A 100.0 mL solution containing 0.965 g of maleic acid (MW=116.072 g/mol) is titrated with 0.339 M KOH. Calculate the pH of the solution after the addition of 49.0 mL of the KOH solution. Maleic acid has p?a 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 H2M, HM−, and M2−, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively.
H2M moles = mass / molar mass
= 0.965 / 116.072
= 8.31 x 10^-3
KOH moles = 0.339 x 49 / 1000
= 0.0166
H2M + 2 KOH ------------------> K2M + 2H2O
moles of KOH neede = 8.31 x 10^-3 x 2
= 0.0166
it is equivalnece point second
M2- formed = 8.31 x 10^-3
total volume = 149 mL = 0.149 L
[M2-] = 8.31 x 10^-3 / 0.149 = 0.0558 M
pKb1 = 14 - pKa2
= 14- 6.27
= 7.73
Kb1 = 10^-7.73 = 1.86 x 10^-8
M2- + H2O ---------------> HM- + OH-
0.0558 0 0
0.0558 -x x x
Kb1 = x^2 / 0.0558-x = 1.86 x 10^-8
x = 3.22 x 10^-5
pOH = -log (3.22 x 10^-5) = 4.49
pH + pOH = 14
pH = 9.51
[H2M] = pKb2 = 14 - 1.92 = 12.08
Kb2 = 8.32 x 10^-13
[H2M] = 8.32 x 10^-13 M
[HM-] = 3.22 x 10^-5
[M2-] = 0.0558 M
pH = 9.51
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