Chapter 15, Problem 15.33QP

(a)

Interpretation Introduction

Interpretation: The concentration of the given ions in the described solution is to be calculated.

Concept introduction: The measure of concentration of ions in a solution gives the ionic strength of the solution. Concentration of ions in a solution depends on the mole ratio between the cation and anions and the dissolved substance.

To determine: The concentration of the given ions in the described solution.

Answer to Problem 15.33QP

Solution

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in the solution is $\underset{\_}{1.190\times 10^{-11}M}$.

Explanation of Solution

Explanation

Given

The concentration of $\text{NaOH}$ is $8.4\times {10}^{-4}\text{M}$.

The auto-dissociation of pure water is shown by the reaction,

${\text{2H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}+{\text{OH}}^{-}$

Concentrations of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ and $\left[{\text{OH}}^{-}\right]$ in the solution are related according to the following relation,

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

The value of $\text{pH}$ is calculated by using the expression,

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

The dissociation of $\text{NaOH}$ in aqueous solution is expressed by the reaction,

$\text{NaOH}\left(aq\right)\stackrel{}{\to }{\text{Na}}^{\text{+}}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

Sodium hydroxide $\left(\text{NaOH}\right)$, is a strong base as it dissociates completely in aqueous solution. Hence, the concentration of ${\text{OH}}^{-}$ ion in the solution is exactly same as the concentration of $\text{NaOH}$.

The concentration of ${\text{OH}}^{-}$ ion in the solution is $8.4\times {10}^{-4}\text{M}$.

The concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ in the solution is calculated by using the formula,

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=\frac{K_w}{\left[{\text{OH}}^{-}\right]}$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above formula to calculate the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$.

$\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=\frac{{10}^{-14}}{8.4\times {10}^{-4}}\text{M}\\ =\underset{\_}{1.190\times 10^{-11}M}\end{array}$

Hence, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $\underset{\_}{1.190\times 10^{-11}M}$.

(b)

Interpretation Introduction

To determine: The concentration of the given ions in the described solution.

Answer to Problem 15.33QP

Solution

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in the solution is $\underset{\_}{7.6\times 10^{-11}M}$.

Explanation of Solution

Explanation

Given

The concentration of $\text{Ca}{\left(\text{OH}\right)}_2$ is $6.6\times {10}^{-5}\text{M}$.

The dissociation of $\text{Ca}{\left(\text{OH}\right)}_2$ in aqueous solution is expressed by the reaction,

$\text{Ca}{\left(\text{OH}\right)}_2\left(aq\right)\stackrel{}{\to }{\text{Ca}}^{\text{+}}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$

Calcium hydroxide $\left(\text{Ca}{\left(\text{OH}\right)}_2\right)$, is a strong base as it dissociates completely in aqueous solution. Hence, the concentration of ${\text{OH}}^{-}$ ion in the solution is exactly double as the concentration of $\text{Ca}{\left(\text{OH}\right)}_2$ as it gives two ${\text{OH}}^{-}$ ions on dissociation.

The concentration of ${\text{OH}}^{-}$ ion in the solution is,

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=2\times 6.6\times {10}^{-5}\text{M}\\ =1.32\times {10}^{-4}\text{M}\end{array}$

The concentration of ${\text{OH}}^{-}$ ion in the solution is $1.32\times {10}^{-4}\text{M}$.

The concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ in the solution is calculated by using the formula,

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=\frac{K_w}{\left[{\text{OH}}^{-}\right]}$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above formula to calculate the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$.

$\begin{array}{c}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=\frac{{10}^{-14}}{1.32\times {10}^{-4}}\text{M}\\ =\underset{\_}{7.6\times 10^{-11}M}\end{array}$

Hence, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ is $\underset{\_}{7.6\times 10^{-11}M}$.

(c)

Interpretation Introduction

To determine: The concentration of the given ions in the described solution.

Answer to Problem 15.33QP

Solution

The concentration of ${\text{OH}}^{-}$ ions in the solution is $\underset{\_}{2.2\times 10^{-12}M}$.

Explanation of Solution

Explanation

Given

The concentration of $\text{HCl}$ is $4.5\times {10}^{-3}\text{M}$.

The dissociation of $\text{HCl}$ in aqueous solution is expressed by the reaction,

$\text{HCl}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$

Hydrochloric acid $\left(\text{HCl}\right)$, is a strong acid as it dissociates completely in aqueous solution. Hence, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion in the solution is exactly same as the concentration of $\text{HCl}$.

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion in the solution is $4.5\times {10}^{-3}\text{M}$.

The concentration of ${\text{OH}}^{-}$ ions in the solution is calculated by using the formula,

$\left[{\text{OH}}^{-}\right]=\frac{K_w}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}$

Substitute the value of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ in the above formula to calculate the concentration of ${\text{OH}}^{-}$.

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{{10}^{-14}}{4.5\times {10}^{-3}}\text{M}\\ =\underset{\_}{2.2\times 10^{-12}M}\end{array}$

Hence, the concentration of ${\text{OH}}^{-}$ is $\underset{\_}{2.2\times 10^{-12}M}$.

(d)

Interpretation Introduction

To determine: The concentration of the given ions in the described solution.

Answer to Problem 15.33QP

Solution

The concentration of ${\text{OH}}^{-}$ ions in the solution is $\underset{\_}{3.4\times 10^{-10}M}$.

Explanation of Solution

Explanation

Given

The concentration of $\text{HCl}$ is $2.9\times {10}^{-5}\text{M}$.

The dissociation of $\text{HCl}$ in aqueous solution is expressed by the reaction,

$\text{HCl}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$

Hydrochloric acid $\left(\text{HCl}\right)$, is a strong acid as it dissociates completely in aqueous solution. Hence, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion in the solution is exactly same as the concentration of $\text{HCl}$.

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ion in the solution is $2.9\times {10}^{-5}\text{M}$.

The concentration of ${\text{OH}}^{-}$ ions in the solution is calculated by using the formula,

$\left[{\text{OH}}^{-}\right]=\frac{K_w}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}$

Substitute the value of $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ in the above formula to calculate the concentration of ${\text{OH}}^{-}$.

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{{10}^{-14}}{2.9\times {10}^{-5}}\text{M}\\ =\underset{\_}{3.4\times 10^{-10}M}\end{array}$

Hence, the concentration of ${\text{OH}}^{-}$ is $\underset{\_}{3.4\times 10^{-10}M}$.

Conclusion

1. a. The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in the solution is $\underset{\_}{1.190\times 10^{-11}M}$.
2. b. The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ ions in the solution is $\underset{\_}{7.6\times 10^{-11}M}$.
3. c. The concentration of ${\text{OH}}^{-}$ ions in the solution is $\underset{\_}{2.2\times 10^{-12}M}$.
4. d. The concentration of ${\text{OH}}^{-}$ ions in the solution is $\underset{\_}{3.4\times 10^{-10}M}$