Question

The dissociation constants of \(H_2A\) are \[ K_{a1}=6\times10^{-2} \quad \text{and} \quad K_{a2}=6\times10^{-5} \] respectively. At equilibrium, \[ [A^{2-}] = [H_2A] \] What is the approximate concentration of \(H^+\) at equilibrium?

• A. \(1.9 \times 10^{-3}\)
• B. \(2 \times 10^{-4}\)
• C. \(1.9 \times 10^{-5}\)
• D. \(1.9 \times 10^{-2}\)

Hint:

For diprotic acids, when: \[ [A^{2-}] = [H_2A] \] directly use: \[ [H^+] = \sqrt{K_{a1}K_{a2}} \] This shortcut saves a lot of calculation time.

Answer 1

The Correct Option is A

Step 1: Write the two dissociation steps.
A diprotic acid loses protons one at a time: \[ H_2A \rightleftharpoons H^+ + HA^-, \qquad HA^- \rightleftharpoons H^+ + A^{2-}. \] Each step has its own acid constant $K_{a1}$ and $K_{a2}$.
Step 2: Write each constant.
\[ K_{a1} = \frac{[H^+][HA^-]}{[H_2A]}, \qquad K_{a2} = \frac{[H^+][A^{2-}]}{[HA^-]}. \]
Step 3: Multiply the two constants.
When we multiply, the $[HA^-]$ cancels neatly: \[ K_{a1}K_{a2} = \frac{[H^+]^2[A^{2-}]}{[H_2A]}. \]
Step 4: Use the special condition.
We are told $[A^{2-}] = [H_2A]$, so their ratio is $1$. That leaves \[ K_{a1}K_{a2} = [H^+]^2 \ \Rightarrow \ [H^+] = \sqrt{K_{a1}K_{a2}}. \]
Step 5: Put in the numbers.
\[ [H^+] = \sqrt{(6\times10^{-2})(6\times10^{-5})} = \sqrt{36\times10^{-7}}. \] Now $\sqrt{36} = 6$ and $\sqrt{10^{-7}} = 10^{-3.5} = 3.16\times10^{-4}$.
Step 6: Finish the multiplication.
\[ [H^+] = 6 \times 3.16\times10^{-4} = 18.96\times10^{-4} \approx 1.9\times10^{-3}. \] So \[ \boxed{[H^+] \approx 1.9\times10^{-3}\ \text{M}} \]