Question

Consider weak monoprotic acid HA(aq) D H⁺(aq) + A^–(aq) with K_a = 4.0 x 10^–5 and pK_a = 4.40. A solution is composed of 900.0 mL of 0.090 M HA and 0.080 M NaA (10 points).

(a) Calculate the pH of this solution.

(b) Calculate the pH of the above solution following the addition of 10.00 mL of 2.0 M NaOH (a strong base).

Answer 1

HA Laq ? Ht cap + Á cap pK a = 4.40 [HA ] = [NaA] = 0.090 M 0.080 M pH = pkat nog [NHAS CHA] pH = 4.40 + voy 100.) pH = 4.348 (b) Holes of Nadh = 2.0m x 10mL 20 mmol Holes of HA = 0.090 M x 900 mL = 81 minol Moles of Nat = 0.080 x 900 mL 72 mmol of HA Addition y Nach will neutralize will decrease and concentration of HA, so concentration Nat will increase Moles of HA = (81 - 20) mund = 61 mmol Moles of NaA = (72 +20) mund = 92 mmol pH = 4.40 + log | 22 = 4.578