Question

Determine the pH (to two decimal places) of the solution that is produced by mixing 28.7 mL of 4.13×10-3 M HI with 38.9 mL of 8.42×10-2 M KH

Answer 1

Given:

M(HI) = 0.0041 M

V(HI) = 28.7 mL

M(KOH) = 0.0842 M

V(KOH) = 38.9 mL

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

mol(HI) = 0.0041 M * 28.7 mL = 0.1185 mmol

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

mol(KOH) = 0.0842 M * 38.9 mL = 3.2754 mmol

We have:

mol(HI) = 0.1185 mmol

mol(KOH) = 3.275 mmol

0.1185 mmol of both will react

remaining mol of KOH = 3.157 mmol

Total volume = 67.6 mL

[OH-]= mol of base remaining / volume

[OH-] = 3.157 mmol/67.6 mL

= 4.67*10^-2 M

use:

pOH = -log [OH-]

= -log (4.67*10^-2)

= 1.3307

use:

PH = 14 - pOH

= 14 - 1.3307

= 12.6693

Answer: 12.67