Chapter 6, Problem 6E.3E

(a)

Interpretation Introduction

Interpretation:

The $pHof0.010M{H}_{2}C{O}_3\left(aq\right)$ has to be calculated.

Explanation of Solution

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

  $\begin{array}{l}{H}_{2}C{O}_3\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+HC{O}_3{}^{-}\left(aq\right),K_{a1}=4.3\times {10}^{-7}\\ \\ HC{O}_3{}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+C{O}_3{}^{2-}\left(aq\right),K_{a2}=5.6\times {10}^{-11}\end{array}$

The second deprotonation can be ignored.

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

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

An equilibrium table can be set up as given below.

  $\begin{array}{ccccc}Composition\left(M\right)& H_{2}C{O}_3& H_{2}O& H_3{O}^{+}& HC{O}_3{}^{-}\\ Initialconcentration& 0.010& -& 0& 0\\ Changeinconcentration& -x& -& +x& +x\\ Equilibriumconcentration& 0.010-x& -& x& x\end{array}$

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

  $K_{a1}=\frac{\left(x\right)\left(x\right)}{\left(0.010-x\right)}$

Now, the above expression can be solved for x.

  $\begin{array}{l}{K}_{a1}=\frac{\left(x\right)\left(x\right)}{\left(0.010-x\right)}\\ \\ 4.3\times {10}^{-7}=\frac{\left(x\right)\left(x\right)}{\left(0.010-x\right)}\\ \\ \left(4.3\times {10}^{-7}\right)\left(0.010-x\right)=x^2\\ \\ x^2+4.3\times {10}^{-7}x-4.3\times {10}^{-9}=0\\ \\ x=\frac{\left(-4.3\times {10}^{-7}\right)\pm \sqrt{{\left(4.3\times {10}^{-7}\right)}^2-\left(4\right)\times \left(1\right)\times \left(-4.3\times {10}^{-9}\right)}}{2}\\ \\ x=6.53\times {10}^{-5}\\ \\ \left[H_3{O}^{+}\right]=x=6.53\times 10^{-5}\end{array}$

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

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(6.53\times {10}^{-5}\right)=4.18$

Therefore, $pHof0.010M{H}_{2}C{O}_3\left(aq\right)$ is $4.18.$

(b)

Interpretation Introduction

Interpretation:

The $pHof0.10M{\left(COOH\right)}_2\left(aq\right)$ has to be calculated.

Explanation of Solution

${\left(COOH\right)}_2$ is a polyprotic acid.  The first and second deprotonation reactions are given below.

The value of $K_{a1}$ is $5.9\times {10}^{-2}$ and the value of $K_{a2}$ is $6.5\times {10}^{-5}.$

The second deprotonation can be ignored.

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

  $K_{a1}=\frac{\left[H_3{O}^{+}\right]\left[COOH-CO{O}^{-}\right]}{\left[COOH-COOH\right]}$

An equilibrium table can be set up as given below.

  $\begin{array}{ccccc}Composition\left(M\right)& {\left(COOH\right)}_2& H_{2}O& H_3{O}^{+}& \left(COOH-CO{O}^{-}\right)\\ Initialconcentration& 0.10& -& 0& 0\\ Changeinconcentration& -x& -& +x& +x\\ Equilibriumconcentration& 0.10-x& -& x& x\end{array}$

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.

  $\begin{array}{l}{K}_{a1}=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}\\ \\ 5.9\times {10}^{-2}=\frac{\left(x\right)\left(x\right)}{\left(0.10-x\right)}\\ \\ \left(5.9\times {10}^{-2}\right)\left(0.10-x\right)=x^2\\ \\ x^2+5.9\times {10}^{-2}x-5.9\times {10}^{-3}=0\\ \\ x=\frac{\left(-5.9\times {10}^{-2}\right)\pm \sqrt{{\left(5.9\times {10}^{-2}\right)}^2-\left(4\right)\times \left(1\right)\times \left(-5.9\times {10}^{-3}\right)}}{2}\\ \\ x=0.052\\ \\ \left[H_3{O}^{+}\right]=x=0.052\end{array}$

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

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(0.052\right)=1.28$

Therefore, $pHof0.10M{\left(COOH\right)}_2\left(aq\right)$ is $1.28.$

(c)

Interpretation Introduction

Interpretation:

The $pHof0.20M{H}_{2}S\left(aq\right)$ has to be calculated.

Explanation of Solution

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

  $\begin{array}{l}{H}_{2}S\left(aq\right)+H_{2}O\left(l\right)\to H_3{O}^{+}\left(aq\right)+H{S}^{-}\left(aq\right),K_{a1}=1.3\times {10}^{-7}\\ \\ H{S}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+S^{2-}\left(aq\right),K_{a2}=7.1\times {10}^{-15}\end{array}$

The second deprotonation can be ignored.

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

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

An equilibrium table can be set up as given below.

  $\begin{array}{ccccc}Composition\left(M\right)& H_{2}S& H_{2}O& H_3{O}^{+}& H{S}^{-}\\ Initialconcentration& 0.20& -& 0& 0\\ Changeinconcentration& -x& -& +x& +x\\ Equilibriumconcentration& 0.20-x& -& x& x\end{array}$

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

  $K_{a1}=\frac{\left(x\right)\left(x\right)}{\left(0.20-x\right)}$

Now, the above expression can be solved for x.

  $\begin{array}{l}{K}_{a1}=\frac{\left(x\right)\left(x\right)}{\left(0.20-x\right)}\\ \\ 1.3\times {10}^{-7}=\frac{\left(x\right)\left(x\right)}{\left(0.20-x\right)}\\ \\ \left(1.3\times {10}^{-7}\right)\left(0.20-x\right)=x^2\\ \\ x^2+1.3\times {10}^{-7}x-2.6\times {10}^{-8}=0\\ \\ x=\frac{\left(-1.3\times {10}^{-7}\right)\pm \sqrt{{\left(1.3\times {10}^{-7}\right)}^2-\left(4\right)\times \left(1\right)\times \left(-2.6\times {10}^{-8}\right)}}{2}\\ \\ x=1.61\times {10}^{-4}\\ \\ \left[H_3{O}^{+}\right]=x=1.61\times 10^{-4}\end{array}$

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

  $pH=-\log \left[H_3{O}^{+}\right]=-\log \left(1.61\times {10}^{-4}\right)=3.80$

Therefore, $pHof0.20M{H}_{2}S\left(aq\right)$ is $3.80.$