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The diprotic acid \( \text{H}_2\text{A} \) has \( \text{pK}_1 = 4.00 \) and \( \text{pK}_2 = 8.00 \). At what pH is \(\text{H}_2\text{A} = \text{HA}^-\)?

A. 4.00
B. 6.00
C. 7.00
D. 8.00

Answer :

At a pH of 4.00 option a, the concentrations of H2A and HA- are equal, as this is the pH value that corresponds to the first dissociation constant (pK1) of the diprotic acid H2A.

The question relates to the point at which the concentrations of H2A and HA- are equal in a diprotic acid dissociation. Given the pK1 value of 4.00, when the pH is equal to the first dissociation constant (pK1), the concentrations of the acid (H2A) and its first deprotonated form (HA-) are equal. This is a direct consequence of the Henderson-Hasselbalch equation. Therefore, at a pH of 4.00, H2A and HA- will be in equal concentrations.

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