Practice Problems with Buffers - ANSWERS

Practice Problems with Buffers - ANSWERS 1. (a) K a  [H  ][OCl ] ( x)( x)   3.0 108 [HOCl] (0.025 M ) x 2  7.5 1010 x  [H  ]  2.7 105 M (b) pH = –log(2.7 × 10–5) = 4.57 (c) % ionization of HOCl = 2. [H  ] 2.7  105 100%   100%  0.11% [HOCl] 0.025 (a) The presence of additional OCl– ions causes the equilibrium position to shift toward the reactants. (b) The value of [H+] in Solution #2 should be less than the value of [H+] in Solution #1. (c) The pH of Solution #2 should be greater than the pH of Solution #1. (d) The % ionization of HOCl in Solution #2 should be less than the % ionization of HOCl in Solution #1. 3. [H  ][OCl ] [H  ](0.015 M ) (a) K a    3.0 108 [HOCl] (0.025 M ) [H+] = 5.0 × 10–8 M (b) pH = –log(5.0 × 10–8) = 7.30 Alternatively, you could have used the Henderson-Hasselbalch equation as follows.  [A  ]   0.015  8 pH  pK a  log     log(3.0 10 )  log    7.52  log  0.60   7.52  0.22  7.30  0.025   [HA]  (c) % ionization of HOCl = 4. [H  ] 5.0  10 8  100%   100%  0.0002% [HOCl] 0.025 Solution B. A mixture of HNO2(aq) and NaNO2(aq) is a better buffer because it consists of a weak acid (HNO2) and its conjugate base (NO2–). A buffer solution that contains a weak acid and its conjugate base is able to resist changes in pH because it can neutralize small quantities of strong acid or strong base [A  ] while still maintaining a ratio of that is approximately the same value. [HA] On the other hand, Buffer A contains HNO3(aq), which is a strong acid. NaNO3(aq) is a neutral salt. This combination would not make a good buffer because it does not consist of a weak acid – conjugate base pair. 5. (a) The value of pKa for HNO2 is equal to –log(4.5 × 10–4) = 3.35 (b) 6. Solution Prediction Justification a mixture of 1.0 M HNO2 and 1.0 M NaNO2 pH = pKa [HNO2] = [NO2–] a mixture of 0.75 M HNO2 and 0.55 M NaNO2 pH < pKa [HNO2] > [NO2–] a mixture of 0.83 M HNO2 and 1.1 M NaNO2 pH > pKa [HNO2] < [NO2–] (a) K a  [H  ][NO2  ] [H  ](1.0 M )   4.5 104 [HNO 2 ] (1.0 M ) [H+] = 4.5 × 10–4 M pH = –log(4.5 × 10–4) = 3.35 Alternatively, you could have used the Henderson-Hasselbalch equation as follows.  [A  ]   1.0  4 pH  pK a  log     log(4.5  10 )  log    3.35  1.0   [HA]  [H  ][NO2  ] [H  ](0.55 M )   4.5 104 (b) K a  [HNO 2 ] (0.75 M )  0.75  –4 [H+] = (4.5 × 10–4)   = 6.1 × 10 M  0.55  pH = –log(6.1 × 10–4) = 3.21 Alternatively, you could have used the Henderson-Hasselbalch equation as follows.  [A  ]   0.55  4 pH  pK a  log     log(4.5 10 )  log    3.35  0.13  3.22  0.75   [HA]  6. [H  ][NO2  ] [H  ](1.1 M )   4.5 104 (c) K a  [HNO 2 ] (0.83 M )  0.83  –4 [H+] = (4.5 × 10–4)   = 3.4 × 10 M 1.1   pH = –log(3.4 × 10–4) = 3.47 Alternatively, you could have used the Henderson-Hasselbalch equation as follows.  [A  ]   1.1  4 pH  pK a  log     log(4.5 10 )  log    3.35  0.12  3.47  0.83   [HA]  7. The value of pKa for HNO2 is equal to –log(4.5 × 10–4) = 3.35 The pH range of an equimolar mixture of HNO2 and NaNO2 is between pH 2.35 and pH 4.35. 8. NO2–(aq) + H+(aq) → HNO2(aq) 9. HNO2(aq) + OH–(aq) → NO2–(aq) + H2O(l) 10. Buffer X is more resistant to changes in pH when a strong acid or a strong base is added to it because it contains a higher concentration of HNO2 and NO2– to react with added H+ or OH– ions. 11. The volume of 1.0 M NaOH needed is equal to 50.0 mL. Moles of HNO2 = 0.100 L × 1.0 mol HNO 2 = 0.10 mol HNO2 L Moles of NaOH needed to react with HALF of the HNO2 = 0.05 mol NaOH When half of the HNO2 is converted to the conjugate base, [HNO2] = [NO2–], and the pH of the solution is equal to the pKa of the weak acid, which is 3.35. Another way to demonstrate this is  [NO 2  ]  – pH  pK a  log   and pH = pKa when [HNO2] = [NO2 ] [HNO ] 2   12. (a) pH = pKa = –log(1.5 × 10–5) = 4.82 (b) pH = pKa = –log(6.2 × 10–8) = 7.21 (c) pH = pKa = –log(5.6 × 10–10) = 9.25 13. Buffer Solution Particulate Diagram of the Buffer Solution Prediction Justification #1 pH < 3.17 [HF] > [F–] #2 pH = 3.17 [HF] = [F–] #3 pH > 3.17 [HF] < [F–]