Question

A student performs a titration in the laboratory. A 20.0 mL sample of 0.080 M HCl is titrated with 0.10 M NaOH. What would be the pH of the solution after 25.0 mL of the base has been added?

__ 13.05

__ 9.80    

__ 10.00

__ 12.30

Answer 1

Given:

M(HCl) = 0.08 M

V(HCl) = 20 mL

M(NaOH) = 0.1 M

V(NaOH) = 25 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.08 M * 20 mL = 1.6 mmol

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

mol(NaOH) = 0.1 M * 25 mL = 2.5 mmol

We have:

mol(HCl) = 1.6 mmol

mol(NaOH) = 2.5 mmol

1.6 mmol of both will react

remaining mol of NaOH = 0.9 mmol

Total volume = 45.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 0.9 mmol/45.0 mL

= 2*10^-2 M

use:

pOH = -log [OH-]

= -log (2*10^-2)

= 1.70

use:

PH = 14 - pOH

= 14 - 1.70

= 12.30

Answer: 12.30