Chemistry: Molecular Approach (4th Edition)

$$[CO{_3}^{2-}] =5.6 \times 10^{-11} \space M$$
According to the solution for exercise 16.17, after the first ionization, there is $1.466 \times 10^{-4} \space M$ of both $HC{O_3}^-$ and $H_3O^+$ in the solution. 1. Draw the ICE table for this equilibrium: $$\begin{vmatrix} Compound& [ HC{O_3}^- ]& [ C{O_3}^{2-} ]& [ H_3O^+ ]\\ Initial& 1.466 \times 10^{-4} & 0 & 1.466 \times 10^{-4} \\ Change& -x& +x& +x\\ Equilibrium& 1.466 \times 10^{-4} -x& 0 +x& 1.466 \times 10^{-4} +x\\ \end{vmatrix}$$ 2. Write the expression for $K_a$, and substitute the concentrations: - The exponent of each concentration is equal to its balance coefficient. $$K_a = \frac{[Products]}{[Reactants]} = \frac{[ C{O_3}^{2-} ][ H_3O^+ ]}{[ HC{O_3}^- ]}$$ $$K_a = \frac{(x)( 1.466 \times 10^{-4} + x)}{[ HC{O_3}^- ]_{initial} - x}$$ 3. Assuming $1.466 \times 10^{-4} \gt\gt x:$ $$K_a = \frac{(x)( 1.466 \times 10^{-4} )}{[ HC{O_3}^- ]_{initial}}$$ $$x = \frac{K_a \times [ HC{O_3}^- ]_{initial}}{ 1.466 \times 10^{-4} } = \frac{ 5.6 \times 10^{-11} \times 1.466 \times 10^{-4} }{ 1.466 \times 10^{-4} }$$ $x = 5.6 \times 10^{-11}$ 4. Test if the assumption was correct: $$\frac{ 5.6 \times 10^{-11} }{ 1.466 \times 10^{-4} } \times 100\% = 3.8 \times 10^{-5} \%$$ 5. The percent is less than 5%. Thus, it is correct to say that $x = 5.6 \times 10^{-11}$ $$[CO{_3}^{2-}] = x = 5.6 \times 10^{-11} \space M$$