Question

What is the pH of a solution made by mixing 200.0 mL of 0.250 M hydrobromic acid and 100.0 mL of 0.150 M potassium hydroxide

Answer 1

Given:

M(HBr) = 0.250 M

V(HBr) = 200.0 mL

M(KOH) = 0.150 M

V(KOH) = 100.0 mL

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.250 M * 200.0 mL = 50.0 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.150 M * 100.0 mL = 15.0 mmol

We have:

mol(HBr) = 50 mmol

mol(KOH) = 15 mmol

15 mmol of both will react

remaining mol of HBr = 35 mmol

Total volume = 300.0 mL

[H+]= mol of acid remaining / volume

[H+] = 35 mmol/300.0 mL

= 0.1167 M

use:

pH = -log [H+]

= -log (0.1167)

= 0.933

Answer: 0.933