Chapter 19, Problem 108A

Interpretation Introduction

Interpretation : The expression of $\text{pH}\text{+}\text{pOH}\text{=}\text{14}$ has to be demonstrated.

Concept Introduction : The negative logarithm of the concentration of hydrogen ions is the pH of a solution. pOH stands for the potential of hydroxide ions. It measures the acidity or basicity of any solution.

Answer to Problem 108A

The expression of $\text{pH}\text{+}\text{pOH}\text{=}\text{14}$ has been demonstrated.

Explanation of Solution

The ion dissociation constant is the product of the concentration of hydrogen ions and the concentration of hydroxide ions.

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

The concentration of ${\text{H}}^{\text{+}}$ ion is $1\times {10}^{-7}\text{M}$ .

The concentration of ${\text{OH}}^{-}$ ion is $1\times {10}^{-7}\text{M}$ .

To calculate the ion dissociation constant, substitute the values in above formula:

  $\begin{array}{c}{\text{K}}_{\text{w}}=\left[{\text{H}}^{\text{+}}\right]\left[{\text{OH}}^{-}\right]\\ =\left(1\times {10}^{-7}\text{M}\right)\left(1\times {10}^{-7}\text{M}\right)\\ =1\times {10}^{-14}{\text{M}}^2\end{array}$

The formula of pH is given as

  $pH=-log\left[H^{+}\right]$

To calculate the pH of the solution, substitute the values in above formula:

  $\begin{array}{c}pH=-log\left[H^{+}\right]\\ =-\log \left(1\times {10}^{-7}\text{M}\right)\\ =7\end{array}$

The formula of pOH is given as

  $pOH=-log\left[O{H}^{-}\right]$

To calculate the pH of the solution, substitute the values in above formula:

  $\begin{array}{c}pOH=-log\left[O{H}^{-}\right]\\ =-\log \left(1\times {10}^{-7}\text{M}\right)\\ =7\end{array}$

The sum of pH and pOH is given as $\text{pH}\text{+}\text{pOH}\text{=}\text{14}$ .