Chapter 6, Problem 6E.12E

Interpretation Introduction

Interpretation:

The molar concentrations of $H_{2}S{O}_3,HS{O}_3{}^{-},S{O}_3{}^{2-},H_3{O}^{+},andO{H}^{-}$ in $0.170MN{a}_{2}S{O}_3\left(aq\right)$ have to be calculated.

Explanation of Solution

The first protonation equilibrium is given below,

  $S{O}_3{}^{2-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons O{H}^{-}\left(aq\right)+HS{O}_3{}^{-}\left(aq\right)$

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

  $\begin{array}{c}{K}_{b1}=\frac{\left[O{H}^{-}\right]\left[HS{O}_3{}^{-}\right]}{\left[S{O}_3{}^{2-}\right]}\\ \\ =\frac{K_w}{K_{a2}}\\ \\ =\frac{1.0\times {10}^{-14}}{1.2\times {10}^{-7}}\\ \\ =8.3\times 10^{-8}\end{array}$

An equilibrium table can be set up as given below.

  $\begin{array}{ccccc}Composition\left(M\right)& S{O}_3{}^{2-}& H_{2}O& O{H}^{-}& HS{O}_3{}^{-}\\ Initialconcentration& 0.170& -& 0& 0\\ Changeinconcentration& -x& -& +x& +x\\ Equilibriumconcentration& 0.170-x& -& x& x\end{array}$

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

  $K_{b1}=\frac{\left(x\right)\left(x\right)}{\left(0.170-x\right)}$

Now, the above expression can be solved for x.

  $\begin{array}{l}{K}_{b1}=\frac{\left(x\right)\left(x\right)}{\left(0.170-x\right)}\\ \\ 8.3\times {10}^{-8}=\frac{\left(x\right)\left(x\right)}{\left(0.170-x\right)}\\ \\ \left(8.3\times {10}^{-8}\right)\left(0.170\right)=x^2\left[\text{Assuming}\text{x}\ll \text{0}\text{.170}\right]\\ \\ x=1.2\times 10^{-4}\\ \\ \left[O{H}^{-}\right]=x=1.2\times 10^{-4}M\\ \\ \left[HS{O}_3{}^{-}\right]=x=1.2\times 10^{-4}M\\ \\ \left[S{O}_3{}^{2-}\right]=0.170-x\approx 0.170M\end{array}$

The second protonation reaction is given below,

  $HS{O}_3{}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons O{H}^{-}\left(aq\right)+H_{2}S{O}_3\left(aq\right)$

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

  $\begin{array}{c}{K}_{b2}=\frac{\left[O{H}^{-}\right]\left[HS{O}_3\right]}{\left[HS{O}_3{}^{-}\right]}\\ \\ =\frac{K_w}{K_{a1}}\\ \\ =\frac{1.0\times {10}^{-14}}{1.5\times {10}^{-2}}\\ \\ =6.7\times 10^{-13}\end{array}$

An equilibrium table can be set up as given below.

  $\begin{array}{ccccc}Composition\left(M\right)& HS{O}_3{}^{-}& H_{2}O& O{H}^{-}& H_{2}S{O}_3\\ Initialconcentration& 1.2\times {10}^{-4}& -& 1.2\times {10}^{-4}& 0\\ Changeinconcentration& -x& -& +x& +x\\ Equilibriumconcentration& 1.2\times {10}^{-4}-x& -& 1.2\times {10}^{-4}+x& x\end{array}$

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

  $K_{b2}=\frac{\left(x\right)\left(1.2\times {10}^{-4}+x\right)}{\left(1.2\times {10}^{-4}-x\right)}$

Now, the above expression can be solved for x.

  $\begin{array}{l}{K}_{b2}=\frac{\left(x\right)\left(1.2\times {10}^{-4}+x\right)}{\left(1.2\times {10}^{-4}-x\right)}\\ \\ 6.7\times {10}^{-13}=\frac{\cancel{\left(1.2\times {10}^{-4}\right)}\left(x\right)}{\cancel{\left(1.2\times {10}^{-4}\right)}}\left[\text{Assuming}x\ll 1.2\times {10}^{-4}\right]\\ \\ x=6.7\times 10^{-13}\\ \\ \left[O{H}^{-}\right]=1.2\times {10}^{-4}+x\approx 1.2\times 10^{-4}M\\ \\ \left[H_{2}S{O}_3\right]=x=6.7\times 10^{-13}M\end{array}$

The concentration of $H_3{O}^{+}$ can be calculated as given below.

  $\left[H_3{O}^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{1.0\times {10}^{-14}}{1.2\times {10}^{-4}}=8.3\times 10^{-11}M$

Therefore, the molar concentrations of $H_{2}S{O}_3,HS{O}_3{}^{-},S{O}_3{}^{2-},H_3{O}^{+},andO{H}^{-}$ in $0.170MN{a}_{2}S{O}_3\left(aq\right)$ can be summarized as follows, in order of decreasing concentration:

  $\begin{array}{cccccc}Species:& S{O}_3{}^{2-}& O{H}^{-}& HS{O}_3{}^{-}& H_3{O}^{+}& H_{2}S{O}_3\\ Concentration\left(M\right)& 0.170& 1.2\times {10}^{-4}& 1.2\times {10}^{-4}& 8.3\times {10}^{-11}& 6.7\times {10}^{-13}\end{array}$