What is the approximate pH at the equivalence point of a weak acid-strong base titration if ...?

Question

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid requires 29.80 mL of 0.0567 M NaOH?

Ka = 1.8x10^(-4) for formic acid

Answer Choices

a. 8.12

b. 5.88

c 11.54

d. 2.46

Answer 1

Moles OH- = 0.02980 L x 0.0567 M =0.00169

HCOOH + OH- >> HCOO- + H2O

[HCOO-] = 0.00169 / 0.0548 L =0.0308

HCOO- + H2O <----> HCOOH + OH-

K = Kw/Ka = 5.56 x 10^-11 = x^2 / 0.0308-x

x = [OH-] =1.31 x 10^-6 M

pOH =5.88

pH = 14 - 5.88 = 8.12