A 100.0 mL solution containing 0.9062 g of maleic acid (MW = 116.072 g/mol) is t

Question

A 100.0 mL solution containing 0.9062 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.3260 M KOH Calculate the pH of the solution after the addition of 46.00 mL of the KOH solution Maleic acid has pK_a values of 1.92 and 6.27. pH = At this pH. calculate the concentration of each form of maleic add In the solution at equilibrium. The three forms of maleic acid are abbreviated H_2 M^-, HM^-, and M^2-, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively. [M^2-] = [HM^-] = [H_2 m] =

Explanation / Answer

Titration

maleic acid pH

[maelic acid] = 0.9082/116.072 x 0.1 = 0.078 M

HA <==> H+ + A-

Ka = [H+][A-]/[HA]

0.012 = x^2/0.078

x = [H+] = 0.0306

pH = -log[H+] = 1.514

[HM-] = 0.0306 M

[H2M] = 0.078 - 0.0306 = 0.0474 M

Ka2 = [M^2-][H+]/[HM-]

5.37 x 10^-7 = x^2/0.0306

x = [M^2-] = 1.28 x 10^-4 M

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