If $38.2\mathrm{mL}$of a $0.163\mathrm{M}\mathrm{KOH}$solution is required to titrate $25.0\mathrm{mL}$of a solution of ${\mathrm{H}}_2{\mathrm{SO}}_4$, what is the molarity of the ${\mathrm{H}}_2{\mathrm{SO}}_4$solution$?$

${\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)+2\mathrm{NaOH}\left(aq\right)⟶2{\mathrm{H}}_2\mathrm{O}\left(l\right)+{\mathrm{Na}}_2{\mathrm{SO}}_4\left(aq\right)$

A conditionthat exists when acidic and basic ions neutralize each other accurately in solution. A state in which the concentrations of acidic ions and basic ions in a solution are neutralized with each other.

Imagine the reaction given as follows:

Think of the given data as follows:

$=0.025\mathrm{L}$

Molarity of ${\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)$is calculate.

Molarity of $\mathrm{KOH}\left(aq\right)=0.163\mathrm{M}$

Volume of $\mathrm{KOH}\left(aq\right)=38.2\mathrm{mL}$

$=0.0382\mathrm{L}$

Find the number of moles of volume

$\text{Find number moles of}\mathrm{KOH}\left(aq\right)=\text{Molarity}x\text{Volume}$

$n=0.163\mathrm{MX}0.0382\mathrm{L}$

$n=0.0062266moles$

At the titration point, $2$moles of $KOH\left(aq\right)$are required to titrate with $1$mole of $H_2{SO}_4$.

So,

$\text{Number of moles of}{\mathrm{H}}_2{\mathrm{SO}}_4=\frac{\mathrm{n}}{2}$

$=\frac{0.0062266}{2}$

$=0.0031$

Molarity of ${\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)=\frac{\mathrm{n}}{\mathrm{v}}$

$=\frac{0.0031\mathrm{moles}}{0.025\mathrm{L}}$

$=0.124\mathrm{M}$

$=0.12\mathrm{M}$