Chapter 6, Problem 6E.15E

(a)

Interpretation Introduction

Interpretation:

The $pHof1.0\times {10}^{-4}M{H}_{3}B{O}_3\left(aq\right)$ has to be calculated.

Explanation of Solution

$H_{3}B{O}_3$ is a monoprotic acid.  The deprotonation reaction is given below.

  $H_{3}B{O}_3\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+H_{2}B{O}_3{}^{-}\left(aq\right),K_a=7.3\times {10}^{-10}$

The expression for equilibrium constant for the deprotonation reaction can be written as shown below,

  $K_a=\frac{\left[H_3{O}^{+}\right]\left[H_{2}B{O}_3{}^{-}\right]}{\left[H_{3}B{O}_3\right]}$

An equilibrium table can be set up as given below.

  $\begin{array}{ccccc}Composition\left(M\right)& H_{3}B{O}_3& H_{2}O& H_3{O}^{+}& H_{2}B{O}_3{}^{-}\\ Initialconcentration& 1.0\times {10}^{-4}& -& 0& 0\\ Changeinconcentration& -x& -& +x& +x\\ Equilibriumconcentration& 1.0\times {10}^{-4}-x& -& x& x\end{array}$

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

  $K_a=\frac{\left(x\right)\left(x\right)}{\left(1.0\times {10}^{-4}-x\right)}$

Now, the above expression can be solved for x.

  $\begin{array}{l}{K}_a=\frac{\left(x\right)\left(x\right)}{\left(1.0\times {10}^{-4}-x\right)}\\ \\ 7.3\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(1.0\times {10}^{-4}-x\right)}\\ \\ \left(7.3\times {10}^{-10}\right)\left(1.0\times {10}^{-4}-x\right)=x^2\\ \\ x^2+7.3\times {10}^{-10}x-7.3\times {10}^{-14}=0\\ \\ x=\frac{\left(-7.3\times {10}^{-10}\right)\pm \sqrt{{\left(7.3\times {10}^{-10}\right)}^2-\left(4\right)\times \left(1\right)\times \left(-7.3\times {10}^{-14}\right)}}{2}\\ \\ x=2.7\times {10}^{-7}\\ \\ \left[H_3{O}^{+}\right]=x=2.7\times 10^{-7}\end{array}$

The $pH$ of the solution can now be calculated.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(2.7\times {10}^{-7}\right)=6.56$

Therefore, $pHof1.0\times {10}^{-4}M{H}_{3}B{O}_3\left(aq\right)$ is $6.56.$

(b)

Interpretation Introduction

Interpretation:

The $pHof0.015M{H}_{3}P{O}_4\left(aq\right)$ has to be calculated.

Explanation of Solution

$H_{3}P{O}_4$ is a polyprotic acid.  The first and second deprotonation reactions are given below.

  $\begin{array}{l}{H}_{3}P{O}_4\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+H_{2}P{O}_4{}^{-}\left(aq\right),K_{a1}=7.6\times {10}^{-3}\\ \\ H_{2}P{O}_4{}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+HP{O}_4{}^{2-}\left(aq\right),K_{a2}=6.2\times {10}^{-8}\end{array}$

The second deprotonation can be ignored.

The expression for equilibrium constant for the first deprotonation reaction can be written as shown below,

  $K_{a1}=\frac{\left[H_3{O}^{+}\right]\left[H_{2}P{O}_4{}^{-}\right]}{\left[H_{3}P{O}_4\right]}$

An equilibrium table can be set up as given below.

  $\begin{array}{ccccc}Composition\left(M\right)& H_{3}P{O}_4& H_{2}O& H_3{O}^{+}& H_{2}P{O}_4{}^{-}\\ Initialconcentration& 0.015& -& 0& 0\\ Changeinconcentration& -x& -& +x& +x\\ Equilibriumconcentration& 0.015-x& -& x& x\end{array}$

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

  $K_{a1}=\frac{\left(x\right)\left(x\right)}{\left(0.015-x\right)}$

Now, the above expression can be solved for x.

  $\begin{array}{l}{K}_{a1}=\frac{\left(x\right)\left(x\right)}{\left(0.015-x\right)}\\ \\ 7.6\times {10}^{-3}=\frac{\left(x\right)\left(x\right)}{\left(0.015-x\right)}\\ \\ \left(7.6\times {10}^{-3}\right)\left(0.015-x\right)=x^2\\ \\ x^2+7.6\times {10}^{-3}x-1.14\times {10}^{-4}=0\\ \\ x=\frac{\left(-7.6\times {10}^{-3}\right)\pm \sqrt{{\left(7.6\times {10}^{-3}\right)}^2-\left(4\right)\times \left(1\right)\times \left(-1.14\times {10}^{-4}\right)}}{2}\\ \\ x=7.53\times {10}^{-3}\\ \\ \left[H_3{O}^{+}\right]=x=7.53\times 10^{-3}\end{array}$

The $pH$ of the solution can now be calculated.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(\text{7}{\text{.53×10}}^{\text{-3}}\right)=2.12$

Therefore, $pHof0.015M{H}_{3}P{O}_4\left(aq\right)$ is $2.12.$

(c)

Interpretation Introduction

Interpretation:

The $pHof0.10M{H}_{2}S{O}_3\left(aq\right)$ has to be calculated.

Explanation of Solution

$H_{2}S{O}_3$ is a polyprotic acid.  The first and second deprotonation reactions are given below.

  $\begin{array}{l}{H}_{2}S{O}_3\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+HS{O}_3{}^{-}\left(aq\right),K_{a1}=1.5\times {10}^{-2}\\ \\ HS{O}_3{}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+S{O}_3{}^{2-}\left(aq\right),K_{a2}=1.2\times {10}^{-7}\end{array}$

The second deprotonation can be ignored.

The expression for equilibrium constant for the first deprotonation reaction can be written as shown below,

  $K_{a1}=\frac{\left[H_3{O}^{+}\right]\left[HS{O}_3{}^{-}\right]}{\left[H_{2}S{O}_3\right]}$

An equilibrium table can be set up as given below.

  $\begin{array}{ccccc}Composition\left(M\right)& H_{2}S{O}_3& H_{2}O& H_3{O}^{+}& HS{O}_3{}^{-}\\ Initialconcentration& 0.10& -& 0& 0\\ Changeinconcentration& -x& -& +x& +x\\ Equilibriumconcentration& 0.10-x& -& x& x\end{array}$

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

  $K_{a1}=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}$

Now, the above expression can be solved for x.

  $\begin{array}{l}{K}_{a1}=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}\\ \\ 1.5\times {10}^{-2}=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}\\ \\ \left(1.5\times {10}^{-2}\right)\left(0.10-x\right)=x^2\\ \\ x^2+1.5\times {10}^{-2}x-1.5\times {10}^{-3}=0\\ \\ x=\frac{\left(-1.5\times {10}^{-2}\right)\pm \sqrt{{\left(1.5\times {10}^{-2}\right)}^2-\left(4\right)\times \left(1\right)\times \left(-1.5\times {10}^{-3}\right)}}{2}\\ \\ x=0.032\\ \\ \left[H_3{O}^{+}\right]=x=0.032\end{array}$

The $pH$ of the solution can now be calculated.

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(0.032\right)=1.49$

Therefore, $pHof0.10M{H}_{2}S{O}_3\left(aq\right)$ is $1.49.$