Chapter 19.4, Problem 38SP

Interpretation Introduction

Interpretation : The molarity of a solution of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is to be calculated.

Concept Introduction : Salt and water are created when bases and acids react.

A strong acid and strong base react completely to form a neutral solution. A neutralization reaction is a name for this kind of reaction.

The formula of molarity is given as

  $\text{molarity}\text{=}\frac{\text{mol}\text{of}\text{solute}}{\text{L}\text{of}\text{solution}}$

Answer to Problem 38SP

The molarity of a solution of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $1.29\text{M}$ .

Explanation of Solution

Given information:

The volume of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is $\text{15}\text{.0}\text{mL}$ .

The concentration of $\text{NaOH}$ is $\text{0}\text{.150}\text{M}$ .

The volume of $\text{NaOH}$ is $38.5\text{mL}$ .

The reaction between sodium hydroxide and phosphoric acid is given as

  $\text{3NaOH}\left(aq\right)\text{+}{\text{H}}_3{\text{PO}}_4\left(aq\right)\to {\text{Na}}_3{\text{PO}}_4\left(aq\right)\text{+}3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The above reaction is balanced.

The mole ratio is $3:1$ for the reaction between $\text{NaOH}$ and ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ .

The moles of sodium hydroxide are calculated as follows:

  $\begin{array}{c}\text{Moles of NaOH}\text{= 0}\text{.385}\cancel{\text{L NaOH}}\times \frac{0.150\text{mol}\text{NaOH}}{1\cancel{\text{L}\text{NaOH}}}\\ =0.058\text{mol}\text{NaOH}\end{array}$

The moles of phosphoric acid are calculated as follows:

  $\begin{array}{c}{\text{Moles of H}}_3{\text{PO}}_4\text{= 0}\text{.058}\cancel{\text{mol NaOH}}\times \frac{1\text{mol}{\text{H}}_3{\text{PO}}_4}{3\cancel{\text{mol}\text{NaOH}}}\\ =0.019\text{mol}{\text{H}}_3{\text{PO}}_4\end{array}$

The molarity of ${\text{H}}_3{\text{PO}}_4$ is given as

  $\begin{array}{c}\text{molarity}\text{=}\frac{\text{mol}\text{of}\text{solute}}{\text{L}\text{of}\text{solution}}\\ =\frac{0.019\text{mol}}{0.015\text{L}}\\ =1.29\text{M}\end{array}$