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emmasim [6.3K]
2 years ago
8

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

Chemistry
1 answer:
Ilia_Sergeevich [38]2 years ago
3 0

Answer:

0.355 N of HF

Explanation:

The titration reaction of HF with KOH is:

HF + KOH → H₂O + KF

<em>Where 1 mole of HF reacts per mole of KOH</em>

<em />

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 =<em> 0.355 N of HF</em>

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