Question

Consider the titration of a 23.0 −mL sample of 0.100 M HC2H3O2 with 0.120 M NaOH. Determine the pH after adding 5.00 mL of base beyond the equivalence point.

Answer 1

find the volume of NaOH used to reach equivalence point

M(HC2H3O2)*V(HC2H3O2) =M(NaOH)*V(NaOH)

0.1 M *23.0 mL = 0.12M *V(NaOH)

V(NaOH) = 19.2 mL

Volume of NaOH added = 19.2 mL + 5.00 mL = 24.2 mL

Given:

M(HC2H3O2) = 0.1 M

V(HC2H3O2) = 23 mL

M(NaOH) = 0.12 M

V(NaOH) = 24.2 mL

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

mol(HC2H3O2) = 0.1 M * 23 mL = 2.3 mmol

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

mol(NaOH) = 0.12 M * 24.2 mL = 2.904 mmol

We have:

mol(HC2H3O2) = 2.3 mmol

mol(NaOH) = 2.904 mmol

2.3 mmol of both will react

excess NaOH remaining = 0.604 mmol

Volume of Solution = 23 + 24.2 = 47.2 mL

[OH-] = 0.604 mmol/47.2 mL = 0.0128 M

use:

pOH = -log [OH-]

= -log (1.28*10^-2)

= 1.8929

use:

PH = 14 - pOH

= 14 - 1.8929

= 12.1071

Answer: 12.11