Question

Determine the pH of the solution that results when 33.0mL of 0.130M HCl is mixed with 26.0mL of 0.160M NaOH

Answer 1

Given:

M(HCl) = 0.13 M

V(HCl) = 33 mL

M(NaOH) = 0.16 M

V(NaOH) = 26 mL

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

mol(HCl) = 0.13 M * 33 mL = 4.29 mmol

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

mol(NaOH) = 0.16 M * 26 mL = 4.16 mmol

We have:

mol(HCl) = 4.29 mmol

mol(NaOH) = 4.16 mmol

4.16 mmol of both will react

remaining mol of HCl = 0.13 mmol

Total volume = 59.0 mL

[H+]= mol of acid remaining / volume

[H+] = 0.13 mmol/59.0 mL

= 2.203*10^-3 M

use:

pH = -log [H+]

= -log (2.203*10^-3)

= 2.6569

Answer: 2.66