Question

What is the pH of a solution prepared by mixing 70.0 mL of 0.110 M HC2H3O2 with 95.0 mL of 0.270 M NaOH?

Answer 1

Given:

M(HC2H3O2) = 0.11 M

V(HC2H3O2) = 70 mL

M(NaOH) = 0.27 M

V(NaOH) = 95 mL

mol(HC2H3O2) = M(HC2H3O2) * V(HC2H3O2)

mol(HC2H3O2) = 0.11 M * 70 mL = 7.7 mmol

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

mol(NaOH) = 0.27 M * 95 mL = 25.65 mmol

We have:

mol(HC2H3O2) = 7.7 mmol

mol(NaOH) = 25.65 mmol

7.7 mmol of both will react

excess NaOH remaining = 17.95 mmol

Volume of Solution = 70 + 95 = 165 mL

[OH-] = 17.95 mmol/165 mL = 0.1088 M

use:

pOH = -log [OH-]

= -log (0.1088)

= 0.96

use:

PH = 14 - pOH

= 14 - 0.96

= 13.04

pH = 13.04