Chapter U4, Problem 10RE

Interpretation Introduction

Interpretation: pH and ${\left[\text{OH}\right]}^{-}$ corresponding to ${\left[\text{H}\right]}^{\text{+}}\text{= 2}{\text{.0×10}}^{\text{-8}}\text{ M}$ is to be calculated.

Concept Introduction :

• pH is a logarithmic scale which specify the acidity or basicity of the solution.

  $pH=-\log [H^{+}]$

• Neutral solution- A solution is neutral if $pH=7$
• Acidic solution- A solution is acidic if $pH<7$
• Basic solution- A solution is basic if $pH>7$

Answer to Problem 10RE

pH and ${\left[\text{OH}\right]}^{-}$ corresponding to ${\left[\text{H}\right]}^{\text{+}}\text{= 2}{\text{.0×10}}^{\text{-8}}\text{ M}$ is 7.7 and $\text{5}\text{.01}\times {\text{10}}^{-7}\text{ M}$ respectively.

Explanation of Solution

As ${\left[\text{H}\right]}^{\text{+}}\text{= 2}{\text{.0×10}}^{\text{-8}}\text{ M}$ , so, the pH of the solution is determined as follows:

  $\begin{array}{l}{\text{pH = -log[H}}^{\text{+}}\text{]}\\ \text{pH = -log[2}{\text{.0×10}}^{\text{-8}}\text{]}\\ \text{pH = 7}\text{.699 }\\ \text{pH }\simeq \text{ 7}\text{.7}\end{array}$

The ${\left[\text{OH}\right]}^{-}$ is calculated as:

  $\text{pH }+\text{ pOH = 14}$

Substituting the values of pH in the equation:

  $\begin{array}{l}\text{7}\text{.7 }+\text{ pOH = 14}\\ \text{pOH = 14 - 7}\text{.7}\\ \text{pOH = 6}\text{.3}\end{array}$

Using relation:

  ${\text{pOH = -log[OH}}^{-}\text{]}$

Substituting the values:

  $\begin{array}{l}{\text{pOH = -log[OH}}^{-}\text{]}\\ \text{6}{\text{.3 = -log[OH}}^{-}\text{]}\\ {\text{[OH}}^{-}\text{] = }{10}^{-6.3}\\ {\text{[OH}}^{-}\text{] = 5}\text{.01}\times {\text{10}}^{-7}\text{ M}\end{array}$

Hence, the pH and ${\left[\text{OH}\right]}^{-}$ corresponding to ${\left[\text{H}\right]}^{\text{+}}\text{= 2}{\text{.0×10}}^{\text{-8}}\text{ M}$ is 7.7 and $\text{5}\text{.01}\times {\text{10}}^{-7}\text{ M}$ respectively.