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

Expert verified
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

01

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.
02

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 \(HCO_3^{-}\)

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