If $32.8\mathrm{mL}$of a $0.162\mathrm{M}\mathrm{NaOH}$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?

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

A state in which acidic and basic ions precisely neutralise each other in solution. In a solution, the concentrations of acidic and basic ions are neutralised by each other.

Consider the following example of a reaction:

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

Consider the following information:

localid="1652714471236" $\text{Volume of}{\mathrm{H}}_2{\mathrm{SO}}_4\left(aq\right)=25.0\mathrm{mL}$

$=0.025\mathrm{L}$

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

localid="1652623407129" $MolarityofNaOH\left(aq\right)=0.162M$

$VolumeofNaOH(aq)=32.8mL$

localid="1652714478436" $=0.0328\mathrm{L}$

Find moles of $\mathrm{NaOH}\left(aq\right)=$Molarity $\times$Volume

$n=0.162\mathrm{M}\times 0.0328\mathrm{L}$

$n=0.0053136moles$

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

So,

localid="1652714488165" $\text{Number of moles of}{\mathrm{H}}_2{\mathrm{SO}}_4=\frac{\mathrm{n}}{2}$

localid="1652714497276" $=\frac{0.0053136}{2}$

localid="1652714503142" $=0.0026$

localid="1652714513679" $\text{Molarity of}{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)=\frac{\mathrm{n}}{\mathrm{v}}$

$=\frac{0.0026\mathrm{moles}}{0.025\mathrm{L}}$

$=0.104\mathrm{M}$

$=0.1\mathrm{M}$