1. 2. 3. A 100.0 ml KOH. Calculate the pH of the solution after the addition of
ID: 578417 • Letter: 1
Question
1. 2. 3. A 100.0 ml KOH. Calculate the pH of the solution after the addition of 49.00 ml of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. n containing 0.9553 g of maleic acid (Mw = 1 16072 gmol) is ith 0.3359 M Number 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 2M, HM, and M, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively. Number NumberExplanation / Answer
Solution-
Mass of maleic acid = 0.9553 g
=> Moles of maleic acid =0.9553 / 116.072
= 8.23 x 10^-3 moles
Conc. of KOH solution = 0.3359 M
Volume of KOH solution = 49 mL = 0.049 L
=> Moles of KOH added = 0.3359 x 0.049
= 0.0164 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
=> 8.23 x 10^-3 moles of H2M reacts with 8.23 x 10^-3 x 2 = 0.01646 moles of KOH
Moles of M2- produced after reaction = 8.23x 10^-3 moles
Total volume of solution = 100 + 49= 149 mL = 0.149 L
=> [M2-] = 8.23 x 10^-3 / 0.149 = 0.055 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-
0.055-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) / (0.055 - X)
=> 1.86 x 10^-8 = X^2 / (0.055 - 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-] = 0.055 - X = 0.055 M (approx.)
[HM-] = X - Y = 3.19 x 10^-5 M
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