10.0 mL of a HF solution was titrated with a 0.120 N solution of KOH; 29.6 mL of

KOH were required to completely neutralize the HF. Calculate the normality of the HF solution.

Question

Chemistry

Zander Dyer

12 December, 18:02

10.0 mL of a HF solution was titrated with a 0.120 N solution of KOH; 29.6 mL of

KOH were required to completely neutralize the HF. Calculate the normality of the HF solution.

+4

Answers (1)

Know the Answer?

Malcolm Friedman · Answer 1

0.355 N of HF

Explanation:

The titration reaction of HF with KOH is:

HF + KOH → H₂O + KF

Where 1 mole of HF reacts per mole of KOH

Moles of KOH are:

0.0296L * (0.120 equivalents / L) = 3.552x10⁻³ equivalents of KOH = equivalents of HF.

As volume of the titrated solution was 10.0mL, normality of HF solution is:

3.552x10⁻³ equivalents of HF / 0.010L = 0.355 N of HF

10.0 mL of a HF solution was titrated with a 0.120 N solution of KOH; 29.6 mL ofKOH were required to completely neutralize the HF. Calculate the normality of the HF solution.

10.0 mL of a HF solution was titrated with a 0.120 N solution of KOH; 29.6 mL of

KOH were required to completely neutralize the HF. Calculate the normality of the HF solution.