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?
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?
Show 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.
Updated On: Jun 17, 2026
- \(1.9 \times 10^{-3}\)
- \(2 \times 10^{-4}\)
- \(1.9 \times 10^{-5}\)
- \(1.9 \times 10^{-2}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
For a diprotic acid:
\[
H_2A \rightleftharpoons H^+ + HA^-
\]
\[
HA^- \rightleftharpoons H^+ + A^{2-}
\]
The dissociation constants are:
\[
K_{a1}
=
\frac{[H^+][HA^-]}{[H_2A]}
\]
\[
K_{a2}
=
\frac{[H^+][A^{2-}]}{[HA^-]}
\]
When expressions are multiplied:
\[
K_{a1}K_{a2}
=
\frac{[H^+]^2[A^{2-}]}{[H_2A]}
\]
This relation becomes extremely useful when:
\[
[A^{2-}] = [H_2A]
\]
Step 1: Write expressions for dissociation constants. First dissociation: \[ K_{a1} = \frac{[H^+][HA^-]}{[H_2A]} \] Second dissociation: \[ K_{a2} = \frac{[H^+][A^{2-}]}{[HA^-]} \]
Step 2: Multiply both expressions. \[ K_{a1}K_{a2} = \left( \frac{[H^+][HA^-]}{[H_2A]} \right) \left( \frac{[H^+][A^{2-}]}{[HA^-]} \right) \] Cancelling \([HA^-]\): \[ K_{a1}K_{a2} = \frac{[H^+]^2[A^{2-}]}{[H_2A]} \]
Step 3: Use the given condition. Given: \[ [A^{2-}] = [H_2A] \] Therefore: \[ \frac{[A^{2-}]}{[H_2A]} = 1 \] Thus: \[ K_{a1}K_{a2} = [H^+]^2 \] Hence: \[ [H^+] = \sqrt{K_{a1}K_{a2}} \]
Step 4: Substitute the values. Given: \[ K_{a1}=6\times10^{-2} \] \[ K_{a2}=6\times10^{-5} \] Thus: \[ [H^+] = \sqrt{ (6\times10^{-2})(6\times10^{-5}) } \] \[ = \sqrt{36\times10^{-7}} \] \[ = 6\times10^{-3.5} \] Now: \[ 10^{-3.5} = 3.16\times10^{-4} \] Therefore: \[ [H^+] = 6\times3.16\times10^{-4} \] \[ = 18.96\times10^{-4} \] \[ = 1.896\times10^{-3} \] Approximating: \[ [H^+] \approx 1.9\times10^{-3} \]
Step 5: Write the final answer. Hence: \[ \boxed{[H^+] = 1.9\times10^{-3}} \] Therefore, the correct option is: \[ \boxed{(A)} \]
Step 1: Write expressions for dissociation constants. First dissociation: \[ K_{a1} = \frac{[H^+][HA^-]}{[H_2A]} \] Second dissociation: \[ K_{a2} = \frac{[H^+][A^{2-}]}{[HA^-]} \]
Step 2: Multiply both expressions. \[ K_{a1}K_{a2} = \left( \frac{[H^+][HA^-]}{[H_2A]} \right) \left( \frac{[H^+][A^{2-}]}{[HA^-]} \right) \] Cancelling \([HA^-]\): \[ K_{a1}K_{a2} = \frac{[H^+]^2[A^{2-}]}{[H_2A]} \]
Step 3: Use the given condition. Given: \[ [A^{2-}] = [H_2A] \] Therefore: \[ \frac{[A^{2-}]}{[H_2A]} = 1 \] Thus: \[ K_{a1}K_{a2} = [H^+]^2 \] Hence: \[ [H^+] = \sqrt{K_{a1}K_{a2}} \]
Step 4: Substitute the values. Given: \[ K_{a1}=6\times10^{-2} \] \[ K_{a2}=6\times10^{-5} \] Thus: \[ [H^+] = \sqrt{ (6\times10^{-2})(6\times10^{-5}) } \] \[ = \sqrt{36\times10^{-7}} \] \[ = 6\times10^{-3.5} \] Now: \[ 10^{-3.5} = 3.16\times10^{-4} \] Therefore: \[ [H^+] = 6\times3.16\times10^{-4} \] \[ = 18.96\times10^{-4} \] \[ = 1.896\times10^{-3} \] Approximating: \[ [H^+] \approx 1.9\times10^{-3} \]
Step 5: Write the final answer. Hence: \[ \boxed{[H^+] = 1.9\times10^{-3}} \] Therefore, the correct option is: \[ \boxed{(A)} \]
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