25.00 mL of a solution of a weak monoprotic acid, HX, was titrated with a 0.0640 M solution of NaOH, — Acid-Base Equilibrium / Titration Chemistry Question
Problem Context
25.00 mL of a solution of a weak monoprotic acid, HX, was titrated with a 0.0640 M solution of NaOH, requiring 18.22 mL. The pH of the solution varied as a function of the percentage of HX titrated. These data were collected.
Calculate the initial concentration of the weak acid in the 25.00 mL of solution.
Determine the value of Ka for two of these three conditions.
Model Answer
With zero % titrated, pH = 3.39; [H+] = 4.07 × 10–4
Assuming negligible dissociation:
Ka = [H+][X–] / [HX] = (4.07 × 10–4)^2 / 0.0466 = 3.56 × 10–6
Assuming significant dissociation:
Ka = [H+][X–] / [HX] = (4.07 × 10–4)^2 / (0.0466 - 4.07 × 10–4) = 3.59 × 10–6
With 33.3 % titrated, pH = 5.14; [H+] = 7.24 × 10–6
Assuming negligible dissociation:
Ka = [H+][X–] / [HX] = (7.24 × 10–6)(0.0155) / 0.0311 = 3.62 × 10–6
Assuming significant dissociation:
Ka = [H+][X–] / [HX] = (7.24 × 10–6)(0.0155 + 7.24 × 10–6) / (0.0311 - 7.24 × 10–6) = 3.61 × 10–6
With 66.7 % titrated, pH = 5.74; [H+] = 1.82 × 10–6
Assuming negligible dissociation:
Ka = [H+][X–] / [HX] = (1.82 × 10–6)(0.0312) / 0.0155 = 3.64 × 10–6
Assuming significant dissociation:
Ka = [H+][X–] / [HX] = (1.82 × 10–6)(0.0312 + 1.82 × 10–6) / (0.0155 - 1.82 × 10–6) = 3.66 × 10–6
Calculate the pH at the equivalence point of this titration and write an equation to account for this pH.
Model Answer
X– + H2O = HX + OH–
Kb = Kw / Ka = 1.00 × 10–14 / 3.60 × 10–6 = 2.78 × 10–9
[X–] = (0.0466 M)(25.00 mL) / 43.22 mL = 0.0270 M
[OH–]^2 / 0.0270 = 2.78 × 10–9 ; [OH–] = 8.66 × 10–6 ; pOH = 5.06 ; pH = 8.94
Calculate the number of moles of a salt, NaX, that must be added to produce a pH of 6.00 in 150.00 mL of the original solution.
Model Answer
pH = 6.0 and [H+] = 1.0 × 10–6
3.60 × 10–6 = (1.0 × 10–6)[X–] / 0.0466 ; [X–] = 0.168 M
Moles X– =