Question

Calculate the concentrations of \(\mathrm{H}^{+}, \mathrm{HCO}_{3}^{-},\) and \(\mathrm{CO}_{3}^{2-}\) in a \(0.025 \mathrm{M} \mathrm{H}_{2} \mathrm{CO}_{3}\) solution.

Short answer

To find the concentrations for \( \mathrm{H}^{+}, \mathrm{HCO}_{3}^{-}\), use the ionization constant and the initial concentration of \( \mathrm{H}_{2}\mathrm{CO}_{3}\). Approximating, the concentration of \( \mathrm{HCO}_{3}^{-}\) can be found using the equation \( \mathrm{HCO}_{3}^{-} = x=\sqrt{K_{a1}[H2CO3]}\), and the concentration of \( \mathrm{H}^{+} = 2x = 2\sqrt{K_{a1}[H2CO3]}\). For \( \mathrm{CO}_{3}^{2-}\), use the equation \( \mathrm{CO}_{3}^{2-} = K_{a2}\) because the contribution from the second ionization is much smaller than from the first ionization and can be approximated to \( \mathrm{CO}_{3}^{2-}\approx K_{a2}\). Note that this assumes the values of ionization constants \( K_{a1}\) and \( K_{a2}\) for the first and second ionization of \( \mathrm{H}_{2}\mathrm{CO}_{3}\) are known.

Step-by-step solution

Write Down the Chemical Equations

The problem refers to two ionizations of carbonic acid (H2CO3). In the first ionization, it forms bicarbonate ion (HCO3-) and a Hydronium ion (H+). \[ \mathrm{H}_{2} \mathrm{CO}_{3} \leftrightarrow \mathrm{H}^{+}+ \mathrm{HCO}_{3}^{-}\] In the second ionization, the bicarbonate ion further ionises to form carbonate ion (CO3^2-) and a Hydronium ion (H+). \[\mathrm{HCO}_{3}^{-}\leftrightarrow \mathrm{H}^{+} + \mathrm{CO}_{3}^{2-}\] The ionization equations are important to determine the concentration of each ion in the solution.

Apply Law of Mass Action

In a chemical equilibrium, the concentrations of the products and reactants are related through the equilibrium constant. For the first ionisation, if we denote the ionisation constant as \(K_{a1}\), and the concentration of \(H^+\) as \(x\), the law of mass action is: \[K_{a1}=\frac{[H^+][HCO_3^-]}{[H2CO3]} = \frac{x^{2}}{0.025 - x}\] Similarly, for the second ionisation, if we denote the ionisation constant as \(K_{a2}\), and the concentration of \(HCO_3^{-} \) = \(x\), and \(H^+ \) = \(2x\) and \(CO_3^{2-}\) = \(y\), we have the second equation: \[K_{a2}=\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]} = \frac{(2x)y}{x}\] However, because the first ionization is dominant, we typically assume \(y<<x\), so it can be approximated that: \[y= K_{a2}\] And the concentration of \(H^+\) is twice that of \(HCO_3^{-}\)