Chapter 16 - Sections 16.1-16.8 - Exercises - Problems by Topic - Page 805: 62a

In both titrations, the volume of added $KOH$ at the equivalence point is equal to 12.5 mL

$0.100 \space M \space HCl$: $$25.0 \space mL \times 0.100 \space M = 0.200 \space M \times V$$ $$\frac{25.0 \space mL \times 0.100 \space M}{0.200 \space M} = V$$ $$V = 12.5 \space mL$$ $0.100 \space M \space HF$: $$25.0 \space mL \times 0.100 \space M = 0.200 \space M \times V$$ $$\frac{25.0 \space mL \times 0.100 \space M}{0.200 \space M} = V$$ $$V = 12.5 \space mL$$