Question

A 100.0 mL solution containing 0.956 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.350 M KOH. Calculate the pH of the solution after the addition of 47.0 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27 pH At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated as H,M, HM, and M2, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively M2- HM- М H,M= MM

Answer 1

The number of Moles of maleic acid taken = 0.956g/116.072g/mol = 0.008236mol

The number of moles of KOH added = 47.0×10^-3L × 0.350mol/L = 0.01645mol

H₂M + 2 KOH ----->​​​​​ K₂M + 2 H₂O

After addition of 47.0ml of KOH , second equivalent point is attained .

[K₂M] = 0.008236mol/(147.0×10^-3L) = 0.05603mol/L

M^2- + H₂O $\leftrightharpoons$ HM^- + OH^{- ,} K_eq= Kw/Ka2 =10^-14/10^-6.27=10^-7.73

pH = 7 + 1/2×(pK_a2 + log c)

= 7 + 1/2× (6.27 + log 0.05603)

= 9.5092

= 9.51 (Answer)

Degree of hydrolysis = √(K_h/c) = √(10^-7.73/0.05603) = 0.000576

Second degree of dissociation would be even smaller .

[M^2-] = 0.0560 M. (Answer)

[HM^-] = [OH^-] = K_w/[H⁺] = 10^-14/10^-9.5092= 10^-4.4908

[HM^-] = 3.23×10^-5 M. (Answer)

K_a1 = [H⁺][HM^-]/[H₂M]

[H₂M] = 10^-9.5092 ×3.23×10^-5/10^-1.92

= 8.32×10^-13 M. (Answer)