Chapter 16.5, Problem 16.7WE

Calculate the pH of an aqueous solution at 25°C that is (a) 0.035 M in HI, (b) 1.2 × 10⁻⁴ M in HNO₃, and (c) 6.7 × 10⁻⁵ M in HClO₄.

Interpretation Introduction

Interpretation:

The pH of the given aqueous solutions at ${\text{25}}^{\circ }\text{C}$ has to be calculated

Concept Information:

Strong acids:

In strong acids, the ionization of acid is complete. This implies that the concentration of the hydrogen ion or hydronium ion will be equal to the initial concentration of the acid at equilibrium.

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

On rearranging, the concentration of hydrogen ion ${\text{[H}}^{+}\text{]}$ can be calculated using pH as follows,

${\text{[H}}^{+}\text{]}\text{=}{\text{10}}^{\text{-pH}}$

To calculate: The pH of an aqueous solution of 0.035 M $\text{HI}$

Answer to Problem 16.7WE

Answer

(a)

The pH of the given aqueous solution (a) is 1.46

Explanation of Solution

$\text{HI}$ is a strong acid.

The concentration of hydronium ion in the $\text{HI}$ is the same as the concentration of the given $\text{HI}$ acid

Concentration of $\text{HI}$ = 0.035 M

Concentration of hydronium ion = 0.035 M

The pH can be calculated as follows,

$\begin{array}{l}\text{pH}\text{= }{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{ = -log(0}\text{.035)}\\ \text{ = 1}\text{.46}\end{array}$

Therefore, the pH of an aqueous solution of 0.035 M $\text{HI}$ is 1.46

Interpretation Introduction

Interpretation:

The pH of the given aqueous solutions at ${\text{25}}^{\circ }\text{C}$ has to be calculated

Concept Information:

Strong acids:

In strong acids, the ionization of acid is complete. This implies that the concentration of the hydrogen ion or hydronium ion will be equal to the initial concentration of the acid at equilibrium.

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

On rearranging, the concentration of hydrogen ion ${\text{[H}}^{+}\text{]}$ can be calculated using pH as follows,

${\text{[H}}^{+}\text{]}\text{=}{\text{10}}^{\text{-pH}}$

To calculate: The pH of an aqueous solution of $1.2\times {10}^{-4}M$ ${\text{HNO}}_{\text{3}}$

Answer to Problem 16.7WE

Answer

The pH of the given aqueous solution (b) is 3.92

Explanation of Solution

${\text{HNO}}_{\text{3}}$ is a strong acid.

The concentration of hydronium ion in the ${\text{HNO}}_{\text{3}}$ is the same as the concentration of the given ${\text{HNO}}_{\text{3}}$ acid

Concentration of ${\text{HNO}}_{\text{3}}$ = $1.2\times {10}^{-4}M$

Concentration of hydronium ion = $1.2\times {10}^{-4}M$

The pH can be calculated as follows,

$\begin{array}{l}\text{pH}\text{= }{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{ = -log(}1.2\times {10}^{-4}\text{)}\\ \text{ = 3}\text{.92}\end{array}$

Therefore, the pH of an aqueous solution of $1.2\times {10}^{-4}M$ ${\text{HNO}}_{\text{3}}$ is 3.92

Interpretation Introduction

Interpretation:

The pH of the given aqueous solutions at ${\text{25}}^{\circ }\text{C}$ has to be calculated

Concept Information:

Strong acids:

In strong acids, the ionization of acid is complete. This implies that the concentration of the hydrogen ion or hydronium ion will be equal to the initial concentration of the acid at equilibrium.

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

On rearranging, the concentration of hydrogen ion ${\text{[H}}^{+}\text{]}$ can be calculated using pH as follows,

${\text{[H}}^{+}\text{]}\text{=}{\text{10}}^{\text{-pH}}$

To calculate: The pH of an aqueous solution of $6.7\times {10}^{-5}M$ ${\text{HClO}}_{\text{4}}$

Answer to Problem 16.7WE

Answer

The pH of the given aqueous solution (c) is 4.17

Explanation of Solution

Explanation

${\text{HClO}}_{\text{4}}$ is a strong acid.

The concentration of hydronium ion in the ${\text{HClO}}_{\text{4}}$ is the same as the concentration of the given ${\text{HClO}}_{\text{4}}$ acid

Concentration of ${\text{HClO}}_{\text{4}}$ = $6.7\times {10}^{-5}M$

Concentration of hydronium ion = $6.7\times {10}^{-5}M$

The pH can be calculated as follows,

$\begin{array}{l}\text{pH}\text{= }{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{ = -log(}6.7\times {10}^{-5}\text{)}\\ \text{ = 4}\text{.17}\end{array}$

Therefore, the pH of an aqueous solution of $6.7\times {10}^{-5}M$ ${\text{HClO}}_{\text{4}}$ is 4.17

Conclusion

The pH of the given aqueous solutions at ${\text{25}}^{\circ }\text{C}$ was calculated