20 ml chlorhydric acid with unknown molarity are titrated by NaOH 0.1 M.
The equivalent point is attained after adding 15.5 ml of the NaOH solution.
Calculate the molarity of the acid solution.
Let $V_{NaOH}$ $=$ $0.0155 L$ be the volume of added sodium hydroxide solution,
$V_{HCl}$ $=$ $0.020L$ the volume of the unknown acid solution.
$n_{NaOH}$ $=$ $V_{NaOH} \cdot [NaOH]$ $=$ $0.0155 \cdot 0.1$ $=$ $0.00155 $ mol
One mole of HCl reacts with one mole NaOH according to:
$HCl$ $+$ $NaOH$ $\longrightarrow$ $NaCl$ $+$ $H_2O$
so the number of added moles $NaOH$ = the number of moles of $HCl$ in the unknown solution =
$0.00155\;mol$
$[HCl]$ $=$ $\frac{n_{HCl}}{V_{HCl}}$ $=$ $\frac{0.00155}{0.020}$ $=$ $0.0775 M$
