What is the pH of a solution made by mixing 30.00 mL of 0.10 M HI with 40.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive.
A) 0.85
B) 1.85
C) 12.15
D) 13.15

Question

Quizplus · Accepted Answer

HI + KOH → KI + H2O

Since HI and KOH are present in equal concentrations, the reaction will result in a neutral solution, with a pH of 7. Adding 30 mL of 0.10 M HI to 40 mL of 0.10 M KOH results in an excess of KOH (10 mL). The excess KOH will determine the pH of the resulting solution.

The moles of KOH present in the solution are:

0.040 L x 0.10 mol/L = 0.004 mol

The moles of HI present in the solution are:

0.030 L x 0.10 mol/L = 0.003 mol

The excess KOH is:

0.004 mol - 0.003 mol = 0.001 mol

Total volume of the solution is:

0.030 L + 0.040 L = 0.070 L

The concentration of the excess KOH is:

0.001 mol / 0.070 L = 0.014 M

pOH = -log(0.014) = 1.85
pH = 14 - pOH = 12.15

What Is the PH of a Solution Made by Mixing