Question

A 100.0 mL solution containing 0.753 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.259 M KOH. Calculate the pH of the solution after the addition of 50.0 mL of the KOH solution. Maleic acid has pKa 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 H2M, HM–, and M2–, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively.

Answer 1

Maleic acid molecular weight, MW = 116.072

Mass of maleic acid used = 0.753 g

Number of moles of maleic acid = mass/MW = 0.753 / 116.072 = 0.0064873 moles

Molarity of maleic acid= moles/V = 0.0064873/0.1 = 0.064873 M

Molarity of KOH, M = 0.259 M

Volume of KOH used, V = 50 mL

mmoles of KOH used = MV = 0.259 x 50 = 12.95mmoles = 0.01295 moles of KOH used

At equilibrium:

H2M + OH^- ======== HM^- + H2O……………..pKa1 = 1.92

HM^- + OH^- ========= M^2- + H2O …………….pKa2 = 6.27

For neutralization for every 1 mole of maleic acid (H2M) two moles of KOH is required.

For complete neutralization of 0.0064873 moles of maleic acid (H2M), number of moles of KOH required is 0.0129746 moles.

Since only 0.01295 moles of KOH is used some amount of conjugate maleic acid (HM^-) will remain unreacted.

Number of moles of conjugate maleic acid (HM^-) remain unreacted = number of moles of KOH required for complete neutralization - number of moles of KOH used for neutralization

= 0.0129746 - 0.01295

= 0.0000246

number of moles of conjugate maleic acid (HM^-) remain unreacted = 0.0000246 moles

Total Volume of solution = volume of maleic acid (H2M) solution + volume of KOH solution

                                               = 100 mL + 50 mL = 150 mL = 0.15 L

Molarity of conjugate maleic acid (HM^-) after neutralization = number of moles of conjugate maleic acid remain unreacted/ Total Volume of solution

Molarity of conjugate maleic acid (HM^-) after neutralization =   0.0000246 / 0.15 = 0.000164 M

Therefore [HM^-] = 0.000164 M

No. of moles of M^2- formed = No. of moles of KOH used / 2 = 0.01295/2 = 0.006475 Moles

Molarity of M^2- = No. of moles of M^2- formed/Total volume of solution

Molarity of M^2- = 0.006475 / 0.15 = 0.043166 M

Therefore [M^2-] = 0.043166 M

pH at equilibrium can be calculated using Henderson–Hasselbalch equation:

pH = pKa2 + log [M^2-/ HM^-]

Substituting the [M^2-] and [HM^-] values into the above equation one will get

pH = 6.27 + log [0.043166/0.000164]

pH = 6.27 + 2.420 = 8.69

Therefore pH of the solution after adding 50 ml of KOH of 0.259 M solution to maleic acid solution = 8.69

The concentration of different forms of maleic acid at pH 8.69 of the solution are as follows:

[H2M] = 0

[HM^-] = 0.000164 M

[M^2-] = 0.043166 M