Question

(13) Hydrogen phthalate ion, HC₈H₄O₄^–, from potassium hydrogen phthalate (abbreviated as “KHP”; formula = KHC₈H₄O₄), is a weak acid. In an acid-base titration, a 50.0-mL solution of 0.100 M “KHP” solution is titrated with NaOH solution of unknown concentration. The following net ionic equation represents the reaction that occurs during titration:

HC₈H₄O₄^–(aq) + OH^– (aq) → C₈H₄O₄^2–(aq) + H₂O(l)

   (a) If the titration requires 34.0 mL of NaOH solution to reach equivalent point, calculate the molar concentration of NaOH solution. What is the molar concentration of phthalate ion, C₈H₄O₄^2–, at equivalent point?

   (b)If the pH of solution at equivalent point is 9.08, determine the K_b of phthalate ion and the K_a of hydrogen phthalate ion, HC₈H₄O₄^–.

   (c) Using the K_a value of hydrogen phthalate ion, HC₈H₄O₄^–, calculated in (b) and the initial concentration of HC₈H₄O₄^–, determine the initial pH of potassium hydrogen phthalate solution.

   (d) What is the expected pH of the solution after 17.0 mL of NaOH solution has been added?

Hint: use a reaction table to determine [HC₈H₄O₄^–] and [C₈H₄O₄^2–] in the mixture after the reaction with NaOH, obtain the molar ratio: [C₈H₄O₄^2–]/HC₈H₄O₄^–].

Answer 1

a) HC8H4O4–(aq) + OH– (aq) → C8H4O42–(aq) + H2O(l)

no of mole of KHP taken = M*V = 50*0.1 = 5 mmol

at equivalence point , 1 mole HC8H4O4– = 1 mole NaOH

no of mole of NaOH reacted = 5 mmol

Molar concentration of NaOH = n/V = 5/34 = 0.147 M

at equivalence point,

Molar concentration of pthalate ion = 50*0.1/(50+34) = 0.06 M

b) at equivalence point, pH = 7+1/2(pka+logC)

                       C = concentration of salt = 0.06 M

                     pka of KHP = ?

                 9.08 = 7+1/2(x+log0.06)

              x = pka = 5.38

               ka= 10^-pka = 10^-5.38 = 4.17*10^-6

              kb = 10^-14/ka = 10^-14/(4.17*10^-6) = 2.4*10^-9

c) initial pH = 1/2(pka-logC)

    c = initial concentration of KHP = 0.1 M

    pka = 5.38

   pH = 1/2(5.38-log0.1)

      = 3.19

d) after 17.0 mL of NaOH solution added

   pH = pka + log(salt/acid)

   pka of KHP = 5.38

   no of mole of NaOH added = 17*0.147 = 2.5 mmol

   no of mole of KHP unreacted = 5- 2.5 = 2.5 mmol

so that, pH = pka = 5.38