is the equation to use is: [H+] = K a ⋅ c weak acid = (1.86 × 10 −5 ) × 0.1 = 0.0013638 M pH = -log[H+] = -log(0.0013638) = 2.865 2. What is the pH of a 0.1 M ammonia solution? Ammonia is a weak base with Kb = 1.75 × 10–5 and in this case cweak base >>> Kb, that is the equation to use is: [OH–] = K b ⋅ c weak base = (1.75 × 10 −5 ) × 0.1 = 0.0013229 M pOH = -log[OH-] = -log(0.0013638) = 2.878 pH = 14.00 – pH = 14.00 – 2.878 = 11.122 3. What is the pH of a 0.1 M sodium acetate solution? Sodium acetate is a weak base, a conjugate base of acetic acid, so: K 10 -14 = 5.376 × 10-10 K aacetic acid × K bacetate = KW that is K bacetate = aceticW acid = Ka 1.86 × 10 -5 In this case cweak base >>> Kb, that is the equation to use is: [OH–] = K b ⋅ c weak base = (5.376 × 10 −10 ) × 0.1 = 7.332 × 10-6 M pOH = -log[OH-] = -log(7.332 × 10-6) = 5.135 pH = 14.00 – pH = 14.00 – 2.878 = 8.865 4. What is the concentration (in g/dm3 units) of an ammonia solution which has a pH of 11.100? Ammonia is a weak base with Kb = 1.75 × 10–5. pH = 11.100 that is pOH = 14.00 – pH = 2.900 [OH-] = 10–pOH = 10–2.9 = 1.2589 × 10-3 M c − [OH − ] [OH − ] = K b ⋅ weak base − [OH ] c weak base − 1.2589 × 10 −3 1.2589 × 10 −3 1.584 × 10-6 = 1.75 × 10-5 × (cweak base – 1.2589 × 10-3) 1.584 × 10-6 = 1.75 × 10-5 cweak base – 2.203 × 10-8 1.606 × 10-6 = 1.75 × 10-5 cweak base cweak base = 0.091773 M So, we have 0.091773 mol ammonia in 1 L solution, that is 1.56 g ammonia (m = n × M.W. = 0.091773 mol × 17 g/mol) in 1 L solution. The concentration is 1.56 g/L. 1.2589 ×10 −3 = 1.75 ×10 −5 5. A monobasic organic acid has a pK of 4.875. The pH of a saturated solution of this acid is 3.700. Calculate the solubility of this organic acid in mol/dm3 units. 1 pH = 3.700, so [H+] = 10–pH = 10–3.7 = 1.9953 × 10-4 M pK = 4.875, so Ka = 10–pK = 10–4.875 = 1.3335 × 10-5 c − [H + ] [H + ] = K a ⋅ weak acid + [H ] − 1.9953 × 10 −4 1.9953 × 10 −4 -8 -5 3.981 × 10 = 1.3335 × 10 (cweak acid – 1.9953 ×10-4) 3.981 × 10-8 = 1.3335 × 10-5 cweak acid – 2.661 ×10-9 4.247 × 10-8 = 1.3335 × 10-5 cweak acid cweak acid = 0.003185 M 1.9953 × 10 −4 = 1.3335 × 10 −5 cweak acid 6. What is the pH and the degree of dissociation in a a) 0.1 M; in a b) 0.01 M and in a c) 0.001 M acetic acid solution, respectively? a) Acetic acid is a weak acid, so: c weak acid − [H + ] + [H ] = K a ⋅ [H + ] However, if the cweak acid = 0.1 M, we can use the simpler form of the formula: [H + ] 2 = K a ⋅ c weak acid [H+] = 1 .86 × 10 −5 × 0 .10 [H+] = 1.3638×10-3 M pH = -log[H+] = -log(1.3638×10-3) = 2.865 [H + ] In this case [H+] = [acetate ion], so α = = 0.0136 = 1.36 % c weak acid b)Acetic acid is a weak acid, so: − [H + ] c [H + ] = K a ⋅ weak acid + [H ] [H + ]2 = K a (c weak acid − [H + ]) [H + ]2 + ( K a × [H + ]) - ( K a × c weak acid ) = 0 By solving the equation, [H + ] = - K a ± K a2 + 4 × ( K a × c weak acid ) =4.22077×10-4 2 M pH = -log[H+] = -log(4.22077×10-4) = 3.375 [H + ] + In this case [H ] = [acetate ion], so α = = 0.0422 = 4.22 % c weak acid c) Here again we do not have enough difference in the order of magnitude to c − [H + ] [H + ] = K a ⋅ weak acid + [H ] [H + ]2 = K a (c weak acid − [H + ]) [H + ]2 + ( K a × [H + ]) - ( K a × c weak acid ) = 0 By solving the equation, [H + ] = - K a ± K a2 + 4 × ( K a × c weak pH = -log[H+] = -log(1.274×10-4) = 3.904 2 2 acid ) = 1.274×10-4 M In this case [H+] = [acetate ion], so α = [H + ] = 0.127 = 12.7 % c weak acid 7. What is the pH in a 0.010 M solution of a moderately weak acid if the Ka = 1.5 × 10–1 ? In the case of this moderately weak acid we do not have enough difference in the c 0.010 order of magnitude as = = 0.066 . So the following formula should K a 1.5 × 10 −1 be used: c weak acid − [H + ] + [H ] = K a ⋅ [H + ] [H + ]2 = K a (c weak acid − [H + ]) [H + ]2 + ( K a × [H + ]) - ( K a × c weak acid ) = 0 [H + ] 2 + (1.5 × 10 −1 × [H + ]) - (1.5 × 10 −1 × 0.010) = 0 By solving the equation, [H + ] = - K a ± K a2 + 4 × ( K a × c weak pH = -log[H+] = -log(9.41× 10–3) = 2.026 2 acid ) = 9.41 × 10–3 M 8. A windscreen washing liquid contains ammonia in 2.00 g/dm3 concentration. What is the pH of this liquid? 1 L solution contains 2.00 g of NH3, that is 0.11737 mol of NH3 (n = 2g m = = 0.11737 mol). So the concentration of ammonia is: M.W. 17 g/mol 0.11737 mol cammonia = = 0.11737 M 1L In this case, cweak base >>> Kb, so [OH–] = K b ⋅ c weak base = 1.75 × 10 −5 ⋅ 0.11737 = 0.001433 M pOH = -log[OH–] = -log(0.001433) = 2.844, pH = 14.00 – pOH = 11.156 9. 20.00 cm3 of 0.1 M ammonia solution is titrated with 0.25 M HClO4. What is the added volume of titrant and the pH at 75% degree of titration? (6 cm3, 8.766) The molar amount of ammonia: n = c × V = 0.1 M × 0.020 L = 0.002 mol At 100 % titration, nHClO4 = nNH3 = 0.002 mol At 75 % titration, nHClO4 = 0.75 × nNH3 = 0.0015 mol HClO4. We know the concentration (0.25 M) of the HClO4 solution, so, the volume can be calculated: n 0.0015 mol V= = = 0.006 L = 6 mL c 0.25 M At 75 % titration, we have a buffer, NH3 and NH4Cl together. 75 % of the original NH3 amount is converted to NH4Cl, and 25 % is remained as NH3. So, using the equation for buffers: n 0.25 × n total 0.25 [OH-] = Kb ⋅ weak base = Kb ⋅ =1.75 × 10-5 ⋅ = 5.83 × 10-6 M 0.75 n salt 0.75 × n total 3 pOH = -log[OH–] = -log(5.83 × 10-6) = 5.234, pH = 14.00 – pOH = 8.766 10. The concentration of a monochloro acetic acid solution is 0.001 M. What are the pH and the degree of dissociation in this solution? Ka = 1.2×10-3 Monochloro acetic acid is a weak acid and we do not have enough difference in the order of magnitude so the following formula should be used: c − [H + ] [H + ] = K a ⋅ weak acid + [H ] The cweak acid = 0.001 M and Ka = 1.2×10-3 [H + ] = 1.2 × 10 −3 ⋅ 0.001 − [H + ] [H + ] [H + ] 2 + (1.2 × 10 −3 × [H + ]) - (1.2 × 10 −3 × 0.001) = 0 By solving the equation, [H + ] = - 1.2 × 10 -3 ± (1.2 × 10 −3 ) 2 + 4 × (1.2 × 10 -3 × 0.001) 2 [H+] = 6.490 × 10–4 M pH = – log[H+] = – log (6.490 × 10–4) pH = 3.188 In this case [H+] = [monochloro acetate ion], so α = [H + ] c weak acid = 0.649 = 64.9 % 11. Calculate the pH when we add: a) 0 mL b) 9 mL c) 20 mL d) 25 mL NaOH solution whose concentration is 0.06 M to a 10 mL sample of acetic acid. The concentration of the acetic acid is unknown (Ka= 1.86 × 10–5). First, calculate the concentration of the acetic acid if we know that 20 mL of NaOH is consumed up to the equivalence point. What kind of indicator would you use for this titration? Results: a) c(acetic acid) = 0.12 mol/dm3 b) at the beginning of the titration: 0 ml of base was added (weak acid) pH = 2.826 c) After the addition: 9 ml of base (buffer system) pH = 4.643 d) After the addition: 20 ml of base (weak base) pH = 8.666 e) After the addition: 25 ml of base (excess of strong base) 4 pH = 11.933 12. 10 mL of acetic acid, whose concentration is unknown is titrated with potassium hidroxide. a) Calculate the concentration of the acetic acid if the volume of potassium hidroxide that is consumed up to the equivalence point is 16 mL. The concentration of KOH is 0.075 M. b) Calculate the pH after the addition of 10 mL of KOH. Results: a) c(acetic acid) = 0.12 M b) It will be a buffer system: pH = 4.952 13. A 10 mL sample of acetic acid with 0.1 M concentration is titrated with sodium hidroxide. The concentration of sodium hidroxide is 0.1 M as well. Decide whether it is right or wrong to use an indicator whose pK=3.8? Support your opinion with calculations! a) At first, we assume that the indicator marked the equivalence point correctly. This would mean that the same amounts of acetic acid and sodium hidroxide are present in the solution. n(acetic acid) = 10 × 0.1 = 1 mmol n(acetic acid) = n(NaOH) = 1 mmol This amount of NaOH is present in 10 mL of solution. Then you should calculate the pH of this solution (which contains a weak base, sodium acetate) to check your assumption about the indicator. Ka = 1.86 × 10–5 1 csalt/weak base = = 0.05 M 20 K 10 −14 = 5.376 × 10–10 Kb= w = −5 K a 1.86 × 10 [OH–] = K b ⋅ c weak base = 5.185 × 10–6 M pOH = -log[OH–] = -log(5.185 × 10–6) = 5.285 pH = 14.00-5.285 = 8.715 From this result you can see that at the equivalent point the pH is 8.715, so the indicator was wrong. b) Now, let’s see where the indicator marked the end-point. Around pH 3.8, we still have a buffer system of acetic acid and sodium acetate. n(acetic acid) = 1 mmol n(NaOH) = y × 0.1 = 0.1y mmol [H+] = 10–3.8 = 1.585 × 10–4 M [H+] = Ka n weak acid n weak base (salt ) = 5 1.585 × 10–4 = 1.86 × 10–5 × 1 − 0.1y 0.1y 8.521 × 0.1y = 1 - 0.1y 0.9521y = 1 y = 1.05 mL Only this amount of sodium hidroxide is given to the acid when the indicator signs! So the mistake that you make by using the incorrect indicator is [(10-1.05)/10] × 100= 89.5 % 14. 20 mL of benzoic acid of unknown concentration is titrated with potassium hydroxide whose concentration is 0.095 M. Calculate the pH at: a) 0 %; b) 25 %; c) 100 %; d) 150 % degree of titration. Up to the equivalence point 13.56 mL of KOH solution is consumed. (Ka = 6.7 × 10–5) a) n(KOH) = 13.56 × 0.095 = 1.2882 mmol n(KOH) = n(benzoic acid) = 1.2882 mmol 1.2882 c(benzoic acid) = = 6.441 × 10–2 M 20 –5 Ka = 6.7 × 10 As we do not have the three orders of magnitude difference between the concentration and the acidic ionization constant, the more complicated formula has to be used for the calculation of the pH. c weak acid − [H + ] [H ] = K a ⋅ [H + ] + [H + ]2 = K a (c weak acid − [H + ]) [H + ]2 + ( K a × [H + ]) - ( K a × c weak acid ) = 0 + [H ] = + [H ] = - K a ± K a2 + 4 × (K a × c weak acid ) 2 − 6.7 × 10 −5 ± (6.7 × 10 ) −5 2 + 4 × 4.31547 × 10 −6 2 = 2.044 × 10–3 M pH = -log[H+] = -log(2.044 × 10–3) = 2.689 b) n(KOH) = 0.25 × 13.56 × 0.095 = 0.32205 mmol n(benzoic acid)sum = 1.2882 mmol We get a basic buffer solution of: n(benzoic acid)rest = 0.96615 mmol n(potassium benzoate) = 0.32205 mmol [H+] = Ka n weak acid n weak base (salt ) = 6.7 × 10–5 × 0.96615 = 2.01 × 10–4 M 0.3205 6 c) n(potassium benzoate) = 1.2882 mmol 1.2882 × 10 −3 c(potassium benzoate) = = 3.838 × 10–2 M 33.56 The salt, potassium benzoate is a weak base, so first the basic ionization constant should be calculated. Kw 10 −14 Kb= = = 1.4925 × 10–10 −5 Ka 6.7 × 10 In this case cweak base >>> Kb, that is the equation to use is: [OH–] = K b ⋅ c weak base = (1.4925 × 10 −10 ) × 0.03838 = 2.395 × 10–6 M pOH = -log[OH–] = -log(2.395 × 10–6) = 5.621 pH = 14.00-5.621 = 8.379 d) n(KOH) = 1.50 × 13.56 × 0.095 = 1.9323 mmol n(benzoic acid)sum = 1.2882 mmol n(potassium benzoate) = 1.2882 mmol n(KOH)excess = 1.9323 - 1.2882 = 0.6441 mmol As we have a weak base and a strong base in the system, it is the strong base, namely the KOH solution in excess that determines the pH of the solution. c(KOH sol.) = 0.6441 = 1.597 × 10–2 M 40.34 pOH = -log[OH–] = -log(1.597 × 10–2) = 1.797 pH = 14.00-1.797 = 12.203 ≈ 12.20 15. 100 mL of formic acid whose concentration is 0.010 M is titrated by sodium hydroxide solution of 0.100 M concentration. Calculate the pH: a) before the titration b) after the addition of 8 mL; c) 9.90 mL of NaOH solution; d) at the equivalence point. (Ka = 1.77 × 10–4) a) c(formic acid) = 0.010 M Ka = 1.77 × 10–4 As we do not have the three orders of magnitude difference between the concentration and the acidic ionization constant, the more complicated formula has to be used for the calculation of the pH. [H + ] = K a ⋅ c weak acid − [H + ] [H + ] [H + ]2 = K a (c weak acid − [H + ]) 7 [H + ]2 + ( K a × [H + ]) - ( K a × c weak acid ) = 0 + [H ] = + [H ] = - K a ± K a2 + 4 × ( K a × c weak acid ) 2 − 1.77 × 10 − 4 ± (1.77 × 10 ) −4 2 ( + 4 × 1.77 × 10 − 4 × 0.010 2 ) = 1.244 ×10–3 M pH = -log[H+] = -log(1.244×10–3) = 2.905 b) n(formic acid) = 100 × 0.010 = 1.00 mmol n(NaOH) = 0.100 × 8 = 0.800 mmol After the addition of 8.00 mL of NaOH solution, we get a basic buffer solution of HCOONa/HCOOH. n(formic acid) = 0.200 mmol n(sodium formiate) = 0.800 mmol [H+] = Ka n weak acid n weak base (salt ) = 1.77 × 10–4× 0.200 = 4.425 × 10–5 M 0.800 pH = -log[H+] = -log(4.425×10–5) = 4.354 c) n(formic acid) = 100 × 0.010 = 1.00 mmol n(NaOH) = 0.100 × 9.90 = 0.990 mmol After the addition of 9.90 mL of NaOH solution, we get a basic buffer solution of HCOONa/HCOOH. n(formic acid) = 0.010 mmol n(sodium formiate) = 0.990 mmol [H+] = Ka n weak acid n weak base (salt ) = 1.77 × 10–4× 0.01 = 1.7523 × 10–6 M 0.990 pH = -log[H+] = -log(1.7523 × 10–6) = 5.756 d) n(formic acid) = 100 × 0.010 = 1.00 mmol n(NaOH) = 0.100 × 10= 1.00 mmol n(sodium formiate) = 1.00 mmol The salt, sodium formiate is a weak base, so first the basic ionization constant should be calculated. Kb= Kw 10 −14 = 5.649 × 10–11 = −4 K a 1.77 × 10 c(sodium formiate) = n (sodiumformiate) 1 = = 0.0091 M V(solution ) 110 In this case cweak base >>> Kb, that is the equation to use is: 8 [OH–] = K b ⋅ c weak base = (5.649 × 10 −11 ) × 0.0091 = 7.170 × 10–7 M pOH = -log[OH–] = -log(7.170 × 10–7) = 6.144 pH = 14.00-6.144 = 7.856 9