How many milliliters of a $9{\mathrm{NH}}_2{\mathrm{SO}}_4$ solution will be required to neutralize completely 20 mL of a $3.6\mathrm{N}$ $\mathrm{NaOH}$ solution?

${\mathrm{V}}_{\mathrm{NaOH}} = 20 \mathrm{ml}$

${\mathrm{N}}_{\mathrm{NaOH} }= 3.6 \mathrm{N}$

${\mathrm{N}}_{{\mathrm{H}}_2{\mathrm{SO}}_4} = 9 \mathrm{N}$

${\mathrm{N}}_1{\mathrm{V}}_1 = {\mathrm{N}}_2{\mathrm{V}}_2$

${\mathrm{V}}_{{\mathrm{H}}_2{\mathrm{SO}}_4}= \frac{{\mathrm{N}}_{\mathrm{NaOH}} \times {\mathrm{V}}_{\mathrm{NaOH}}}{{\mathrm{N}}_{{\mathrm{H}}_2{\mathrm{SO}}_4}}= \frac{3.6\times 20}{9}$= 8.0 ml

8.0 mL milliliters of a $9{\mathrm{NH}}_2{\mathrm{SO}}_4$ solution will be required to neutralize completely 20 mL of a $3.6\mathrm{N}$ $\mathrm{NaOH}$ solution.