Chapter 16, Problem 16.111AP

(a)

Interpretation Introduction

Interpretation: The two equilibrium constant for the given equilibrium reactions are given. The species that is more likely present in the drinking water, the equilibrium constant of the overall reaction of the two reactions, the $\text{pH}$ and the equilibrium concentration of ${\text{HF}}_{\text{2}}{}^{-}$ is to be calculated.

Concept introduction: The solution is supersaturated if the ionic product is more than the solubility product for a salt.

The overall equilibrium constant of the two equilibrium reaction is calculated as,

$K_{\text{a}}\left(\text{net}\text{reaction}\right)=K_{{\text{a}}_1}\times K_{{\text{a}}_2}$

The $\text{pH}$ of the solution is calculated by the formula,

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

To determine: The fluorine containing species that is more likely to be present in drinking water.

Answer to Problem 16.111AP

Solution:

The fluorine containing species that is more likely to be present in drinking water is ${\text{HF}}_{\text{2}}{}^{-}$

Explanation of Solution

Given

The equilibrium of hydrofluoric acid in water is given as,

$\begin{array}{l}\text{HF}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)+{\text{F}}^{-}\left(aq\right)\\ {\text{F}}^{-}\left(aq\right)+\text{HF}\left(aq\right)\underset{}{\rightleftharpoons }{\text{HF}}_{\text{2}}{}^{-}\left(aq\right)\end{array}$

The equilibrium of the first equilibrium reaction, $K_{{\text{a}}_1}$ is $1.1\times {10}^{-3}$.

The equilibrium of the first equilibrium reaction, $K_{{\text{a}}_2}$ is $2.6\times {10}^{-1}$.

The equilibrium constant of first reaction is less than that of the second reaction, therefore, the amount of the reactant of first equilibrium is present in large amount and the amount of product of second equilibrium is present in large amount.

Thus, undissociated $\text{HF}$ is present in large amount and this is utilized in second equilibrium to form ${\text{HF}}_{\text{2}}{}^{-}$. Therefore, the concentration of ${\text{HF}}_{\text{2}}{}^{-}$ is maximum in the drinking water.

(b)

Interpretation Introduction

To determine: The equilibrium constant of the given reaction.

Answer to Problem 16.111AP

Solution:

The equilibrium constant of the given reaction is $\underset{\_}{2.86\times 10^{-4}}$.

Explanation of Solution

Given

The equilibrium of hydrofluoric acid in water is given as,

$\begin{array}{l}\text{HF}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)+{\text{F}}^{-}\left(aq\right)\\ {\text{F}}^{-}\left(aq\right)+\text{HF}\left(aq\right)\underset{}{\rightleftharpoons }{\text{HF}}_{\text{2}}{}^{-}\left(aq\right)\end{array}$

The equilibrium of the first equilibrium reaction, $K_{{\text{a}}_1}$ is $1.1\times {10}^{-3}$.

The equilibrium of the first equilibrium reaction, $K_{{\text{a}}_2}$ is $2.6\times {10}^{-1}$.

The overall reaction of the above two equilibriums is,

$\text{2HF}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)+{\text{HF}}_{\text{2}}{}^{-}\left(aq\right)$

The equilibrium constant of the overall reaction is the product of the equilibrium constant of the individual reaction.

Therefore, the overall equilibrium constant of the two equilibrium reaction is calculated as,

$K_{\text{a}}\left(\text{net}\text{reaction}\right)=K_{{\text{a}}_1}\times K_{{\text{a}}_2}$

Substitute the value of the equilibrium constant of the individual reactions in the above formula,

$\begin{array}{c}{K}_{\text{a}}\left(\text{net}\text{reaction}\right)=1.1\times {10}^{-3}\times 2.6\times {10}^{-1}\\ =2.86\times {10}^{-4}\end{array}$

Therefore, the equilibrium constant of the given reaction is $\underset{\_}{2.86\times 10^{-4}}$.

(c)

Interpretation Introduction

To determine: The $\text{pH}$ of the solution of $\text{HF}$ and the equilibrium concentration of ${\text{HF}}_{\text{2}}{}^{-}$

Answer to Problem 16.111AP

Solution:

The $\text{pH}$ of the solution of $\text{HF}$ and the equilibrium concentration of ${\text{HF}}_{\text{2}}{}^{-}$ is $\underset{\_}{2.6}$ and $\underset{\_}{2.54\times 10^{-3}M}$, respectively.

Explanation of Solution

Given

The concentration of $\text{HF}$ is $0.150\text{M}$.

The equilibrium reaction of $\text{HF}$ is,

$\text{2HF}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)+{\text{HF}}_{\text{2}}{}^{-}\left(aq\right)$

The equilibrium constant, $K_{\text{a}}$, of the above reaction is $2.86\times {10}^{-4}$.

The concentration of hydronium ion and ${\text{HF}}_{\text{2}}{}^{-}$ is assumed to be $x$.

The initial and the equilibrium concentrations of the species involved in the above equilibrium reaction is given as,

$\begin{array}{cccccccc}& \text{2HF}\left(aq\right)& +& {\text{H}}_{\text{2}}\text{O}\left(l\right)& \underset{}{\rightleftharpoons }& {\text{H}}_{\text{3}}{\text{O}}^{+}\left(aq\right)& +& {\text{HF}}_{\text{2}}{}^{-}\left(aq\right)\\ \text{Initial}\left(\text{M}\right):& 0.150& & & & 0& & 0\\ \text{Change}\left(\text{M}\right):& -x& & & & +x& & +x\\ \text{Equilibrium}\left(\text{M}\right):& 0.150-x& & & & x& & x\end{array}$

The equilibrium constant is given by the formula,

$K_{\text{a}}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\times \left[{\text{HF}}_{\text{2}}{}^{-}\right]}{{\left[\text{HF}\right]}^2}$

Substitute the value of $K_{\text{a}}$, $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$, $\left[{\text{HF}}_{\text{2}}{}^{-}\right]$ and $\left[\text{HF}\right]$ in the above equation.

$2.86\times {10}^{-4}=\frac{x\times x}{{\left(0.150-x\right)}^2}$

Since, the value of $K_{\text{a}}$ is very small, the value of $x$ is neglected compared to $0.150$.

$\begin{array}{c}2.86\times {10}^{-4}=\frac{x\times x}{{\left(0.150\right)}^2}\\ x^2=2.86\times {10}^{-4}\times 2.25\times {10}^{-2}\\ =6.432\times {10}^{-6}\\ x=2.54\times {10}^{-3}\text{M}\end{array}$

Therefore, the concentration of hydronium ion and ${\text{HF}}_{\text{2}}{}^{-}$ at equilibrium is $\underset{\_}{2.54\times 10^{-3}M}$.

The $\text{pH}$ of the solution is calculated by the formula,

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

Substitute the concentration of hydronium ion in the above equation.

$\begin{array}{c}\text{pH}=-\log \left(2.54\times {10}^{-3}\text{M}\right)\\ =2.6\end{array}$

Thus, the $\text{pH}$ of the solution of $\text{HF}$ is $\underset{\_}{2.6}$.

Conclusion:

1. a. The fluorine containing species that is more likely to be present in drinking water is ${\text{HF}}_{\text{2}}{}^{-}$.
2. b. The equilibrium constant of the given reaction is $\underset{\_}{2.86\times 10^{-4}}$.
3. c. The $\text{pH}$ of the solution of $\text{HF}$ and the equilibrium concentration of ${\text{HF}}_{\text{2}}{}^{-}$ is $\underset{\_}{2.6}$ and $\underset{\_}{2.54\times 10^{-3}M}$, respectively