Animated Solution for Chemistry - Ionic Equilibrium: In aqueous solution, the ionisation constants for carbonic acid are $K_1=4.2\times 10^{-7}$ and $K_2=4.8\times 10^{-11}$. Select the correct statement for a saturated $0.034\text{ M}$ solution of the carbonic acid.

Imagine you have a beaker filled with a saturated solution of carbonic acid ($H_{2}C{O}_3$). Carbonic acid is a weak diprotic acid, meaning it has two protons ($H^{+}$) it can potentially give up. However, it doesn't release them both at once. It's a step-by-step process, and understanding the dynamics of these steps is the key to unlocking this problem.

The first step is the primary dissociation: This step is governed by the first ionization constant, $K_1=4.2\times 10^{-7}$. While it's a weak acid, it still produces a measurable amount of protons and bicarbonate ions ($HC{O}_3^{-}$).

Now, the bicarbonate ion formed in the first step can also act as an acid and release a second proton: This second step is governed by $K_2=4.8\times 10^{-11}$.

Notice the massive difference in magnitude! $K_1$ is roughly 10,000 times larger than $K_2$. This is a crucial physical insight. It tells us that the first dissociation is far more favorable than the second. The bicarbonate ion is holding onto that second proton very tightly.

Because $K_1\gg K_2$, almost all the $H^{+}$ ions in the solution come from the first dissociation step. The amount of $H^{+}$ contributed by the second step is so minuscule that it's practically negligible.

Therefore, for every $H_{2}C{O}_3$ molecule that dissociates in the first step, we get exactly one $H^{+}$ and one $HC{O}_3^{-}$. This leads us to our first major conclusion:

Now, let's look at the equilibrium expression for the second step to find the concentration of the carbonate ion ($C{O}_3^{2-}$):

Here is where the magic happens. We just established that $[H^{+}]\approx [HC{O}_3^{-}]$. If we substitute this approximation into our $K_2$ expression, the terms beautifully cancel each other out!

Armed with these insights, let's evaluate the given statements. The concentration of $C{O}_3^{2-}$ is incredibly small ($4.8\times 10^{-11}\text{ M}$), nowhere near the initial $0.034\text{ M}$ of the acid, and certainly not greater than the bicarbonate concentration.

However, our derivation perfectly aligns with statement (c): The concentration of $H^{+}$ and $HC{O}_3^{-}$ are approximately equal. This elegant approximation is a standard and powerful tool for analyzing weak polyprotic acids where successive ionization constants differ by several orders of magnitude.