Chapter 9, Problem 9.36E

Interpretation Introduction

(a)

Interpretation:

The $\text{pH}$ of the water solutions with given characteristics is to be calculated. Also the solution is to be classified as acidic, basic, or neutral.

Concept introduction:

The $\text{pH}$ of a solution is represented as the negative logarithm of the concentration of protons.

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

This scale is defined from $1$ to $14$. It helps to determine whether the given solution is acidic, basic or neutral. The solutions with $\text{pH}<\text{7}$ are acidic in nature, with $\text{pH}=\text{7}$ are neutral in nature and with $\text{pH}>\text{7}$ are basic in nature.

Answer to Problem 9.36E

The $\text{pH}$ of the given water solution is $8.39$. Since the $\text{pH}>\text{7}$, the given solution is basic.

Explanation of Solution

The $\text{pH}$ of the solution is calculated by the formula,

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

The given $\left[{\text{H}}^{+}\right]$ value of the solution is $4.1\times {10}^{-9}\text{mol}/\text{L}$.

Substitute the $\left[{\text{H}}^{+}\right]$ value in the given formula.

$\begin{array}{l}\text{pH}=-\log \left(4.1\times {10}^{-9}\text{mol}/\text{L}\right)\\ \text{pH}=-\left(-8.39\right)\\ \text{pH}=8.39\end{array}$

Thus, the $\text{pH}$ of the given water solution is $8.39$. Since the $\text{pH}>\text{7}$, the given solution is basic.

Conclusion

The $\text{pH}$ of the given water solution is $8.39$. Since the $\text{pH}>\text{7}$, the given solution is basic.

Interpretation Introduction

(b)

Interpretation:

The $\text{pH}$ of the water solutions with given characteristics is to be calculated. Also the solution is to be classified as acidic, basic, or neutral.

Concept introduction:

The $\text{pH}$ of a solution is represented as the negative logarithm of the concentration of protons.

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

This scale is defined from $1$ to $14$. It helps to determine whether the given solution is acidic, basic or neutral. The solutions with $\text{pH}<\text{7}$ are acidic in nature, with $\text{pH}=\text{7}$ are neutral in nature and with $\text{pH}>\text{7}$ are basic in nature.

Answer to Problem 9.36E

The $\text{pH}$ of the given water solution is $10.97$. Since the $\text{pH}>\text{7}$, the given solution is basic.

Explanation of Solution

The given $\left[{\text{OH}}^{-}\right]$ value of the solution is $9.4\times {10}^{-4}\text{mol}/\text{L}$.

The concentration of ${\text{H}}^{+}$ is calculated by,

$K_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$

Substitute the value of ionization constant and $\left[{\text{OH}}^{-}\right]$ value in the given formula.

$\begin{array}{rll}1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2& =& \left[{\text{H}}^{+}\right]9.4\times {10}^{-4}\text{mol}/\text{L}\\ \left[{\text{H}}^{+}\right]& =& \frac{1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2}{9.4\times {10}^{-4}\text{mol}/\text{L}}\\ \left[{\text{H}}^{+}\right]& =& 1.06\times {10}^{-11}\text{mol}/\text{L}\end{array}$

Thus, the $\left[{\text{H}}^{+}\right]$ value for the given solution is $1.06\times {10}^{-11}\text{mol}/\text{L}$.

The $\text{pH}$ of the solution is calculated by the formula,

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

Substitute the $\left[{\text{H}}^{+}\right]$ value in the given formula.

$\begin{array}{l}\text{pH}=-\log \left(1.06\times {10}^{-11}\text{mol}/\text{L}\right)\\ \text{pH}=-\left(-10.97\right)\\ \text{pH}=10.97\end{array}$

Thus, the $\text{pH}$ of the given water solution is $10.97$. Since the $\text{pH}>\text{7}$, the given solution is basic.

Conclusion

The $\text{pH}$ of the given water solution is $10.97$. Since the $\text{pH}>\text{7}$, the given solution is basic.

Interpretation Introduction

(c)

Interpretation:

The $\text{pH}$ of the water solutions with given characteristics is to be calculated. Also the solution is to be classified as acidic, basic, or neutral.

Concept introduction:

The $\text{pH}$ of a solution is represented as the negative logarithm of the concentration of protons.

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

This scale is defined from $1$ to $14$. It helps to determine whether the given solution is acidic, basic or neutral. The solutions with $\text{pH}<\text{7}$ are acidic in nature, with $\text{pH}=\text{7}$ are neutral in nature and with $\text{pH}>\text{7}$ are basic in nature.

Answer to Problem 9.36E

The $\text{pH}$ of the given water solution is $7.5$. Since the $\text{pH}>\text{7}$, the given solution is basic.

Explanation of Solution

It is given that $\left[{\text{OH}}^{-}\right]=10.\left[{\text{H}}^{+}\right]$. From ionic product of water,

$K_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}$. Substitute the value of $\left[{\text{OH}}^{-}\right]$ in terms of $\left[{\text{H}}^{+}\right]$ in the given formula of ionic product of water.

$\begin{array}{l}{K}_{\text{w}}=10.\left[{\text{H}}^{+}\right]\left[{\text{H}}^{+}\right]=1.0\times {10}^{-14}\\ 10.{\left[{\text{H}}^{+}\right]}^2=1.0\times {10}^{-14}\\ {\left[{\text{H}}^{+}\right]}^2=1.0\times {10}^{-15}\\ \left[{\text{H}}^{+}\right]=3.16\times {10}^{-8}\end{array}$

The $\left[{\text{H}}^{+}\right]$ value is $3.16\times {10}^{-8}$.

Substitute the $\left[{\text{H}}^{+}\right]$ value in the given formula.

$\begin{array}{l}\text{pH}=-\log \left(3.16\times {10}^{-8}\text{mol}/\text{L}\right)\\ \text{pH}=-\left(-7.5\right)\\ \text{pH}=7.5\end{array}$

Thus, the $\text{pH}$ of the given water solution is $7.5$. Since the $\text{pH}>\text{7}$, the given solution is basic.

Conclusion

The $\text{pH}$ of the given water solution is $7.5$. Since the $\text{pH}>\text{7}$, the given solution is basic.

Interpretation Introduction

(d)

Interpretation:

The $\text{pH}$ of the water solutions with given characteristics is to be calculated. Also the solution is to be classified as acidic, basic, or neutral.

Concept introduction:

The $\text{pH}$ of a solution is represented as the negative logarithm of the concentration of protons.

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

This scale is defined from $1$ to $14$. It helps to determine whether the given solution is acidic, basic or neutral. The solutions with $\text{pH}<\text{7}$ are acidic in nature, with $\text{pH}=\text{7}$ are neutral in nature and with $\text{pH}>\text{7}$ are basic in nature.

Answer to Problem 9.36E

The $\text{pH}$ of the given water solution is $1.64$. Since the $\text{pH}<\text{7}$, the given solution is acidic.

Explanation of Solution

The $\text{pH}$ of the solution is calculated by the formula,

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

The given $\left[{\text{H}}^{+}\right]$ value of the solution is $2.3\times {10}^{-2}\text{mol}/\text{L}$.

Substitute the $\left[{\text{H}}^{+}\right]$ value in the given formula.

$\begin{array}{l}\text{pH}=-\log \left(2.3\times {10}^{-2}\text{mol}/\text{L}\right)\\ \text{pH}=-\left(-1.64\right)\\ \text{pH}=1.64\end{array}$

Thus, the $\text{pH}$ of the given water solution is $1.64$. Since the $\text{pH}<\text{7}$, the given solution is acidic.

Conclusion

The $\text{pH}$ of the given water solution is $1.64$. Since the $\text{pH}<\text{7}$, the given solution is acidic.

Interpretation Introduction

(e)

Interpretation:

The $\text{pH}$ of the water solutions with given characteristics is to be calculated. Also the solution is to be classified as acidic, basic, or neutral.

Concept introduction:

The $\text{pH}$ of a solution is represented as the negative logarithm of the concentration of protons.

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

This scale is defined from $1$ to $14$. It helps to determine whether the given solution is acidic, basic or neutral. The solutions with $\text{pH}<\text{7}$ are acidic in nature, with $\text{pH}=\text{7}$ are neutral in nature and with $\text{pH}>\text{7}$ are basic in nature.

Answer to Problem 9.36E

The $\text{pH}$ of the given water solution is $4.7$. Since the $\text{pH}<\text{7}$, the given solution is acidic.

Explanation of Solution

The given $\left[{\text{OH}}^{-}\right]$ value of the solution is $5.1\times {10}^{-10}\text{mol}/\text{L}$.

The concentration of ${\text{H}}^{+}$ is calculated by,

$K_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$

Substitute the value of ionization constant and $\left[{\text{OH}}^{-}\right]$ value in the given formula.

$\begin{array}{rll}1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2& =& \left[{\text{H}}^{+}\right]5.1\times {10}^{-10}\text{mol}/\text{L}\\ \left[{\text{H}}^{+}\right]& =& \frac{1.0\times {10}^{-14}{\left(\text{mol}/\text{L}\right)}^2}{5.1\times {10}^{-10}\text{mol}/\text{L}}\\ \left[{\text{H}}^{+}\right]& =& 1.96\times {10}^{-5}\text{mol}/\text{L}\end{array}$

Thus, the $\left[{\text{H}}^{+}\right]$ value for the given solution is $1.96\times {10}^{-5}\text{mol}/\text{L}$.

The $\text{pH}$ of the solution is calculated by the formula,

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

Substitute the $\left[{\text{H}}^{+}\right]$ value in the given formula.

$\begin{array}{l}\text{pH}=-\log \left(1.96\times {10}^{-5}\text{mol}/\text{L}\right)\\ \text{pH}=-\left(-4.7\right)\\ \text{pH}=4.7\end{array}$

Thus, the $\text{pH}$ of the given water solution is $4.7$. Since the $\text{pH}<\text{7}$, the given solution is acidic.

Conclusion

The $\text{pH}$ of the given water solution is $4.7$. Since the $\text{pH}<\text{7}$, the given solution is acidic.