Chapter 16.4, Problem 1PPB

Practice ProblemBUILD

Determine the pH of a solution at $\text{25°C}$ in which the hydroxide ion concentration is $\text{(a) }8.3\times {10}^{-15}M\text{, (b) }3.3\times {10}^{-4}M,\text{ and (c) 1}\text{.2}\times {\text{10}}^{-3}M$.

Interpretation Introduction

Interpretation:

The pH of the solution for each of the given concentrations of hydroxide ions is to be determined.

Concept introduction:

pH is a measure of the acidity of a solution, which depends on the concentration of hydronium ions and the temperature of the solution.

Use the equilibrium expression $K_w$ as follows:

$K_w={\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}=1.0\times {10}^{-14}$

Rearrange the above equation for hydronium ions as:

${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\frac{1.0\times {10}^{-14}}{{\text{[OH}}^{\text{-}}\text{]}}$ ……(1)

The formula to calculate the pH of a solution is:

$\text{pH}=-\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=-\text{log}\left[{\text{H}}^{+}\right]$ as $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=\left[{\text{H}}^{+}\right]$ …… (2)

Answer to Problem 1PPB

Solution:

a) $-0.08$

b) $10.52$

c) $11.08$

Explanation of Solution

$8.3\times {10}^{-15}M$

Use the equilibrium expression $K_w$ as follows:

$K_w={\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}=1.0\times {10}^{-14}$

Rearrange the above equation for hydronium ions as:

${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\frac{1.0\times {10}^{-14}}{{\text{[OH}}^{\text{-}}\text{]}}$

Use equation (1) to calculate the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in the solution at $25{}^{\circ }\text{C}$ and substitute the value of ${\text{OH}}^{-}$ as $8.3\times {10}^{-15}M$ to get:

$\begin{array}{c}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\frac{1.0\times {10}^{-14}}{8.3\times {10}^{-15}}\\ =1.2\end{array}$

Now, substitute the value of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ as $1.2$ in equation (2) as:

$\text{pH}=-\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=-\text{log}\left[{\text{H}}^{+}\right]$

$\begin{array}{c}\text{pH}=-\text{log}\left[1.2\right]\\ =-\left(0.08\right)\\ =-0.08\end{array}$

$3.3\times {10}^{-4}M$

Use the equilibrium expression $K_w$ as follows:

$K_w={\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}=1.0\times {10}^{-14}$

Rearrange the above equation for hydronium ions as:

${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\frac{1.0\times {10}^{-14}}{{\text{[OH}}^{\text{-}}\text{]}}$

Use equation (1) to calculate the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in the solution at ${25}^{\circ }\text{C}$ and substitute the value of ${\text{OH}}^{-}$ as $3.3\times {10}^{-4}M$ to get:

$\begin{array}{c}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\frac{1.0\times {10}^{-14}}{3.3\times {10}^{-4}}\\ =3.0\times {10}^{-11}\end{array}$

Now, substitute the value of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ as $3.0\times {10}^{-11}$ in equation (2) as:

$\text{pH}=-\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=-\text{log}\left[{\text{H}}^{+}\right]$

$\begin{array}{c}\text{pH}=-\text{log}\left[3.0\times {10}^{-11}\right]\\ =-\left(-10.52\right)\\ =10.52\end{array}$

The pH of the solution is $10.52$.

$1.2\times {10}^{-3}M$

Use the equilibrium expression $K_w$ as follows:

$K_w={\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][OH}}^{\text{-}}\text{]}=1.0\times {10}^{-14}$

Rearrange the above equation for hydronium ions as:

${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\frac{1.0\times {10}^{-14}}{{\text{[OH}}^{\text{-}}\text{]}}$

Use equation (1) to calculate the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in the solution at ${25}^{\circ }\text{C}$ and substitute the value of ${\text{OH}}^{-}$ as $1.2\times {10}^{-3}M$ to get:

$\begin{array}{c}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}=\frac{1.0\times {10}^{-14}}{1.2\times {10}^{-3}}\\ =8.3\times {10}^{-12}\end{array}$

Now, substitute the value of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ as $8.3\times {10}^{-12}$ in equation (2) as:

$\text{pH}=-\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=-\text{log}\left[{\text{H}}^{+}\right]$

$\begin{array}{c}\text{pH}=-\text{log}\left[8.3\times {10}^{-12}\right]\\ =-\left(-11.08\right)\\ =11.08\end{array}$

The pH of the solution is $11.08$.