If 67.0 mL of 0.119 mol/L NaOH(aq) and 35.0 mL of 0.193 mol/L HA(aq) are mixed,

Question

If 67.0 mL of 0.119 mol/L NaOH(aq) and 35.0 mL of 0.193 mol/L HA(aq) are mixed, then what is [HA] at equilibrium? Assume that HA is a weak monoprotic acid with pKa=6.23 at 25oC. Give your answer with three significant figures.

Explanation / Answer

The reaction is: NaOH + HA -----> NaA + H2O M = mol / L mol = M x L mol NaOH = 0.119 mol/L x 0.0670 L = 0.0079 mol NaOH mol HA = 0.193 mol/L x 0.0350 L = 0.00675 mol HA All the HA is neutralized forming 0.00662 mol of A-. The total volume is 0.0350 + 0.0670 = 0.102 L M of A- = mol / L = 0.006675 mol / 0.102 L = 0.0661 M A- The excess NaOH is 0.0079 - 0.00675 = 0.00115 mol OH- M of excess OH- = mol / L = 0.00115 mol / 0.102 L = 0.0112. The anion of the weak acid undergoes hydrolysis: A- + H2O -----> HA + OH- Ka(HA) x Kb(A-) = Kw Ka(HA) = 10^-6.09 = 8.13 x 10^-7 Kb(A-) = Kw / Ka(HA) = 1.00 x 10^-14 / 8.13 x 10^-7 = 1.23 x 10^-8 Kb = [HA][OH-] / [A-] 1.23 x 10^-8 = [HA] x 0.0112 / 0.0661 [HA] = 7.47 x 10^-8 M

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