# Question #1355e

Mar 26, 2014

This is a comparison of molarity by concentration and volume using the equation ${n}_{b} {M}_{a} {V}_{a} = {n}_{a} {M}_{b} {V}_{B}$.

Where
${n}_{b}$ = OH ions in the base
${M}_{a}$ = molarity of the acid
${V}_{a}$ = volume of the acid
${n}_{a}$ = H ions in the acid
${M}_{b}$ = molarity of the base
${V}_{b}$ = volume of the base

$H C l {O}_{4}$
${n}_{a}$ = 1
${M}_{a}$ = ?
${V}_{a}$ = 25.00 mL

$N a O H$
${n}_{b}$ = 1
${M}_{b}$ = 0.2000M
${V}_{b}$ = 22.62 mL

${n}_{b} {M}_{a} {V}_{a} = {n}_{a} {M}_{b} {V}_{B}$
$\left(1\right) \left(x M\right) \left(25.00 m L\right) = \left(1\right) \left(0.2000 M\right) \left(22.62 m L\right)$
$\left(x M\right) \left(25.00 m L\right) = \left(4.524 M m L\right)$
$\left(x M\right) = \frac{4.524 M m L}{25.00 m L}$
$\left(x M\right) = 0.1809 M$