Chapter 15, Problem 46A

Interpretation Introduction

Interpretation: Molarity and normality of a phosphoric acid, ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution consisting of $35.2g$ of it per liter of solution is to be calculated.

Concept introduction: One normal solution is a measure of the number of moles of hydrogen ions furnished by one mol of an acid. One molar solution is a measure of the number of moles of molecules of acid in one mol of an acid. One molecule of phosphoric acid, ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ furnishes 3 hydrogen ions. So, 1 mol of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ will furnish 3 mol of hydrogen ions. Relation between molarity and normality of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is as follows: ${\text{Normality of H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{=3× Molarity of H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Molarity is a defined as the number of moles of solute dissolved per liter of the solution. It is calculated as follows:

  ${\text{Molarity of H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{=}\frac{{\text{number of moles of H}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{Volume of solution in liters}}$

Answer to Problem 46A

Molarity and normality of the given ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $\text{0}\text{.359 M}$ and $\text{1}\text{.077 N}$ respectively.

Explanation of Solution

One normal solution is a measure of the number of moles of hydrogen ions furnished by one mol of an acid. One molar solution is a measure of the number of moles of molecules of acid in one mol of an acid. One molecule of phosphoric acid, ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ furnishes 3 hydrogen ions. So, 1 mol of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ will furnish 3 mol of hydrogen ions. Relation between molarity and normality of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ is as follows: ${\text{Normality of H}}_{\text{3}}{\text{PO}}_{\text{4}}{\text{=3× Molarity of H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Molarity is a defined as the number of moles of solute dissolved per liter of the solution. It is calculated as follows:

  ${\text{Molarity of H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{=}\frac{{\text{number of moles of H}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{Volume of solution in liters}}$

Or,

  $\begin{array}{l}{\text{Number of moles of H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{=}\frac{{\text{mass of H}}_{\text{3}}{\text{PO}}_{\text{4}}}{{\text{Molar mass of H}}_{\text{3}}{\text{PO}}_{\text{4}}}\\ \text{ =}\frac{\text{35}\text{.2 g}}{\text{97}\text{.994 g/mol}}\\ \text{ = 0}\text{.359 mol}\\ {\text{Molarity of H}}_{\text{3}}{\text{PO}}_{\text{4}}\text{=}\frac{\text{0}\text{.359 mol}}{\text{1 L}}\\ \text{ =0}\text{.359 M}\end{array}$

Or,

  $\begin{array}{l}{\text{Normality of H}}_{\text{3}}{\text{PO}}_{\text{4}}=3\times {\text{Molarity of H}}_{\text{3}}{\text{PO}}_{\text{4}}\\ {\text{Normality of H}}_{\text{3}}{\text{PO}}_{\text{4}}=(3\times \text{0}\text{.359)N}\\ \text{ =1}\text{.077 N}\end{array}$

Conclusion

Molarity and normality of the given ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $\text{0}\text{.359 M}$ and $\text{1}\text{.077 N}$ respectively.