Chapter 17, Problem 9PE

Interpretation Introduction

Interpretation:

For a solution with $\text{pOH}=6$, the values of $\text{pH}$, $\left[{\text{H}}^{\text{+}}\right]$, and $\left[{\text{OH}}^{-}\right]$ are to be calculated.

The correctness of the answer is to be verified.

Concept introduction:

The $\text{pH}$ scale tells about the strength of an acid and a base. It measures the concentration of the hydrogen ions. There is an inverse relation between the $\text{pH}$ value and the concentration of the hydrogen ion. Higher the concentration of hydrogen ions, lesser the value of $\text{pH}$. There is an inverse relation between the $\text{pOH}$ value and the concentration of the hydroxide ion. Higher the concentration of hydroxide ions, lesser the value of $\text{pOH}$.

Answer to Problem 9PE

The value of $\text{pH}$ is $8$.

The value of $\left[{\text{OH}}^{-}\right]$ is ${10}^{-6}\text{M}$.

The value of $\left[{\text{H}}^{\text{+}}\right]$ is ${10}^{-8}\text{M}$.

Explanation of Solution

The $\text{pOH}$ of the solution is $6$.

The relationship between $\text{pOH}$ and $\left[{\text{OH}}^{-}\right]$ is shown below.

$\left[{\text{OH}}^{-}\right]={10}^{-\text{pOH}}\text{M}$…(1)

Substitute the value of $\text{pOH}$ in equation (1).

$\left[{\text{OH}}^{-}\right]={10}^{-\text{6}}\text{M}$

From the above calculation, it is predicted that the value of $\left[{\text{OH}}^{-}\right]$ is ${10}^{-\text{6}}\text{M}$.

The equilibrium reaction for water is shown below.

${\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}^{+}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)$

For the above reaction, the ionization constant for water is shown below.

${\text{K}}_{\text{W}}=\frac{\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_{\text{2}}\text{O}\right]}$

Since, the concentration of ${\text{H}}_{\text{2}}\text{O}$ is very high, so it is ignored and the ionization constant for water is shown below.

${\text{K}}_{\text{W}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$…(2)

Where,

• $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ion.

• $\left[{\text{OH}}^{-}\right]$ is the concentration of hydroxide ion.

The value of ${\text{K}}_{\text{W}}$ is ${10}^{-14}\text{M}$.

Substitute the values of ${\text{K}}_{\text{W}}$ and $\left[{\text{OH}}^{-}\right]$ in equation (2).

$\begin{array}{rll}{\text{10}}^{-14}& =& \left[{\text{H}}^{+}\right]\times {10}^{-6}\\ \left[{\text{H}}^{+}\right]& =& \frac{{\text{10}}^{-14}\text{M}}{{10}^{-6}\text{M}}\\ & =& {10}^{-8}\text{M}\end{array}$

The relationship between $\text{pH}$ and $\text{pOH}$ is shown below.

$\text{pH}+\text{pOH}=14$…(3)

Substitute the value of $\text{pOH}$ in equation (3).

$\text{pH}+\text{6}=14$

So, the value of $\text{pH}$ is calculated as shown below.

$\begin{array}{rll}\text{pH}& =& 14-6\\ & =& 8\end{array}$

The answer can be verified as shown below.

The $\text{pH}$ is calculated as $8$.

Substitute the value of $\text{pH}$ in equation (3).

$\begin{array}{rll}\text{8}+\text{pOH}& =& 14\\ \text{pOH}& =& 14-8\\ & =& 6\end{array}$

The $\left[{\text{H}}^{+}\right]$ is calculated as ${10}^{-8}\text{M}$.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in equation (2).

$\begin{array}{rll}{\text{10}}^{-14}& =& {10}^{-8}\times \left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]& =& \frac{{\text{10}}^{-14}\text{M}}{{10}^{-8}\text{M}}\\ & =& {10}^{-6}\text{M}\end{array}$

The calculated value of $\left[{\text{OH}}^{-}\right]$ is ${10}^{-\text{6}}\text{M}$.

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in equation (1).

$\left[{10}^{-\text{6}}\right]={10}^{-\text{pOH}}$

From the above calculation, it is predicted that the value of $\text{pOH}$ is $6$.

Therefore, all the calculated answers are verified.

Conclusion

The values of $\text{pH}$, $\left[{\text{OH}}^{-}\right]$, and $\left[{\text{H}}^{\text{+}}\right]$ are $8,{10}^{-6}\text{M},$ and ${\text{10}}^{-8}\text{M}$ respectively.