Chapter 2, Problem 25RE

MATHEMATICAL Calculate the hydroxide ion concentration, $\left[{\text{OH}}^{-}\right]$, for each of the materials used in Question 24.

Interpretation Introduction

Interpretation:

The hydroxide ion concentration for each material used in Question 24 is to be calculated.

Concept introduction:

$\text{pH}$ is a scale that is based on hydrogen ion concentration and is used to express the acidity or basicity of the solutions.

$\text{pH}$ of a solution is defined as the negative of the log of the hydronium ion concentration.

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

Ion product constant for water is defined as the product of concentration of hydronium and hydroxide ion.

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

Answer to Problem 25RE

Solution:

(a) $3.2\times {10}^{-8}\text{M}$

(b) $\text{7}\text{.9}\times {10}^{-8}\text{M}$

(c) $\text{2}\text{.0}\times {10}^{-10}\text{M}$

(d) $\text{1}\text{.6}\times {10}^{-11}\text{M}$

Explanation of Solution

a) Saliva, $\text{pH}\text{=}6.5$

Calculate the hydrogen concentration as follows:

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

Or

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

Substitute $6.5$ for $\text{pH}$ in the above expression as follows:

$-6.\text{5}=\log \left[{\text{H}}^{+}\right]$

Take antilog on both sides as follows:

${10}^{-6.\text{5}}={10}^{\log \left[{\mathrm{H}}^{+}\right]}$

And

$\left[{\mathrm{H}}^{+}\right]=3.16\times {10}^{-7}\text{ M}$

Calculate the concentration of ${\text{OH}}^{-}$ as follows:

$\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]=K_{\text{w}}=1.0\times {10}^{-14}$

Substitute $3.16\times {\text{10}}^{-7}\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in the above equation as follows:

$3.16\times {\text{10}}^{-7}\text{ M}\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}$

Rearrange the above equation for $\left[{\text{OH}}^{-}\right]$ as follows:

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{3.16\times {\text{10}}^{-7}\text{ M}}\\ =3.2\times {10}^{-8}\text{M}\end{array}$

b) Intracellular fluid of liver, $\text{pH}\text{=}6.9$

Calculate the hydrogen concentration as follows:

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

Or

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

Substitute $6.9$ for $\text{pH}$ in the above expression as follows:

$-6.\text{9}=\log \left[{\text{H}}^{+}\right]$

Take antilog on both sides as follows:

${10}^{-6.\text{9}}={10}^{\log \left[{\mathrm{H}}^{+}\right]}$

And

$\left[{\mathrm{H}}^{+}\right]=1.26\times {10}^{-7}\text{ M}$

Calculate the concentration of ${\text{OH}}^{-}$ as follows:

$\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]=K_{\text{w}}=1.0\times {10}^{-14}$

Substitute $1.26\times {\text{10}}^{-7}\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in the above equation as follows:

$1.26\times {\text{10}}^{-7}\text{ M}\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}$

Rearrange the above equation for $\left[{\text{OH}}^{-}\right]$ as follows:

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{1.26\times {\text{10}}^{-7}\text{ M}}\\ =7.9\times {10}^{-8}\text{M}\end{array}$

c) Tomato juice, $\text{pH}\text{=}4.3$

Calculate the hydrogen concentration as follows:

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

Or

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

Substitute $4.3$ for $\text{pH}$ in the above expression as follows:

$-4.\text{3}=\log \left[{\text{H}}^{+}\right]$

Take antilog on both sides as follows:

${10}^{-4.3}={10}^{\log \left[{\mathrm{H}}^{+}\right]}$

And

$\left[{\mathrm{H}}^{+}\right]=5.0\times {10}^{-5}\text{ M}$

Calculate the concentration of ${\text{OH}}^{-}$ as follows:

$\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]=K_{\text{w}}=1.0\times {10}^{-14}$

Substitute $5.0\times {\text{10}}^{-5}\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in the above equation as follows:

$5.0\times {\text{10}}^{-5}\text{ M}\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}$

Rearrange the above equation for $\left[{\text{OH}}^{-}\right]$ as follows:

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{5.0\times {\text{10}}^{-5}\text{ M}}\\ =2.0\times {10}^{-10}\text{M}\end{array}$

d) Grapefruit juice, $\text{pH}\text{=}3.2$

Calculate the hydrogen concentration as follows:

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

Or

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

Substitute $3.2$ for $\text{pH}$ in the above expression as follows:

$-3.\text{2}=\log \left[{\text{H}}^{+}\right]$

Take antilog on both sides as follows:

${10}^{-3.2}={10}^{\log \left[{\mathrm{H}}^{+}\right]}$

And

$\left[{\mathrm{H}}^{+}\right]=6.3\times {10}^{-4}\text{ M}$

Calculate the concentration of ${\text{OH}}^{-}$ as follows:

$\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]=K_{\text{w}}=1.0\times {10}^{-14}$

Substitute $6.3\times {\text{10}}^{-4}\text{ M}$ for $\left[{\text{H}}^{+}\right]$ in above equation as follows:

$6.3\times {\text{10}}^{-4}\text{ M}\left[{\text{OH}}^{-}\right]=1.0\times {10}^{-14}$

Rearrange the above equation for $\left[{\text{OH}}^{-}\right]$ as follows:

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{6.3\times {\text{10}}^{-4}\text{ M}}\\ =1.6\times {10}^{-11}\text{M}\end{array}$