Question

What is the pH of a 200ml solution of 0.25M HCl combined with a 100ml solution of 0.25 NaOH?

Answer 1

Given:

M(HCl) = 0.25 M

V(HCl) = 200 mL

M(NaOH) = 0.25 M

V(NaOH) = 100 mL

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

mol(HCl) = 0.25 M * 200 mL = 50 mmol

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

mol(NaOH) = 0.25 M * 100 mL = 25 mmol

We have:

mol(HCl) = 50 mmol

mol(NaOH) = 25 mmol

25 mmol of both will react

remaining mol of HCl = 25 mmol

Total volume = 300.0 mL

[H+]= mol of acid remaining / volume

[H+] = 25 mmol/300.0 mL

= 8.333*10^-2 M

use:

pH = -log [H+]

= -log (8.333*10^-2)

= 1.0792

Answer: 1.08