10.0 mL of a HF solution was titrated with a 0.120 N solution of KOH; 29.6 mL of
1 answer:
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>
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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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