Question

If 5.25 mL of HCl requires 4.96 mL of 0.9845 M NaOH to reach the equivalence point,

what is the concentration of the HCI?

Answer 1

`M_(A)V_(A)=M_(B)V_(B)\\(5.25)M_(A)=(4.96)(0.9845)\\M_(A)=((4.96)(0.9845))/(5.25) \approx \boxed{0.93 \text{ M}}`