Question

Problem: Use the data in Table 15.2 to determine the equilibrium

constant for the reaction

${\mathbf{H}}_{\mathbf{2}}{\mathbf{PO}}_{\mathbf{4}}^{\mathbf{-}}\left(\mathbf{aq}\right){\mathbf{+}\mathbf{2}\mathbf{CO}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\left(\mathbf{aq}\right)\rightleftharpoons {\mathbf{PO}}_{\mathbf{4}}^{\mathbf{3}\mathbf{-}}\left(\mathbf{aq}\right){\mathbf{+}\mathbf{2}\mathbf{HCO}}_{\mathbf{3}}^{\mathbf{-}}\left(\mathbf{aq}\right)$

Short answer

The equilibrium constant for the given reaction is$2.2\times {10}^{-3}$.

Step-by-step solution

Equilibrium reaction

The given equilibrium reaction is represented as follows:

${\mathrm{H}}_2{\mathrm{PO}}_4^{-}\left(\mathrm{aq}\right){+2\mathrm{CO}}_3^{2-}\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{PO}}_4^{3-}\left(\mathrm{aq}\right){+2\mathrm{HCO}}_3^{-}\left(\mathrm{aq}\right)$

The equilibrium expression for the given reaction is written as follows:
$\mathrm{K}=\frac{\left[{\mathrm{PO}}_4^{3-}\right]{\left[{\mathrm{HCO}}_3^{-}\right]}^2}{\left[{\mathrm{H}}_2{\mathrm{PO}}_4^{-}\right]{\left[{\mathrm{CO}}_3^{2-}\right]}^2}$

Construction of reaction

The given equilibrium reaction is constructed by subtracting the second acid ionization constant from the first ionization constant.

$$\begin{gathered} {\mathrm{HPO}}_4^{2-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{PO}}_4^{2-}+{\mathrm{H}}_3{\mathrm{O}}^{+} \\ \frac{\left[{\mathrm{PO}}_4^{2-}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[{\mathrm{HPO}}_4^{2-}\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}={\mathrm{K}}_{{\mathrm{a}}_1} \end{gathered}$$

$$\begin{gathered} {2\mathrm{HCO}}_3^{2-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {2\mathrm{CO}}_3^{2-}+{\mathrm{H}}_3{\mathrm{O}}^{+} \\ \frac{{\left[{\mathrm{CO}}_3^{2-}\right]}^2\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{{\left[{\mathrm{HCO}}_3^{2-}\right]}^2\left[{\mathrm{H}}_2\mathrm{O}\right]}=2{\mathrm{K}}_{{\mathrm{a}}_2} \end{gathered}$$

Determining equilibrium constant

• The first acid ionization constant is $\left({\text{K}}_{\text{a1}}\right){\text{= 2.2×10}}^{-13}$
• The second acid ionization constant is$\left({\text{2K}}_{\text{a2}}\right){\text{= 9.6×10}}^{-11}$

The equilibrium constant fr the net reaction is the ratio of the equilibrium constant of separate reactions determined as follows:

$\mathrm{K}=\frac{{\mathrm{K}}_{{\mathrm{a}}_1}}{{\mathrm{K}}_{{\mathrm{a}}_2}}$

$=\frac{2.3\times {\text{10}}^{\text{- 13}}}{9.6\times {\text{10}}^{\text{- 11}}}$

$=2.2\times {\text{10}}^{\text{- 3}}$