Chapter 16, Problem 77E

Interpretation Introduction

(a)

Interpretation:

The molarity of $\text{0}\text{.965}\text{N}$ sodium hydroxide solution is to be calculated.

Concept introduction:

Solution is a homogeneous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. The normality of a solution is defined as the number of equivalents per liter of the solution. One equivalent of an acid is the quantity that gives $1$ mole of hydrogen ions in a chemical reaction. One equivalent of a base is the quantity that reacts with one mole of hydrogen ions. The molarity of a solution is defined as the moles of a solute per liter of the solution.

Answer to Problem 77E

The molarity of $\text{0}\text{.965}\text{N}$ sodium hydroxide solution is $\text{0}\text{.965}\text{M}$.

Explanation of Solution

The formula to calculate the molarity is given below.

$\text{Molarity}=\frac{\text{Normality}}{\text{Number}\text{of}\text{equivalents}}$ …(1)

The normality of the solution is $\text{0}\text{.965}\text{N}$.

The relation between $\text{N}$ and $\text{eq}/\text{L}$ as shown below.

$1\text{N}=1\text{eq}/\text{L}$

The probable conversion factors are given below.

$\frac{1\text{N}}{1\text{eq}/\text{L}}\text{and}\frac{1\text{eq}/\text{L}}{1\text{N}}$

The conversion factor to determine $\text{eq}/\text{L}$ from $\text{N}$ is given below.

$\frac{1\text{eq}/\text{L}}{1\text{N}}$

Therefore, $\text{0}\text{.965}\text{N}$ can be written as shown below.

$\begin{array}{rll}\text{Normality}& =& \text{0}\text{.965}\text{N}\text{NaOH}\times \frac{1\text{eq}/\text{L}}{1\text{N}}\\ & =& \frac{\text{0}\text{.965}\text{eq}\text{NaOH}}{\text{L}}\end{array}$

The reaction is given below.

${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}+\text{2NaOH}\to {\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}+2{\text{H}}_{\text{2}}\text{O}$

In this reaction, $1$ mole of $\text{NaOH}$ gives $1$ mole of ${\text{OH}}^{-}$ ions during the reaction. Therefore, the number of equivalents per mole is $1$.

Substitute the number of equivalents and the normality of $\text{NaOH}$ in equation (1).

$\begin{array}{rll}\text{Molarity}& =& \frac{\text{0}\text{.965}\text{eq}\text{NaOH}}{\text{L}}\times \frac{1}{\text{1}\text{eq}/\text{mol}}\\ & =& \text{0}\text{.965}\text{mol}/\text{L}\end{array}$

The relation between $\text{M}$ and $\text{mol}/\text{L}$ as shown below.

$1\text{M}=1\text{mol}/\text{L}$

The probable conversion factors are given below.

$\frac{1\text{M}}{1\text{mol}/\text{L}}\text{and}\frac{1\text{mol}/\text{L}}{1\text{M}}$

The conversion factor to determine $\text{M}$ from $\text{mol}/\text{L}$ is given below.

$\frac{1\text{M}}{1\text{mol}/\text{L}}$

Therefore, $\text{0}\text{.965}\text{mol}/\text{L}$ can be written as shown below.

$\begin{array}{rll}\text{Molarity}& =& \text{0}\text{.965}\text{mol}/\text{L}\text{NaOH}\times \frac{1\text{M}}{1\text{mol}/\text{L}}\\ & =& \text{0}\text{.965}\text{M}\text{NaOH}\end{array}$

Therefore, the molarity of $\text{0}\text{.965}\text{N}$ sodium hydroxide solution is $\text{0}\text{.965}\text{M}$.

Conclusion

The molarity of $\text{0}\text{.965}\text{N}$ sodium hydroxide solution is $\text{0}\text{.965}\text{M}$.

Interpretation Introduction

(b)

Interpretation:

The molarity of $\text{0}\text{.237}\text{N}$ ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is to be calculated.

Concept introduction:

Solution is a homogeneous mixture of two or more components. A sample taken from any part of the solution will have the same composition as the rest of the solution. The normality of a solution is defined as the number of equivalents per liter of the solution. One equivalent of an acid is the quantity that gives $1$ mole of hydrogen ions in a chemical reaction. One equivalent of a base is the quantity that reacts with one mole of hydrogen ions. The molarity of a solution is defined as the moles of a solute per liter of the solution.

Answer to Problem 77E

The molarity of $\text{0}\text{.237}\text{N}$ ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $\text{0}\text{.119}\text{M}$.

Explanation of Solution

The formula to calculate the molarity is given below.

$\text{Molarity}=\frac{\text{Normality}}{\text{Number}\text{of}\text{equivalents}}$ …(1)

The normality of the solution is $\text{0}\text{.237}\text{N}$.

The relation between $\text{N}$ and $\text{eq}/\text{L}$ as shown below.

$1\text{N}=1\text{eq}/\text{L}$

The probable conversion factors are given below.

$\frac{1\text{N}}{1\text{eq}/\text{L}}\text{and}\frac{1\text{eq}/\text{L}}{1\text{N}}$

The conversion factor to determine $\text{eq}/\text{L}$ from $\text{N}$ is given below.

$\frac{1\text{eq}/\text{L}}{1\text{N}}$

Therefore, $\text{0}\text{.237}\text{N}$ can be written as shown below.

$\begin{array}{rll}\text{Normality}& =& \text{0}\text{.237}\text{N}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\times \frac{1\text{eq}/\text{L}}{1\text{N}}\\ & =& \frac{\text{0}\text{.237}\text{eq}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{L}}\end{array}$

The reaction is given below.

${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}+\text{2NaOH}\to {\text{Na}}_{\text{2}}{\text{HPO}}_{\text{4}}+2{\text{H}}_{\text{2}}\text{O}$

In this reaction, $1$ mole of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ gives $2$ moles of ${\text{H}}^{+}$ ions during the reaction. Therefore, the number of equivalents per mole is $2$.

Substitute the number of equivalents and the normality of $\text{NaOH}$ in equation (1).

$\begin{array}{rll}\text{Molarity}& =& \frac{\text{0}\text{.237}\text{eq}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}}{\text{L}}\times \frac{1}{2\text{eq}/\text{mol}}\\ & =& \text{0}\text{.119}\text{mol}/\text{L}\end{array}$

The relation between $\text{M}$ and $\text{mol}/\text{L}$ as shown below.

$1\text{M}=1\text{mol}/\text{L}$

The probable conversion factors are given below.

$\frac{1\text{M}}{1\text{mol}/\text{L}}\text{and}\frac{1\text{mol}/\text{L}}{1\text{M}}$

The conversion factor to determine $\text{M}$ from $\text{mol}/\text{L}$ is given below.

$\frac{1\text{M}}{1\text{mol}/\text{L}}$

Therefore, $\text{0}\text{.119}\text{mol}/\text{L}$ can be written as shown below.

$\begin{array}{rll}\text{Molarity}& =& \text{0}\text{.119}\text{mol}/\text{L}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\times \frac{1\text{M}}{1\text{mol}/\text{L}}\\ & =& \text{0}\text{.119}\text{M}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\end{array}$

Therefore, the molarity of $\text{0}\text{.237}\text{N}$ ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $\text{0}\text{.119}\text{M}$.

Conclusion

The molarity of $\text{0}\text{.237}\text{N}$ ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $\text{0}\text{.119}\text{M}$.