Question

A 25.0 mL sample of 0.105 M HCl was titrated to the equivalence point with 31.5 mL of NaOH. What is the concentration of the NaOH

Answer 1

Answer:  The concentration of $NaOH$ is 0.08 M.

Explanation:-

According to the neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$M_1$ = molarity of $HCl$ solution = 0.105 M

$V_1$ = volume of $HCl$ solution = 25 ml

$M_2$ = molarity of $NaOH$ solution = ?

$V_2$ = volume of $NaOH$ solution = 31.5 ml

$n_1$ = valency of $HCl$ = 1

$n_2$ = valency of $NaOH$ = 1

$1\times 0.105M\times 25=1\times M_2\times 31.5$

$M_2=0.08M$

Therefore, the concentration of $NaOH$ is 0.08 M.