Chapter 16, Problem 16.161AP

(a)

Interpretation Introduction

To determine: If fluoride in drinking water with $\text{pH}=7.00$ is present in the form of r ${\text{F}}^{-}$ or in the form of ${\text{HF}}_{\text{2}}{}^{-}$.

Interpretation: The element, fluoride in drinking water with $\text{pH}=7.00$ is present in which form, ${\text{F}}^{-}$ or in the form of ${\text{HF}}_{\text{2}}{}^{-}$ is to be predicted.

Concept introduction: In equilibrium state the rate of forward reaction is equal to the rate of back ward reaction.

The ratio of the concentration of product to the concentration of reactant is termed as termed as equilibrium constant. It is denoted by $K$.

Answer to Problem 16.161AP

The element, fluoride in drinking water with $\text{pH}=7.00$ is present in the form of ${\text{F}}^{-}$.

Explanation of Solution

The element, fluoride in drinking water with $\text{pH}=7.00$ is present in the form of ${\text{F}}^{-}$. The concentration of ${\text{F}}^{-}$ in drinking water ranges from $0.7\text{ppm}$ to $1.22\text{ppm}$.

Excess concentration of ${\text{F}}^{-}$ in drinking water can cause dental fluorosis, renal and thyroid toxicity.

(b)

Interpretation Introduction

To determine: The equilibrium constant for the given reaction at equilibrium.

Interpretation: The equilibrium constant for the given reaction at equilibrium is to be predicted.

Concept introduction: In equilibrium state the rate of forward reaction is equal to the rate of back ward reaction.

The ratio of the concentration of product to the concentration of reactant is termed as termed as equilibrium constant. It is denoted by $K$.

Answer to Problem 16.161AP

The equilibrium constant $\left(K_{{\text{a}}_1}\right)$ for the given reaction is $\underset{\_}{2.86\times 10^{-4}}$.

Explanation of Solution

The first equilibrium reaction is,

$\text{HF}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(aq\right)+{\text{F}}_{}{}^{-}\left(aq\right)$

The equilibrium constant $\left(K_{\text{a}}\right)$ for this reaction is $1.1\times {10}^{-3}$.

The second equilibrium reaction is,

${\text{F}}^{-}\left(aq\right)+\text{HF}\left(g\right)\rightleftharpoons {\text{HF}}_2{}^{-}\left(aq\right)$

The equilibrium constant $\left(K\right)$ for this reaction is $2.6\times {10}^{-1}$.

The resultant reaction at equilibrium is,

$\text{2HF}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(aq\right)+{\text{HF}}_{\text{2}}{}^{-}\left(aq\right)$

The equilibrium constant $\left(K_{a_1}\right)$ for this reaction is calculated by the expression,

$K_{a_1}=K_{\text{a}}\times K$

Substitute the value of $K$ and $K_{\text{a}}$ in the above expression.

$\begin{array}{c}{K}_{{\text{a}}_1}=1.1\times {10}^{-3}\times 2.6\times {10}^{-1}\\ =\underset{\_}{2.86\times 10^{-4}}\end{array}$

The equilibrium constant $\left(K_{{\text{a}}_1}\right)$ for the given reaction is $\underset{\_}{2.86\times 10^{-4}}$.

(c)

Interpretation Introduction

To determine: The $\text{pH}$ and the equilibrium concentration of ${\text{HF}}_2{}^{-}$ in a $0.150\text{M}$ solution of $\text{HF}$.

Interpretation: The $\text{pH}$ and the equilibrium concentration of ${\text{HF}}_2{}^{-}$ in a $0.150\text{M}$ solution of $\text{HF}$ are to be calculated.

Concept introduction: In equilibrium state the rate of forward reaction is equal to the rate of back ward reaction.

The ratio of the concentration of product to the concentration of reactant is termed as termed as equilibrium constant. It is denoted by $K$.

Answer to Problem 16.161AP

The concentration and $\text{pH}$ of ${\text{HF}}_2{}^{-}$ in a $0.150\text{M}$ solution of $\text{HF}$ are $\underset{\_}{0.00245M}$ and $\underset{\_}{2.611}$ respectively.

Explanation of Solution

The reaction at equilibrium given by the solution of $\text{HF}$ in water,

$\text{2HF}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(aq\right)+{\text{HF}}_{\text{2}}{}^{-}\left(aq\right)$

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

The concentration of ${\text{HF}}_2{}^{-}$ is assumed to be $x$.

The equilibrium constant $\left(K_{{\text{a}}_1}\right)$ for this reaction is $2.86\times {10}^{-4}$.

The ICE table for the reaction is,

$\begin{array}{cccccccc}& \text{2HF}\left(aq\right)& +& {\text{H}}_2\text{O}\left(l\right)& \rightleftharpoons & {\text{H}}_3{\text{O}}^{+}\left(aq\right)& +& {\text{HF}}_{\text{2}}{}^{-}\left(aq\right)\\ \text{Initial}\left(\text{M}\right)& 0.150& & & & 0& & 0\\ \text{Change}\left(\text{M}\right)& -2x& & & & +x& & +x\\ \text{Equilibrium}\left(\text{M}\right)& 0.150-2x& & & & x& & x\end{array}$

The equilibrium constant $\left(K_{{\text{a}}_1}\right)$ for the reaction is,

$K_{{\text{a}}_1}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{HF}}_{\text{2}}{}^{-}\right]}{\left[\text{HF}\right]}$

Substitute the respective values of concentrations at equilibrium from the ICE table and the value of equilibrium constant $\left(K_{{\text{a}}_1}\right)$ value in the above expression.

$\begin{array}{c}2.86\times {10}^{-4}=\frac{\left[x\right]\left[x\right]}{{\left[\text{0}\text{.150}-2x\right]}^2}\\ 2.86\times {10}^{-4}={\left[\frac{x}{\left[\text{0}\text{.150}-2x\right]}\right]}^2\\ x=\left(1.69\times {10}^{-2}\right)\left(\text{0}\text{.150}-2x\right)\\ x=\underset{\_}{0.00245M}\end{array}$

Therefore, the concentration of $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]=\left[{\text{HF}}_{\text{2}}{}^{-}\right]$ is $\underset{\_}{0.00245M}$.

The $\text{pH}$ of ${\text{HF}}_2{}^{-}$ in the solution is calculated by the formula,

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

Substitute the concentration of ${\text{H}}^{+}$ in the solution in the above expression.

$\begin{array}{c}\text{pH}=-\log \left[0.00245\text{M}\right]\\ =\underset{\_}{2.611}\end{array}$

Therefore, the $\text{pH}$ of ${\text{HF}}_2{}^{-}$ in the solution is $\underset{\_}{2.611}$.

Conclusion

1. a. The element, fluoride in drinking water with $\text{pH}=7.00$ is present in the form of ${\text{F}}^{-}$.
2. b. The equilibrium constant $\left(K_{{\text{a}}_1}\right)$ for the given reaction is $\underset{\_}{2.86\times 10^{-4}}$.
3. c. The concentration and $\text{pH}$ of ${\text{HF}}_2{}^{-}$ in a $0.150\text{M}$ solution of $\text{HF}$ are $\underset{\_}{0.00245M}$ and $\underset{\_}{2.611}$ respectively.