For the equilibrium below, $ \text{p}K_a1 = 7.9 $ and $ \text{p}K_a2 = 10.1 $. At what pH is $[\text{H}_2\text{A}]$ equal to $[\text{HA}^-]$?

\[ \text{H}_2\text{A} \leftrightarrow \text{H}^+ + \text{HA}^- \leftrightarrow 2\text{H}^+ + \text{A}^{2-} \]

A. $ \text{pH} < 7.9 $

B. $ \text{pH} = 7.9 $

C. Between $ \text{pH} 7.9 $ and $ 10.1 $

D. $ \text{pH} = 10.1 $

E. $ \text{pH} > 10.1 $

The pH at which [H₂A] is equal to [HA⁻] for the given equilibrium H₂A <=> H⁺ + HA⁻ <=> 2H⁺ + A²⁻, with pKa1 = 7.9 and pKa2 = 10.1, is between pH 7.9 and 10.1.Thus the correct option is C.

Explanation:

The question asks at what pH [H₂A] is equal to [HA⁻] for the given equilibrium H₂A <=> H⁺ + HA⁻ <=> 2H⁺ + A²⁻, with pKa1 = 7.9 and pKa2 = 10.1.

First, let's consider the dissociation of H₂A into H⁺ and HA⁻:

H₂A <=> H⁺ + HA⁻

At equilibrium, the Henderson-Hasselbalch equation can be applied:

pH = pKa + log([A⁻]/[HA])

If [A⁻] equals [HA⁻], the equation simplifies to:

pH = pKa

So, at pH = pKa1 = 7.9, [H₂A] will be equal to [HA⁻] for this step of the equilibrium.

Next, consider the second dissociation:

HA⁻ <=> H⁺ + A²⁻

Again, using the Henderson-Hasselbalch equation:

pH = pKa + log([A²⁻]/[HA⁻])

At pH = pKa2 = 10.1, [HA⁻] will be equal to [A²⁻] for this step.

Now, looking at the overall equilibrium, we have two consecutive steps. In the first step, [H₂A] equals [HA⁻] at pH 7.9, and in the second step, [HA⁻] equals [A²⁻] at pH 10.1. Therefore, between pH 7.9 and 10.1, [H₂A] will be equal to [HA⁻], and the equilibrium will favor the formation of HA⁻. Thus the correct option is C.

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