What is the pH at the equivalence point of a weak acid, strong base titration if 25.0 mL of formic acid requires 29.80 mL of 0.1567 M NaOH. Ka for formic acid is 1.8 x 10-4.

pH = 8.34

Explanation:

Formic acid reacts with a strong base producing formate ion, HCOO⁻, the equilibrium is:

Kb = [HCOOH] [OH⁻] / [HCOO⁻]

Where Kb is basic equilibrium constant = Kw / Ka = 1x10⁻¹⁴ / 1.8x10⁻⁴ = 5.56x10⁻¹¹

[HCOOH] and [OH⁻] comes both from formate ion, that means  [OH⁻] = [HCOOH] = X, our incognite

And [HCOO⁻] is molarity of the solution. We need to determine this:

Moles of formate = Moles of NaOH added until equivalence point:

29.80mL = 0.02980L * (0.1567mol / L) = 4.670x10⁻³ moles NaOH = Moles Formate

And total volume of the solution is:

29.80mL + 25mL = 54.8mL = 0.0548L. Molarity is:

4.670x10⁻³ moles / 0.0548L = 0.08522M

Replacing in Kb expression:

Kb = [HCOOH] [OH⁻] / [HCOO⁻]

5.56x10⁻¹¹ = [X] [X] / [0.08522M]

4.738x10⁻¹² = X²

X = [OH⁻] = 2.177x10⁻⁶M

pOH = -log [OH⁻] = 5.66

And pH = 14 - pOH is:

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