Chapter 16, Problem 16.46QP

(a)

Interpretation Introduction

Interpretation:

The percent ionization of the given $0.60M$ hydrofluoric acid solution has to be calculated.

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

Acids ionize in water. Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

The ionization of a weak acid $\text{HA}$ can be given as follows,

  ${\text{HA}}_{(aq)}\to {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the above reaction is given below.

  $K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where,

  $K_a$ is acid ionization constant,

  $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion

  $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion

  $[\text{HA]}$ is concentration of the acid

Percent ionization:

A quantitative measure of the degree of ionization is percent ionization.

For a weak, monoprotic acid $\text{HA}$ percent ionization can be calculated as follows,

$\text{percent}\text{ionization}=\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%$

Answer to Problem 16.46QP

The percent ionization of given $0.60M$ hydrofluoric acid solution is $3.5\%$

Explanation of Solution

From the given concentrations of hydrofluoric acid solution and its $K_a$ value, the hydrogen ion concentration can be found out.

Then the percent ionization can be calculated from the obtained hydrogen ion concentration as follows,

Construct an equilibrium table and express the equilibrium concentration of each species in terms of $x$

$\begin{array}{cccccc}& H{F}_{\left(aq\right)}& \underset{}{\rightleftharpoons }& H_{\left(aq\right)}^{+}& +& F_{\left(aq\right)}^{-}\\ Initial\left(M\right)& 0.60& & 0& & 0\\ Change\left(M\right)& -x& & +x& & +x\\ Equilibrium\left(M\right)& \left(0.60-x\right)& & x& & x\end{array}$

The $K_a$ of hydrofluoric acid is $7.1\times {10}^{-4}$

  $\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][F}}^{\text{-}}\text{]}}{\left[HF\right]}\\ \text{7}{\text{.1×10}}^{\text{-4}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.60-x)}}\end{array}$

Assuming that x is small compared to $0.60$ , we neglect it in the denominator:

  $\begin{array}{c}\text{7}{\text{.1×10}}^{\text{-4}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.60)}}\\ \text{x}\text{=}\sqrt{\text{(0}\text{.60)}\text{×7}{\text{.1×10}}^{\text{-4}}}\\ \text{=}\text{0}\text{.021}\text{M}\\ \left[{\text{H}}^{\text{+}}\right]\text{=}\text{0}\text{.021}\text{M}\end{array}$

The percent ionization can be calculated as follows,

  $\begin{array}{l}\text{percent}\text{ionization=}\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\text{×100}\text{\%}\\ \text{=}\frac{{\text{[H}}^{\text{+}}\text{]}}{\left[HF\right]}\text{×100}\text{\%}\\ \text{=}\frac{\text{0}\text{.021}\text{M}}{\text{0}\text{.60}\text{M}}\text{×100}\text{\%}\\ \text{=}3.5\%\end{array}$

Therefore, the percent ionization of given $0.60M$ hydrofluoric acid solution is $3.5\%$

(b)

Interpretation Introduction

Interpretation:

The percent ionization of the given $0.080M$  hydrofluoric acid solution has to be calculated.

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

Acids ionize in water.  Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

The ionization of a weak acid $\text{HA}$ can be given as follows,

  ${\text{HA}}_{(aq)}\to {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the above reaction is given below.

  $K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where,

  $K_a$ is acid ionization constant,

  $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion

  $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion

  $[\text{HA]}$ is concentration of the acid

Percent ionization:

A quantitative measure of the degree of ionization is percent ionization.

For a weak, monoprotic acid $\text{HA}$ percent ionization can be calculated as follows,

  $\text{percent}\text{ionization}=\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%$

Answer to Problem 16.46QP

The percent ionization of given $0.080M$ hydrofluoric acid solution is $3.5\%$

Explanation of Solution

From the given concentrations of hydrofluoric acid solution and its $K_a$ value, the hydrogen ion concentration can be found out.

Then the percent ionization can be calculated from the obtained hydrogen ion concentration as follows,

Construct an equilibrium table and express the equilibrium concentration of each species in terms of $x$

$\begin{array}{cccccc}& H{F}_{\left(aq\right)}& \underset{}{\rightleftharpoons }& H_{\left(aq\right)}^{+}& +& F_{\left(aq\right)}^{-}\\ Initial\left(M\right)& 0.080& & 0& & 0\\ Change\left(M\right)& -x& & +x& & +x\\ Equilibrium\left(M\right)& \left(0.080-x\right)& & x& & x\end{array}$

The $K_a$ of hydrofluoric acid is $7.1\times {10}^{-4}$

  $\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][F}}^{\text{-}}\text{]}}{\left[HF\right]}\\ \text{7}{\text{.1×10}}^{\text{-4}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.080-x)}}\end{array}$

Assuming that x is small compared to $0.60$, we neglect it in the denominator:

  $\begin{array}{c}\text{7}{\text{.1×10}}^{\text{-4}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.080)}}\\ \\ x^2\text{=}0.0000568\\ x\text{=}\sqrt{0.0000568}\\ \left[{\text{H}}^{\text{+}}\right]\text{=}\text{0}\text{.0075}\text{M}\end{array}$

The percent ionization can be calculated as follows,

  $\begin{array}{l}\text{percent}\text{ionization=}\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\text{×100}\text{\%}\\ \text{=}\frac{{\text{[H}}^{\text{+}}\text{]}}{\left[HF\right]}\text{×100}\text{\%}\\ \text{=}\frac{\text{0}\text{.0075}\text{M}}{\text{0}\text{.080}\text{M}}\text{×100}\text{\%}\\ \text{=9}\text{.4}\%\end{array}$

Therefore, the percent ionization of given $0.080M$ hydrofluoric acid solution is $\text{9}\text{.4}\%$

(c)

Interpretation Introduction

Interpretation:

The percent ionization of the given $0.0046M$  hydrofluoric acid solution has to be calculated.

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

Acids ionize in water.  Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

The ionization of a weak acid $\text{HA}$ can be given as follows,

  ${\text{HA}}_{(aq)}\to {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the above reaction is given below.

  $K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where,

  $K_a$ is acid ionization constant,

  $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion

  $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion

  $[\text{HA]}$ is concentration of the acid

Percent ionization:

A quantitative measure of the degree of ionization is percent ionization.

For a weak, monoprotic acid $\text{HA}$ percent ionization can be calculated as follows,

$\text{percent}\text{ionization}=\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%$

Answer to Problem 16.46QP

The percent ionization of given $0.0046M$ hydrofluoric acid solution is $33\%$

Explanation of Solution

From the given concentrations of hydrofluoric acid solution and its $K_a$ value, the hydrogen ion concentration can be found out.

Then the percent ionization can be calculated from the obtained hydrogen ion concentration as follows,

Construct an equilibrium table and express the equilibrium concentration of each species in terms of $x$

$\begin{array}{cccccc}& H{F}_{\left(aq\right)}& \rightleftarrows & H_{\left(aq\right)}^{+}& +& F_{\left(aq\right)}^{-}\\ Initial\left(M\right)& 0.0046& & 0& & 0\\ Change\left(M\right)& -x& & +x& & +x\\ Equilibrium\left(M\right)& \left(0.0046-x\right)& & x& & x\end{array}$

The $K_a$ of hydrofluoric acid is $7.1\times {10}^{-4}$

  $\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][F}}^{\text{-}}\text{]}}{\left[HF\right]}\\ \text{7}{\text{.1×10}}^{\text{-4}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(}0.0046M\text{-x)}}\end{array}$

Solving the above quadratic equation,

  $\begin{array}{c}{\text{x}}^{\text{2}}\text{+}\left(\text{7}{\text{.1×10}}^{\text{-4}}\right)\text{x}\text{-}\left(\text{3}\text{.3×}{\text{10}}^{\text{-6}}\right)\text{=0}\\ \text{x}\text{=}\text{1}\text{.5}{\text{×10}}^{\text{-3}}\text{M}\\ \left[{\text{H}}^{\text{+}}\right]\text{=}\text{1}\text{.5}{\text{×10}}^{\text{-3}}\text{M}\end{array}$

The percent ionization can be calculated as follows,

  $\begin{array}{l}\text{percent}\text{ionization=}\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\text{×100}\text{\%}\\ \text{=}\frac{{\text{[H}}^{\text{+}}\text{]}}{\left[HF\right]}\text{×100}\text{\%}\\ \text{=}\frac{\text{1}\text{.5}{\text{×10}}^{\text{-3}}\text{M}}{\text{0}\text{.0046}\text{M}}\text{×100}\text{\%}\\ \text{=}33\%\end{array}$

Therefore, the percent ionization of given $0.0046M$ hydrofluoric acid solution is $33\%$

(d)

Interpretation Introduction

Interpretation:

The percent ionization of the given $0.00028M$  hydrofluoric acid solution has to be calculated.

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

Acids ionize in water.  Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

The ionization of a weak acid $\text{HA}$ can be given as follows,

  ${\text{HA}}_{(aq)}\to {\text{H}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the above reaction is given below.

  $K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where,

  $K_a$ is acid ionization constant,

  $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion

  $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion

  $[\text{HA]}$ is concentration of the acid

Percent ionization:

A quantitative measure of the degree of ionization is percent ionization.

For a weak, monoprotic acid $\text{HA}$ percent ionization can be calculated as follows,

$\text{percent}\text{ionization}=\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{[HA]}}\times 100\%$

Answer to Problem 16.46QP

The percent ionization of given $0.00028M$ hydrofluoric acid solution is $79\%$

Explanation of Solution

From the given concentrations of hydrofluoric acid solution and its $K_a$ value, the hydrogen ion concentration can be found out.

Then the percent ionization can be calculated from the obtained hydrogen ion concentration as follows,

Construct an equilibrium table and express the equilibrium concentration of each species in terms of $x$

$\begin{array}{cccccc}& H{F}_{\left(aq\right)}& \rightleftarrows & H_{\left(aq\right)}^{+}& +& F_{\left(aq\right)}^{-}\\ Initial\left(M\right)& 0.00028& & 0& & 0\\ Change\left(M\right)& -x& & +x& & +x\\ Equilibrium\left(M\right)& \left(0.00028-x\right)& & x& & x\end{array}$

The $K_a$ of hydrofluoric acid is $7.1\times {10}^{-4}$

  $\begin{array}{c}{\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}^{\text{+}}{\text{][F}}^{\text{-}}\text{]}}{\left[HF\right]}\\ \text{7}{\text{.1×10}}^{\text{-4}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(}0.00028M\text{-x)}}\end{array}$

Solving the above quadratic equation,

  $\begin{array}{c}{\text{x}}^{\text{2}}\text{+}\left(\text{7}{\text{.1×10}}^{\text{-4}}\right)\text{x}\text{-}\left(\text{2}\text{.0×}{\text{10}}^{\text{-7}}\right)\text{=0}\\ \text{x}\text{=}\text{2}\text{.2}{\text{×10}}^{\text{-4}}\text{M}\\ \left[{\text{H}}^{\text{+}}\right]\text{=}\text{2}\text{.2}{\text{×10}}^{\text{-4}}\text{M}\end{array}$

The percent ionization can be calculated as follows,

  $\begin{array}{l}\text{percent}\text{ionization=}\frac{{\text{[H}}^{\text{+}}\text{]}}{\text{[HA]}}\text{×100}\text{\%}\\ \text{=}\frac{{\text{[H}}^{\text{+}}\text{]}}{\left[HF\right]}\text{×100}\text{\%}\\ \text{=}\frac{\text{2}\text{.2}{\text{×10}}^{\text{-4}}}{0.00028\text{M}}\text{×100}\text{\%}\\ \text{=}79\%\end{array}$

The percent ionization of given $0.00028M$ hydrofluoric acid solution is $79\%$

As the solution becomes more dilute, the percent ionization increases.