Question

You are titrating 50.00 mL of a 0.161 M formic acid solution with 0.2949 M sodium hydroxide. Calculate the pH of the solution after adding 46.94 mL of titrant.

Answer 1

Ka of HCOOH = 1.8*10^-4

Given:

M(HCOOH) = 0.161 M

V(HCOOH) = 50 mL

M(NaOH) = 0.2949 M

V(NaOH) = 46.94 mL

mol(HCOOH) = M(HCOOH) * V(HCOOH)

mol(HCOOH) = 0.161 M * 50 mL = 8.05 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.2949 M * 46.94 mL = 13.8426 mmol

We have:

mol(HCOOH) = 8.05 mmol

mol(NaOH) = 13.8426 mmol

8.05 mmol of both will react

excess NaOH remaining = 5.7926 mmol

Volume of Solution = 50 + 46.94 = 96.94 mL

[OH-] = 5.7926 mmol/96.94 mL = 0.0598 M

use:

pOH = -log [OH-]

= -log (5.975*10^-2)

= 1.2236

use:

PH = 14 - pOH

= 14 - 1.2236

= 12.7764

Answer: 12.78