Calculate the normality of the HF solution.

Question

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.

Answer 1

HF + KOH ==> H2O + KF ... balanced equation

moles KOH required = 29.6 ml x 1 L/1000 ml x 0.120 mol/L = 0.003552 moles

moles HF present = 0.003552 moles b/c they react in a 1:1 mol ratio

The volume of HF = 10.0 ml x 1 L/1000 ml = 0.0100 L

Normality = 0.003552 moles / 0.010 L = 0.3552 N (note: normality isn't frequently used today in most chemistry courses). In this case N is the same as M (molarity) because the equivalents is 1.