Question

What volume of 0.0522 M $\mathbf{Ba}{\left(\mathrm{OH}\right)}_2$is required to neutralize exactly 14.20 mL of 0.141 M ${\mathbf{H}}_3{\mathbf{PO}}_4$? Phosphoric acid contains three acidic hydrogen.

Short answer

The required volume is 57.6 mL.

Step-by-step solution

Given reaction

The balanced chemical equation for the given chemical reaction is as shown below:

$2H_{3}P{O}_4\left(aq\right)+3Ba{\left(OH\right)}_2\left(aq\right)\to B{a}_3{\left(P{O}_4\right)}_2\left(aq\right)+6H_{2}O\left(l\right)$

Calculation of moles

$\begin{array}{l}\text{Molesof}{H}_{3}P{O}_4=\frac{0.141mol{H}_{3}P{O}_4}{1.00L{H}_{3}P{O}_4}\times \frac{1.00L{H}_{3}P{O}_4}{1000mL{H}_{3}P{O}_4}\times 14.20mL{H}_{3}P{O}_4\\ \text{Molesof}{H}_{3}P{O}_4=2.00\times {10}^{-3}mol{H}_{3}P{O}_4\end{array}$

Hence, thenumber of moles is$2.00\times {10}^{-3}mol{H}_{3}P{O}_4$.

Calculation of volume

$\begin{array}{l}\text{Molarityof}Ba{\left(OH\right)}_2=2.00\times {10}^{-3}mol{H}_{3}P{O}_4\times \frac{3.0molBa{\left(OH\right)}_2}{2.0mol{H}_{3}P{O}_4}\times \frac{1.00LBa{\left(OH\right)}_2}{0.0521molBa{\left(OH\right)}_2}\\ \text{Molarityof}Ba{\left(OH\right)}_2=0.0576LBa{\left(OH\right)}_2\times \frac{1.00LBa{\left(OH\right)}_2}{0.0521molBa{\left(OH\right)}_2}\\ \text{Molarityof}Ba{\left(OH\right)}_2=57.6mLBa{\left(OH\right)}_2\end{array}$

Hence, the required volumeis$57.6mLBa{\left(OH\right)}_2$ .