Derive the Henderson-Hasselbalch equation using activity coefficients. And why is the pH of buffer nearly independent of concentration
The correct answer and explanation is :
Derivation of the Henderson-Hasselbalch Equation using Activity Coefficients
The Henderson-Hasselbalch equation describes the pH of a buffer solution and relates the pH to the ratio of the concentrations of an acid and its conjugate base. To derive it with activity coefficients, let’s begin with the acid dissociation equilibrium:
[
HA \rightleftharpoons H^+ + A^-
]
For this equilibrium, the expression for the acid dissociation constant ( K_a ) in terms of the activities of the species is:
[
K_a = \frac{a_{H^+} \cdot a_A^-}{a_{HA}}
]
where ( a_{H^+} ), ( a_A^- ), and ( a_{HA} ) are the activities of ( H^+ ), ( A^- ), and ( HA ), respectively. The activity of a substance is related to its concentration ( [\text{species}] ) and its activity coefficient ( \gamma ):
[
a_{X} = \gamma_X [X]
]
where ( X ) refers to any species in the system.
Therefore, we can rewrite the dissociation constant as:
[
K_a = \frac{\gamma_{H^+} [H^+] \cdot \gamma_A^- [A^-]}{\gamma_{HA} [HA]}
]
Taking the logarithm of both sides:
[
\ln K_a = \ln \left( \frac{\gamma_{H^+} [H^+] \cdot \gamma_A^- [A^-]}{\gamma_{HA} [HA]} \right)
]
This simplifies to:
[
\ln K_a = \ln (\gamma_{H^+}) + \ln ([H^+]) + \ln (\gamma_A^-) + \ln ([A^-]) – \ln (\gamma_{HA}) – \ln ([HA])
]
Now, rearrange this expression to isolate ( \ln [H^+] ):
[
\ln [H^+] = \ln K_a – \ln (\gamma_{H^+}) – \ln (\gamma_A^-) – \ln (\gamma_{HA}) + \ln ([HA]) – \ln ([A^-])
]
For dilute solutions, the activity coefficients ( \gamma_{H^+} ), ( \gamma_A^- ), and ( \gamma_{HA} ) are nearly equal to 1 (i.e., ideal behavior), so we simplify this to the familiar form:
[
\ln [H^+] = \ln K_a + \ln \left( \frac{[HA]}{[A^-]} \right)
]
Exponentiating both sides gives the final form of the equation:
[
[H^+] = K_a \cdot \frac{[HA]}{[A^-]}
]
Taking the negative logarithm to get pH:
[
\text{pH} = -\log[H^+] = -\log \left( K_a \cdot \frac{[HA]}{[A^-]} \right)
]
This simplifies to the Henderson-Hasselbalch equation:
[
\text{pH} = \text{pK}_a + \log \left( \frac{[A^-]}{[HA]} \right)
]
Why pH of a Buffer is Nearly Independent of Concentration
The pH of a buffer solution is primarily determined by the ratio of the concentrations of the weak acid ( [HA] ) and its conjugate base ( [A^-] ), rather than their absolute concentrations. When the buffer concentration is increased, both ( [HA] ) and ( [A^-] ) increase proportionally, and the ratio ( [A^-]/[HA] ) remains constant.
Since the pH is governed by this ratio, not by the actual concentrations of the acid and base, the pH remains nearly constant as long as the ratio of the concentrations is maintained. The concentration only affects the buffer’s capacity (how much acid or base it can neutralize), but not the pH itself.
Therefore, for a buffer to work effectively, the concentrations of the acid and base should be sufficient to absorb added ( H^+ ) or ( OH^- ), but once this ratio is established, the pH remains stable despite changes in the total concentration of the buffer components. This is why buffer solutions exhibit nearly constant pH over a range of concentrations.