Chapter 12, Problem 12.21PAE

(a)

Interpretation Introduction

To write: the equilibrium expressions for ${\text{K}}_{\text{1}}$ and ${\text{K}}_2$

Explanation of Solution

Equilibrium constant, K_eq,for a reaction is actually the concentration of the products raised to the power of number of moles divided by the concentration of the reactants raised to the power of number of moles.

For the reaction ${\text{CO}}_3{}^{2-}\left(aq\right){\text{+H}}^{\text{+}}(aq)\rightleftarrows {\text{HCO}}_{\text{3}}{}^{\text{-}}\left(aq\right)$, the equilibrium constant ${\text{K}}_{\text{1}}$ is written as:

${\text{K}}_{\text{1}}=\frac{\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]}{\left[{\text{CO}}_3{}^{2-}\right]\left[{\text{H}}^{\text{+}}\right]}$

For the reaction ${\text{HCO}}_{\text{3}}{}^{\text{-}}\left(aq\right){\text{+H}}^{\text{+}}(aq)\rightleftarrows {\text{H}}_2{\text{CO}}_{\text{3}}\left(aq\right)$, the equilibrium constant ${\text{K}}_2$ is written as:

${\text{K}}_2=\frac{\left[{\text{H}}_2{\text{CO}}_{\text{3}}\right]}{\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]\left[{\text{H}}^{\text{+}}\right]}$

Interpretation Introduction

(b) To write: the equation for the reaction that is the sum of two reactions

Explanation of Solution

Add the two equations to obtain their sum Adding ${\text{CO}}_3{}^{2-}\left(aq\right){\text{+H}}^{\text{+}}(aq)\rightleftarrows {\text{HCO}}_{\text{3}}{}^{\text{-}}\left(aq\right)$ and ${\text{HCO}}_{\text{3}}{}^{\text{-}}\left(aq\right){\text{+H}}^{\text{+}}(aq)\rightleftarrows {\text{H}}_2{\text{CO}}_{\text{3}}\left(aq\right)$,

$\begin{array}{l}{\text{ CO}}_3{}^{2-}\left(aq\right){\text{+H}}^{\text{+}}(aq)\rightleftarrows {\text{HCO}}_{\text{3}}{}^{\text{-}}\left(aq\right)\\ \underset{\_}{{\text{+ HCO}}_{\text{3}}{}^{\text{-}}\left(aq\right){\text{+H}}^{\text{+}}(aq)\rightleftarrows {\text{H}}_2{\text{CO}}_{\text{3}}\left(aq\right)}\\ {\text{CO}}_3{}^{2-}\left(aq\right){\text{ + 2H}}^{\text{+}}(aq)\rightleftarrows {\text{H}}_2{\text{CO}}_{\text{3}}\left(aq\right)\end{array}$

(c)

Interpretation Introduction

To write: The equilibrium expression for the reaction obtained on summation of the two reactions.

Explanation of Solution

The reaction obtained on summation of the two reactions${\text{CO}}_3{}^{2-}\left(aq\right){\text{+H}}^{\text{+}}(aq)\rightleftarrows {\text{HCO}}_{\text{3}}{}^{\text{-}}\left(aq\right)$ and ${\text{HCO}}_{\text{3}}{}^{\text{-}}\left(aq\right){\text{+H}}^{\text{+}}(aq)\rightleftarrows {\text{H}}_2{\text{CO}}_{\text{3}}\left(aq\right)$ is ${\text{CO}}_3{}^{2-}\left(aq\right){\text{ + 2H}}^{\text{+}}(aq)\rightleftarrows {\text{H}}_2{\text{CO}}_{\text{3}}\left(aq\right)$

When two reactions are added, their equilibrium constants get multiplied.

Therefore, the equilibrium constant${\text{K}}_3$ is obtained as $\begin{array}{l}\\ {\text{K}}_1\times {\text{K}}_2=\frac{\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]}{\left[{\text{CO}}_3{}^{2-}\right]\left[{\text{H}}^{\text{+}}\right]}\times \frac{\left[{\text{H}}_2{\text{CO}}_{\text{3}}\right]}{\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]\left[{\text{H}}^{\text{+}}\right]}\\ =\frac{\left[{\text{H}}_2{\text{CO}}_{\text{3}}\right]}{\left[{\text{CO}}_3{}^{2-}\right]{\left[{\text{H}}^{\text{+}}\right]}^2}\end{array}$

Conclusion

${\text{K}}_{\text{1}}=\frac{\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]}{\left[{\text{CO}}_3{}^{2-}\right]\left[{\text{H}}^{\text{+}}\right]}$, ${\text{K}}_2=\frac{\left[{\text{H}}_2{\text{CO}}_{\text{3}}\right]}{\left[{\text{HCO}}_{\text{3}}{}^{\text{-}}\right]\left[{\text{H}}^{\text{+}}\right]}$, the equation for the reaction that is the sum of two reactions ${\text{CO}}_3{}^{2-}\left(aq\right){\text{ + 2H}}^{\text{+}}(aq)\rightleftarrows {\text{H}}_2{\text{CO}}_{\text{3}}\left(aq\right)$ and the equilibrium expression is ${\text{K}}_{\text{3}}\text{=}\frac{\left[{\text{H}}_2{\text{CO}}_{\text{3}}\right]}{\left[{\text{CO}}_3{}^{2-}\right]{\left[{\text{H}}^{\text{+}}\right]}^2}$.