Question

A 50.0 mL sample of 0.25 M formic acid (HCOOH) aqueous solution is titrated with 0.125 M NaOH solution. Ka of HCOOH = 1.7 x 10−4.

a. Calculate the pH of the solution after 50 mL of NaOH solution has been added.

b. How many mL of 0.125 M NaOH need to be added to the sample to reach the equivalence point? What is the pH at the equivalence point?

Answer 1

a.

50.0 mL of 0.25 M of formic acid = 50 x 0.25 = 12.5 mol formic acid

50.0 mL of 0.125 M of NaOH solution = 50 x 0.125 = 6.25 mol NaOH

6.25 mol NaOH will react with 6.25 mol of HCOOH to give 6.26 mol of HCOONa.

Therefore, the moles of remaining HCOOH = (12.5 - 6.25) = 6.25 mol

Total volume of the solution = (50 + 50) = 100 mL

Concentration of HCOOH = 6.25/100 = 0.0625 M

Concentration of HCOONa = 6.25/100 = 0.0625 M

K_a of HCOOH = 1.7 x 10⁻⁴

From Henderson-Hasselbalch equation,

pH = pK_a + log [base]/[acid]

      = - log(1.7 x 10⁻⁴) + log 0.0625/0.0625

      = 3.77

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b.

Volume of formic acid, V₁ = 50.0 mL

Concentration of formic acid, S₁ = 0.25 M

Volume of NaOH = V₂

Concentration of NaOH, S₂ = 0.125 M

Now, equating the moles of acid and the base we get,

V₁S₁ = V₂S₂

or, 50.0 x 0.25 = V₂ x 0.125

or, V₂ = 100 mL

Therefore, the volume of the NaOH solution need to be added = 100 mL

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Moles of HCOONa forms at equivalence point = 100 x 0.125 = 12.5 mol

Total volume of the solution = 100 + 50 = 150 mL

Concentration of HCOONa = 12.5/150 = 0.083 M

K_a of HCOOH = 1.7 x 10⁻⁴

Therefore, K_b of HCOO^- = 10^-14/(1.7 x 10⁻⁴) = 5.88 x 10^-11

                                                                       HCOONa + H₂O HCOOH + OH^-

Initial concentration (M)                                     0.083    0 0

Change in concentration (M)                              -X                            X               X

Equilibrium concentration (M)                       (0.083 - X)                    X               X

Now,

K_b = [HCOOH][OH^-]/[HCOONa]

or, 5.88 x 10^-11 = X²/(0.083 - X)       [(0.083 - X) 0.083 as X << 0.083]

or, 5.88 x 10^-11 = X²/0.083

or, X² = 5.88 x 10^-12

or, X = 2.21 x 10^-6

or, [OH^-] = 2.21 x 10^-6

or, - log[OH^-] = - log(2.21 x 10^-6)    [Taking -log on both sides of the equation]

or, pOH = 5.66

or, 14 - pH = 5.66       [As pH + pOH = 14]

or, pH = 8.34

Therefore, the pH at the equivalence point = 8.34