Chapter 17, Problem 57E

Interpretation Introduction

Interpretation:

The values of pOH, $\left[{\text{H}}^{\text{+}}\right]$, and $\left[{\text{OH}}^{-}\right]$ of a $\text{0}\text{.20}\text{M}$ ammonium nitrate solution is to be calculated.

Concept introduction:

The pH of a solution can be determined by taking the negative logarithm of the concentration of hydrogen ions in a solution. The pOH of a solution can be determined by taking the negative logarithm of the concentration of ${\text{OH}}^{-}$ ions in a solution. The sum of pH and pOH is equal to $14$.

Answer to Problem 57E

The values of pOH, $\left[{\text{H}}^{\text{+}}\right]$, and $\left[{\text{OH}}^{-}\right]$ of a $\text{0}\text{.20}\text{M}$ ammonium nitrate solution are equal to $\text{9}\text{.04}$, $\text{1}\text{.1}\times {10}^{-5}\text{M}$ and $\text{9}\text{.1}\times {10}^{-10}\text{M}$ respectively.

Explanation of Solution

The pH of a $\text{0}\text{.20}\text{M}$ ammonium nitrate solution is $4.96$.

The value of pOH is calculated by the formula given below.

$\text{pOH}=14-\text{pH}$

Substitute the value of pH in the above equation.

$\begin{array}{rll}\text{pOH}& =& 14-\text{pH}\\ & =& \text{14}-\text{4}\text{.96}\\ & =& \text{9}\text{.04}\end{array}$

Therefore, the value of pOH of the solution is equal to $\text{9}\text{.04}$.

The concentration of ${\text{H}}^{\text{+}}$ ions is calculated by the formula given below.

$\left[{\text{H}}^{\text{+}}\right]={\text{10}}^{-\text{pH}}$

Substitute the value of pH in the above expression.

$\begin{array}{rll}\left[{\text{H}}^{\text{+}}\right]& =& {\text{10}}^{-\text{pH}}\\ & =& {10}^{-4.96}\\ & =& 1.09\times {10}^{-5}\text{M}\\ & \approx & \text{1}\text{.1}\times {10}^{-5}\text{M}\end{array}$

Therefore, the value of concentration of ${\text{H}}^{\text{+}}$ ions is equal to $\text{1}\text{.1}\times {10}^{-5}\text{M}$.

The value of ionic product of water $\left({\text{K}}_{\text{w}}={10}^{-14}\right)$ is calculated by the formula given below.

$\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]={\text{K}}_{\text{w}}$

Rearrange the above equation to obtain the value of $\left[{\text{OH}}^{-}\right]$.

$\begin{array}{rll}\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]& =& {\text{K}}_{\text{w}}\\ \left[{\text{OH}}^{-}\right]& =& \frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}^{\text{+}}\right]}\end{array}$

Substitute the values of ${\text{K}}_{\text{w}}$ and $\left[{\text{H}}^{\text{+}}\right]$ in the above expression.

$\begin{array}{rll}\left[{\text{OH}}^{-}\right]& =& \frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}^{\text{+}}\right]}\\ & =& \frac{{10}^{-14}}{\text{1}\text{.1}\times {10}^{-5}}\\ & =& 9.09\times {10}^{-10}\text{M}\\ & \approx & \text{9}\text{.1}\times {10}^{-10}\text{M}\end{array}$

Therefore, the value of concentration of ${\text{OH}}^{-}$ ions is equal to $\text{9}\text{.1}\times {10}^{-10}\text{M}$.

Conclusion

The value of pOH of $\text{0}\text{.20}\text{M}$ ammonium nitrate solution is equal to $\text{9}\text{.04}$, the value of ${\text{H}}^{\text{+}}$ ions is equal to $\text{1}\text{.1}\times {10}^{-5}\text{M}$ and the value of $\left[{\text{OH}}^{-}\right]$ is equal to $\text{9}\text{.1}\times {10}^{-10}\text{M}$.