Select the correct option if pH=pKa in the Henderson-Hasselbalch equation.

Correct option is A

Correct Answer is (A) [HA] = [A⁻]

Explanation:

The Henderson-Hasselbalch equation is given by:

​$$\text{pH} = \text{pKa} + \log \left( \frac{[A^-]}{[HA]} \right)$$​​

When pH = pKa, the equation simplifies to:

​$$\log \left( \frac{[A^-]}{[HA]} \right) = 0$$​​

Since log(1) = 0, this implies:

​$$\frac{[A^-]}{[HA]} = 1$$​​

which means:

​$$[A^-] = [HA]$$​​

Thus, when pH = pKa, the concentrations of the acid ([HA]) and its conjugate base ([A⁻]) are equal.

Information Booster:

• The Henderson-Hasselbalch equation is used to calculate pH of buffer solutions.
• When pH < pKa, the acidic form ([HA]) is dominant.
• When pH > pKa, the basic form ([A⁻]) is dominant.
• Buffers work best when pH ≈ pKa since both acid and conjugate base are present in nearly equal amounts.
• Biological relevance: The equation is crucial in blood pH regulation (e.g., bicarbonate buffer system).