A 100.0 mL solution containing 0.9359 g of maleic acid (MW = 116.072 g/mol) is t

Question

A 100.0 mL solution containing 0.9359 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.2880 M KOH. Calculate the pH of the solution after the addition of 56.00 mL of the KOH solution. Maleic acid has pK_a values of 1.92 and 6.27. 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 H_2M, HM^-, and M^2", which represent the fully protonated, intermediate, and fully deprotonated forms, respectively. [M^2-] = [HM^-] = [H_2M] =

Explanation / Answer

maleic acid concentration = 0.9359 g/116.072 g/mol x 0.1 L

                                          = 0.0806 M

Titrated with 56 ml of 0.2880 M KOH

moles of KOH = 0.2880 M x 56 ml = 16.128 mmol

moles of maleic acid = 0.081 M x 100 ml = 8.06 mmol

So we would have,

[HM-] = 0.008 mmol/156 ml = 5.13 x 10^-5 M

[M^2-] = 8.06 mmol/156 ml = 0.052 M

pH = pKa2 + log(base/acid)

      = 6.27 + log(0.052/5.13 x 10^-5)

      = 9.27

At this pH concentration of,

[H2M] = 8.32 x 10^-13 M

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