Question

Calculate the pH when 70 mL of 0.20 M HI are titrated with 100 mL of 0.20 M LiOH.

Answer 1

Given:

M(HI) = 0.2 M

V(HI) = 70 mL

M(LiOH) = 0.2 M

V(LiOH) = 100 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.2 M * 70 mL = 14 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.2 M * 100 mL = 20 mmol

We have:

mol(HI) = 14 mmol

mol(LiOH) = 20 mmol

14 mmol of both will react

remaining mol of LiOH = 6 mmol

Total volume = 170.0 mL

[OH-]= mol of base remaining / volume

[OH-] = 6 mmol/170.0 mL

= 3.529*10^-2 M

use:

pOH = -log [OH-]

= -log (3.529*10^-2)

= 1.4523

use:

PH = 14 - pOH

= 14 - 1.4523

= 12.5477

Answer: 12.55