The Henderson-Hasselbalch equation for the $\mathrm{pH}$ of a buffer solution of the monoprotic acid $\ce{HA}$ is given by $$\mathrm{pH}=\mathrm pK_\mathrm a+\log{\frac{[\ce{A-}]}{[\ce{HA}]}}$$ Since concentration appears in both the numerator and denominator of the fraction $\frac{[\ce{A-}]}{[\ce{HA}]}$ and $\mathrm pK_\mathrm a$ is constant (at a fixed temperature), it appears that dilution of the solution with pure $\ce{H2O}$ would not change the $\mathrm{pH}$. However, since $$\mathrm{pH}=-\log{[\ce{H+}]}$$ the amount of substance of $\ce{H+}$ must increase in order for $\mathrm{pH}$ to stay constant upon dilution.
Where is this additional $\ce{H+}$ coming from? I know that diluting an acid causes it to dissociate to a greater extent. But at the same time, you would be diluting its conjugate base and causing it to associate more, cancelling the dissociation of the acid.
Answer
In the Henderson-Hasselbalch equation, $K_\mathrm{a}$ is a product of concentrations and considered a constant.
In reality, $K_\mathrm{a}$, when defined as a product of concentrations, is not a constant:
Upon dilution (decrease in ionic strength) the $\mathrm{p}K_\mathrm{a}$ will change, and therefore the pH of the solution will change.
In addition to the above reason, pH will always approach 7 at extreme dilution as it approaches being pure water.
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