Answer:
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
