Answer Key to Acid/Base Problem Set This is a strong base problem. 1. Ba(OH)2 (aq) Ba2+(aq) + 2 OH‾(aq) → pOH = - log[OH‾] = - log(0.040) = 1.40 pH = 14 pOH = 14 1.40 = 12.60 – – 2. This is an acidic buffer problem because a weak acid and a salt of the weak acid are present in appreciable quantities. From 16.3, the Ka of HCN = 4.9 x 10-10. HCN(aq) + H2O (l) CN‾(aq) + H3O+(aq) init 0.30 0.40 0 change -x x x equil 0.30 x 0.40 + x x – Simplifies to: [HCN] = 0.30 (3.7 x 10-10 ) = 0.30 M – [CN‾] = 0.40 + x = 0.40 + (3.7 x 10-10 ) = 0.40 M [Na+] = 0.40 M [H2O] = 55.5 M 3. This is a strong acid problem. HNO3(aq) + H2O(l) NO3‾(aq) + H3O+(aq) [HNO3] = 0 M
Answer Key to Acid/Base Problem Set 1. This is a strong base problem. Ba(OH)2 (aq) → Ba2+(aq) + 2 OH‾(aq) pOH = - log[OH‾] = - log(0.040) = 1.40 pH = 14 – pOH = 14 – 1.40 = 12.60 2. This is an acidic buffer problem because a weak acid and a salt of the weak acid are present in appreciable quantities. From 16.3, the Ka of HCN = 4.9 x 10-10. HCN(aq) + H2O (l) CN‾(aq) + H3O+(aq) init 0.30 0.40 0 change -x x x equil 0.30 – x 0.40 + x x Simplifies to: [HCN] = 0.30 – (3.7 x 10-10 ) = 0.30 M [CN‾] = 0.40 + x = 0.40 + (3.7 x 10-10 ) = 0.40 M [Na+] = 0.40 M [H2O] = 55.5 M 3. This is a strong acid problem. HNO3(aq) + H2O(l) NO3‾(aq) + H3O+(aq) [HNO3] = 0 M
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