Chapter 14, Problem 58E

Interpretation Introduction

(a)

Interpretation:

The $\text{pH}$ of the given ${\text{OH}}^{-}$ ion with concentration $\text{0}\text{.000 031 }M$ is to be calculated.

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 relationship between the $\text{pH}$ value and the concentration of the hydrogen ion. Higher the hydrogen ions concentration is lesser the value of $\text{pH}$. There is an inverse relationship 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 58E

The value of $\text{pH}$ with ${\text{OH}}^{-}$ ion concentration $\text{0}\text{.000 031 }M$ is $9.49$.

Explanation of Solution

The relationship between $\text{pOH}$ and hydroxide ion concentration is equal to $10$ raised to the power negative value of $\text{pOH}$. It is represented as shown below.

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

The above equation (1) is rearranged as shown below.

$\text{pOH=}-\text{log}\left[{\text{OH}}^{-}\right]$ …(2)

The value of $\left[{\text{OH}}^{-}\right]$ ion is $\text{0}\text{.000 031 }M$.

Substitute the value of $\left[{\text{OH}}^{-}\right]$ ion in equation (2) as shown below.

$\begin{array}{rll}\text{pOH}& =& -\text{log}\left[{\text{OH}}^{-}\right]\\ & =& -\text{log}\left(\text{0}\text{.000 031 }M\right)\\ & =& -\text{log}\left(3.1\times {10}^{-5}\right)\end{array}$

Therefore,

$\begin{array}{rll}\text{pOH}& =& -\left[\text{log}\left({10}^{-5}\right)+\text{log}\left(3.1\right)\right]\\ & =& 5-0.49\\ & =& 4.51\end{array}$

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

$\text{pH}=\text{14}-\text{pOH}$ …(3)

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

$\begin{array}{rll}\text{pH}& =& \text{14}-\text{pOH}\\ & =& 14-4.51\\ & =& 9.49\end{array}$

Therefore, the value of $\text{pH}$ with ${\text{OH}}^{-}$ ion concentration $\text{0}\text{.000 031 }M$ is $9.49$.

Conclusion

The value of $\text{pH}$ with ${\text{OH}}^{-}$ ion concentration $\text{0}\text{.000 031 }M$ is calculated as $9.49$.

Interpretation Introduction

(b)

Interpretation:

The $\text{pH}$ of the given ${\text{OH}}^{-}$ ion with concentration $\text{0}\text{.000 000 000 66 }M$ is to be calculated.

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 relationship 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 relationship 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 58E

The value of $\text{pH}$ with ${\text{OH}}^{-}$ ion concentration $\text{0}\text{.000 000 000 66 }M$ is $4.82$.

Explanation of Solution

The relationship between $\text{pH}$ and hydrogen ion concentration is equal to $10$ raise to the power negative value of $\text{pH}$. It is represented as shown below.

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

The above equation (1) is rearranged as shown below.

$\text{pOH=}-\text{log}\left[{\text{OH}}^{-}\right]$ …(2)

The value of $\left[{\text{OH}}^{-}\right]$ ion is $\text{0}\text{.000 000 000 66 }M$.

Substitute the value of $\left[{\text{OH}}^{-}\right]$ ion in equation (2) as shown below.

$\begin{array}{rll}\text{pOH}& =& -\text{log}\left[{\text{OH}}^{-}\right]\\ & =& -\text{log}\left(\text{0}\text{.000 000 000 66 }M\right)\\ & =& -\text{log}\left(6.6\times {10}^{-10}\right)\end{array}$

Therefore,

$\begin{array}{rll}\text{pOH}& =& -\left[\text{log}\left({10}^{-10}\right)+\text{log}\left(6.6\right)\right]\\ & =& 10-0.82\\ & =& 9.18\end{array}$

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

$\text{pH}=\text{14}-\text{pOH}$ …(3)

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

$\begin{array}{rll}\text{pH}& =& \text{14}-\text{pOH}\\ & =& 14-9.18\\ & =& 4.82\end{array}$

Hence, the value of $\text{pH}$ with ${\text{OH}}^{-}$ ion concentration $\text{0}\text{.000 000 000 66 }M$ is $4.82$.

Conclusion

The value of $\text{pH}$ with ${\text{OH}}^{-}$ ion concentration $\text{0}\text{.000 000 000 66 }M$ is calculated as $4.82$