The methanoate ion, HCO2–, reacts with water to form methanoic acid and hydroxide ion, as shown in the following equation. ⇄ HCO2–(aq) + H2O(l) HCO2H(aq) + OH–(aq) Given that [OH–] is 4.18 x 10–6 in a 0.309 solution of sodium methanoate, calculate M M each of the following. The value of for the methanoate ion, HCO2–(aq). (1) Kb ⇄ HCO2–(aq) + H2O(l) OH–(aq) + HCO2H(aq) I 0.309 0 0 M M M C 4.18 x 10–6 + 4.18 x 10–6 + 4.18 x 10–6 – E 0.309 4.18 x 10–6 4.18 x 10–6 M M M = [HCO2H][ OH–] = (4.18 x 10–6)2 = 5.65 x 10–11 Kb Kb Kb [HCO2-] (0.309) The value of for the methanoic acid, HCO2H(aq). (1) Ka = x Kw Ka Kb = / Ka Kw Kb = 1.00 x 10–14 = 1.77 x 10–4 Ka 5.65 x 10–11 Which acid is stronger, propanoic acid or methanoic acid? Justify your answer. (1) for propanoic acid is 1.34 x 10–5, and for methanoic acid is 1.77 x 10–4. For acids, the larger the value Ka Ka of the greater the strength; Ka, therefore methanoic acid is the stronger acid because is 1.77 x 10–4 > 1.34 x 10–5.
The methanoate ion, HCO2–, reacts with water to form methanoic acid and hydroxide ion, as shown in the following equation. HCO2–(aq) + H2O(l) ⇄ HCO2H(aq) + OH–(aq) Given that [OH–] is 4.18 x 10–6 M in a 0.309 M solution of sodium methanoate, calculate each of the following. The value of Kb for the methanoate ion, HCO2–(aq). (1) HCO2–(aq) + H2O(l) ⇄ OH–(aq) + HCO2H(aq) I 0.309 M 0 M 0 M C – 4.18 x 10–6 + 4.18 x 10–6 + 4.18 x 10–6 E 0.309 M 4.18 x 10–6 M 4.18 x 10–6 M Kb = [HCO2H][ OH–] Kb = (4.18 x 10–6)2 Kb = 5.65 x 10–11 [HCO2-] (0.309) The value of Ka for the methanoic acid, HCO2H(aq). (1) Kw = Ka x Kb Ka = Kw / Kb Ka = 1.00 x 10–14 = 1.77 x 10–4 5.65 x 10–11 Which acid is stronger, propanoic acid or methanoic acid? Justify your answer. (1) Ka for propanoic acid is 1.34 x 10–5, and Ka for methanoic acid is 1.77 x 10–4. For acids, the larger the value of Ka, the greater the strength; therefore methanoic acid is the stronger acid because is 1.77 x 10–4 > 1.34 x 10–5.
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