Chapter 13, Problem 8PP

(a)

Interpretation Introduction

Interpretation:

The concentration of ${\text{H}}_3{\text{O}}^{+}$ ions for $\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-8}\text{ M}$ is to be determined.

Explanation of Solution

Ionic product of water is defined as the product of concentrations of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions present in the water.

Calculation of the concentration of ${\text{H}}_3{\text{O}}^{+}$ ions is as follows:

$K_{\text{w}}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$

Here, $K_{\text{w}}$ is the ionic product of water, $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is the concentration of ${\text{H}}_3{\text{O}}^{+}$ ions, and $\left[{\text{OH}}^{-}\right]$ is the concentration of ${\text{OH}}^{-}$ ions.

Substitute $1.0\times {10}^{-14}$ for $K_{\text{w}}$ and $\text{1}\text{.0}\times {\text{10}}^{-8}\text{ M}$ for $\left[{\text{OH}}^{-}\right]$ in the above equation.

$\begin{array}{c}1.0\times {10}^{-14}=\left[{\text{H}}_3{\text{O}}^{+}\right]\times 1.0\times {10}^{-8}\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{1.0\times {10}^{-14}}{1.0\times {10}^{-8}}\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=1.0\times {10}^{-6}\text{ M}\end{array}$

Therefore, the concentration of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is $1.0\times {10}^{-6}\text{ M}$ . For acidic solutions, $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is more than $1.0\times {10}^{-7}\text{ M}$ . So, the solution is acidic in nature.

(b)

Interpretation Introduction

Interpretation:

The concentration of ${\text{H}}_3{\text{O}}^{+}$ ions for $\left[{\text{OH}}^{-}\right]=0.010\text{ M}$ is to be determined.

Explanation of Solution

Ionic product of water is defined as the product of the concentrations of ${\text{H}}^{+}$ and ${\text{OH}}^{-}$ ions present in the water.

Calculation of the concentration of ${\text{H}}_3{\text{O}}^{+}$ ions is as follows:

$K_{\text{w}}=\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$

Here, $K_{\text{w}}$ is the ionic product of water, $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is the concentration of ${\text{H}}_3{\text{O}}^{+}$ ions, and $\left[{\text{OH}}^{-}\right]$ is the concentration of ${\text{OH}}^{-}$ ions.

Substitute $1.0\times {10}^{-14}$ for $K_{\text{w}}$ and $0.010\text{ M}$ for $\left[{\text{OH}}^{-}\right]$ in the above equation.

$\begin{array}{c}1.0\times {10}^{-14}=\left[{\text{H}}_3{\text{O}}^{+}\right]\times 0.010\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=\frac{1.0\times {10}^{-14}}{0.010}\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=1.0\times {10}^{-12}\text{ M}\end{array}$

Therefore, the concentration of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is $1.0\times {10}^{-12}\text{ M}$ . For basic solutions, $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is less than $1.0\times {10}^{-7}\text{ M}$ . So, the solution is basic in nature.