Homework Sourse1 A 1000 mL solution containing 09797 g of maleic acid MW 1
1) A 100.0 mL solution containing 0.9797 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.2911 M KOH. Calculate the pH of the solution after the addition of 58.00 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27.
2) 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 H2M, HM–, and M2–, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively.
Solution
Mass of maleic acid = 0.9797 g
=> Moles of maleic acid =amount in g/ molar mass
= 0.9797 / 116.072 = 8.44 x 10^-3 moles
Conc. of KOH solution = 0.2911 M
Volume of KOH solution = 58 mL = 0.058 L
=> Moles of KOH added = 0.2911 x 0.058= 0.01688 moles
The reactions are as follows,
H2M + KOH -----> HM- + H2O
HM- + KOH -------> M2- + H2O
Overall: H2M + 2KOH -----> M2- + 2H2O
According to the stoichiometry of the reaction 1 mole of H2M reacts with 2 moles of KOH
=> 8.44 x 10^-3 moles of H2M reacts with 8.44 x 10^-3 x 2 = 0.01688 moles of KOH
Moles of M2- produced after reaction = 8.44 x 10^-3 moles
Total volume of solution = 100 + 58= 158 mL = 0.158 L
=> [M2-] = 8.44 x 10^-3 / 0.158 = 0.0534 M
M2- + H2O --------> HM- + OH-
pKb2 for this reaction = 14 - 6.27 = 7.73
=> Kb2 = 1.86 x 10^-8
HM- + H2O --------> H2M + H2O
pKb1 for this reaction = 14 - 1.92 = 12.08
=> Kb1 = 8.32 x 10^-13
M2- + H2O --------> HM- + OH-
0.0534-X.................X........X+Y
HM- + H2O --------> H2M + OH-
X-Y..........................Y.......X+Y
Since Kb2 >> Kb1, X >> Y, therefore we can neglect Y with reapect to X
Kb2 = 1.86 x 10^-8 = X (X+Y) / (0.0534 - X)
=> 1.86 x 10^-8 = X^2 / (0.0534 - X)
=> X = 3.2 x 10^-5 M
Kb1 = 8.32 x 10^-13 = Y (X+Y) / (X-Y) = Y
=> Y = 8.32 x 10^-13 M
[M2-] = 0.0534 - X = 0.0534 M (approx.)
[HM-] = X - Y = 3.2 x 10^-5 M
[H2M] = Y = 8.32 x 10^-13 M
[OH-] = X + Y = 3.2 x 10^-5
=> pOH = -log [OH] = 4.49
=> pH = 14 - 4.49 = 9.51
