Chapter 16, Problem 16.33QP

(a)

Interpretation Introduction

To determine: The hydrogen ion concentration of the given solution.

Interpretation: The hydrogen ion concentration of the given solution is to be calculated.

Concept introduction: The hydroxide and hydrogen ion concentration is calculated by the formula,

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

Answer to Problem 16.33QP

The hydrogen ion concentration of the given solution is $\underset{\_}{1.187\times 10^{-11}M}$.

Explanation of Solution

Given

The concentration of $\text{NaOH}$ solution is $8.42\times {10}^{-4}\text{M}$.

Sodium hydroxide is a strong base. It is completely dissociates in water.

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

In the above reaction, $\text{NaOH}$ dissociates to give one mole of ${\text{Na}}^{+}$ and one mole of ${\text{OH}}^{-}$ ions in the solution.

So, the concentration of ${\text{OH}}^{-}$ in the solution is $8.42\times {10}^{-4}\text{M}$.

The concentration of ${\text{H}}^{+}$ in the given solution is calculated by the formula,

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

Substitute the concentration of ${\text{OH}}^{-}$ in the above expression.

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\left(\frac{1.0\times {10}^{-14}}{8.42\times {10}^{-4}}\right)\text{M}\\ =\underset{\_}{1.187\times 10^{-11}M}\end{array}$

Therefore, hydrogen ion concentration of the given solution is $\underset{\_}{1.187\times 10^{-11}M}$.

(b)

Interpretation Introduction

To determine: The hydrogen ion concentration of the given solution.

Interpretation: The hydrogen ion concentration of the given solution is to be calculated.

Concept introduction: The hydroxide and hydrogen ion concentration is calculated by the formula,

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

Answer to Problem 16.33QP

The hydrogen ion concentration of the given solution is $\underset{\_}{1.25\times 10^{-10}M}$.

Explanation of Solution

Given

The concentration of $\text{Ca}{\left(\text{OH}\right)}_2$ solution is $3.97\times {10}^{-5}\text{M}$.

Calcium hydroxide is a strong base. It is completely dissociates in the water.

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

In the above reaction, $\text{Ca}{\left(\text{OH}\right)}_2$ dissociates to give one mole of ${\text{Ca}}^{2+}$ and two moles of ${\text{OH}}^{-}$ ions in the solution.

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

$\left[{\text{OH}}^{-}\right]=2\times \left[\text{Ca}{\left(\text{OH}\right)}_2\right]$

Substitute the concentration of $\text{Ca}{\left(\text{OH}\right)}_2$ solution in the above expression.

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=2\times \left(3.97\times {10}^{-5}\text{M}\right)\\ =7.94\times {10}^{-5}\text{M}\end{array}$

The concentration of ${\text{OH}}^{-}$ in the solution is $7.94\times {10}^{-5}\text{M}$.

The concentration of ${\text{H}}^{+}$ in the given solution is calculated by the formula,

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

Substitute the concentration of ${\text{OH}}^{-}$ in the above expression.

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\left(\frac{1.0\times {10}^{-14}}{7.94\times {10}^{-5}}\right)\text{M}\\ =\underset{\_}{1.25\times 10^{-10}M}\end{array}$

Therefore, the hydrogen ion concentration of the given solution is $\underset{\_}{1.25\times 10^{-10}M}$.

(c)

Interpretation Introduction

Interpretation: The hydroxide ion concentration of the given solution is to be calculated.

Concept introduction: The hydroxide and hydrogen ion concentration is calculated by the formula,

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

Answer to Problem 16.33QP

The hydroxide ion concentration of the given solution is $\underset{\_}{2.22\times 10^{-12}M}$.

Explanation of Solution

Given

The concentration of $\text{HCl}$ solution is $4.51\times {10}^{-3}\text{M}$.

Hydrochloric acid is a strong acid. It is completely dissociates in the water.

$\text{HCl}\left(aq\right)\to {\text{H}}^{+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$

In the above reaction, $\text{HCl}$ dissociates to give one mole of ${\text{H}}^{+}$ and one mole of ${\text{Cl}}^{-}$ ions in the solution.

So, the hydrogen ion concentration of the given solution is $4.51\times {10}^{-3}\text{M}$.

The hydroxide ion concentration of the given solution is calculated by the formula,

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

Substitute the hydrogen ion concentration of the given solution in the above expression.

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\left(\frac{1.0\times {10}^{-14}}{4.51\times {10}^{-3}}\right)\text{M}\\ =\underset{\_}{2.22\times 10^{-12}M}\end{array}$

Therefore, the hydroxide ion concentration of the given solution is $\underset{\_}{2.22\times 10^{-12}M}$.

(d)

Interpretation Introduction

To determine: The hydroxide ion concentration of the given solution.

Interpretation: The hydroxide ion concentration of the given solution is to be calculated.

Concept introduction: The hydroxide and hydrogen ion concentration is calculated by the formula,

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

Answer to Problem 16.33QP

The hydroxide ion concentration of the given solution is $\underset{\_}{1.44\times 10^{-6}M}$.

Explanation of Solution

Given

The concentration of $\text{HCl}$ solution is $6.92\times {10}^{-5}\text{M}$.

Hydrochloric acid is a strong acid. It is completely dissociates in the water.

$\text{HCl}\left(aq\right)\to {\text{H}}^{+}\left(aq\right)+{\text{Cl}}^{-}\left(aq\right)$

In the above reaction, $\text{HCl}$ dissociates to give one mole of ${\text{H}}^{+}$ and one mole of ${\text{Cl}}^{-}$ ions in the solution.

So, the hydrogen ion concentration of the given solution is $6.92\times {10}^{-5}\text{M}$.

The hydroxide ion concentration of the given solution is calculated by the formula,

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

Substitute the hydrogen ion concentration of the given solution in the above expression.

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\left(\frac{1.0\times {10}^{-14}}{6.92\times {10}^{-5}}\right)\text{M}\\ \text{=}\underset{\_}{1.44\times 10^{-6}M}\end{array}$

Therefore, the hydroxide ion concentration of the given solution is $\underset{\_}{1.44\times 10^{-6}M}$.

Conclusion

1. a. The hydrogen ion concentration of the given solution is $\underset{\_}{1.187\times 10^{-11}M}$.
2. b. The hydrogen ion concentration of the given solution is $\underset{\_}{1.25\times 10^{-10}M}$.
3. c. The hydroxide ion concentration of the given solution is $\underset{\_}{2.22\times 10^{-12}M}$.
4. d. The hydroxide ion concentration of the given solution is $\underset{\_}{1.44\times 10^{-6}M}$.