Question

A 100.0 mL solution containing 0.990 g of maleic acid (MW = 116,072 g/mol) is titrated with 0.363 M KOH. Calculate the pH of the solution after the addition of 47.0 mL of the KOH solution. Maleic acid has pK, 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 M-, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively. M-) = HM] =

Answer 1

pH = 9.25

[M^2-] = 0.0580 M

[HM^-] = 0 M

[H₂M] = 0 M

Explanation

mass maleic acid = 0.990 g

moles maleic acid = (mass maleic acid) / (molar mass maleic acid)

moles maleic acid = (0.990 g) / (116.072 g/mol)

moles maleic acid = 0.008529 mol

moles H⁺ = 2 * (moles maleic acid)

moles H⁺ = 2 * (0.008529 mol)

moles H⁺ = 0.0170584 mol

moles OH^- added = (concentration NaOH) * (volume NaOH in Liter)

moles OH^- added = (0.363 M) * (0.0470 L)

moles OH^- added = 0.017061 mol

Since moles OH^- > moles H⁺, therefore, solution will be basic

excess moles OH^- = (moles OH^- added) - (moles H⁺)

excess moles OH^- = (0.017061 mol) - (0.0170584 mol)

excess moles OH^- = 2.62 x 10^-6 mol

Total volume = 100.0 mL + 47.0 mL

Total volume = 147.0 mL

Total volume = 0.147 L

[OH^-] = (excess moles OH^-) / (Total volume)

[OH^-] = (2.62 x 10^-6 mol) / (0.147 L)

[OH^-] = 1.784 x 10^-5 M

pOH = -log[OH^-]

pOH = -log(1.784 x 10^-5 M)

pOH = 4.75

pH = 14 - pOH

pH = 14 - 4.75

pH = 9.25