Question

A 100.0 mL solution containing 0.914g of maleic acid (MW=116.072 g/mol) is titrated with 0.281M KOH. Calculate the pH of the solution after the addition of 56.0 mL of the KOH solution. Maleic acid has pKa 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 M-, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively.

Answer 1

Answer:

Given 100 mL of solution containing 0.914 g maleic acid

                                                  = 0.914 / 116.072 mole of maleic acid

                                                  = 0.00787 mole of maleic acid. (H₂M)

It is titrated with 56 mL of 0.281 M KOH = 56 mL x 0.281 mole / 1000 mL

                                                          = 0.01574 moe of KOH

Given the pka1 for maleic acid = 1.92           H₂M H⁺ + HM^-

              pka2 for maleic acid = 6.27            HM^- H⁺ + M^2-

Consider the equation

                      H₂M + KOH        HM^- + H₂O + K⁺

                     HM^- + KOH       M^2- + H₂O + K⁺

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overall reaction                    H₂M + 2KOH M^2- + 2K⁺ + 2 H₂O

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In the titration the moles of H₂M = 0.00787

                    the moles of KOH = 0.01574

The moles of KOH added is just twice the moles of H₂M, which is as per the overall reaction is the second equivalence point.

      H₂M +    2KOH             M^2-    +      2K⁺   +   2 H₂O

initial moles                 0.00787                0.1574               0                 0            

moles at second

equivalence point            0                            0                 0.00787       0.1574       0.1574

Total volume = 100 + 56 mL = 0.156 L

Therefore concentration of M^2- at the second equivalence point = 0.00787 moles / 0.156 L

                                                                                           = 0.05045 M

At this point all the maleic acid will be present in M^2- form, which is the conjugate base of the maleic acid.

Thus the concentration of H₂M at this point is zero.

The pKb of this conjugate base = 14 - pKa2 of maleic acid

                                             = 14 - 6.27 = 7.73

Therefore K_b for M^2- = 10^-7.73 = 1.86 x 10^-8 which has the following equiliibrium

                        M^2-       + H₂O              MH^-    +      OH^-

                                                0.05045 -x                                       x                  x

where x is the concentration of OH^- / MH^- at equillibrium

                           K_b = x² / [0.05045 -x]

                             1.86 x 10^-8 = x² / 0.05045                 [ x in the denominator is neglected as it is very small]

                             x² = 9.38 x 10^-10

                            x = 3.063 x 10^-5

Thus the equliibrium concentrations of M^2- = 0.05045 = 0.00003063

                                                             = 0.05042 M

The equillibrium concentration of MH^- = 3.063 x 10^-5

The equillibrium concentration of OH^- = 3.063 x 10^-5

pOH = - log [3.063 x 10^-5 ]

        = 4.5

Therefore pH = 14 - 4.5 = 9.5

Thus after the addition of 56 mL of 0.281 M KOH, the final solution will have

pH = 9.5

[H₂M] = 0

[HM^-] = 3.063 x 10^-5 M

[M^2-] = 0.05042 M