7
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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.

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    $\begingroup$ Your intuition is right. Strictly speaking, dilution does affect the pH of the buffer because it affects the position of the equilibrium $\ce{HA + H2O <=> A- + H3O+}$. However the effect is really very small which is why it is commonly said that the pH is unchanged. $\endgroup$ – orthocresol Sep 7 '16 at 7:48
  • $\begingroup$ Related-chemistry.stackexchange.com/questions/32176/… $\endgroup$ – JM97 Sep 7 '16 at 11:53
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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:

https://commons.wikimedia.org/wiki/File:PK_acetic_acid.png

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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  • $\begingroup$ So if I did dilute a buffer solution with water, in a 10 to 1 ratio, would that affect the pH or not? I'm still unable to understand... My book says that the pH will remain unchanged, but as per your answer, the pH should change, and approach 7 $\endgroup$ – AbhigyanC Sep 16 '17 at 1:24
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    $\begingroup$ @Abhihyan it won't charge much, but it could change on the order of 0.1 pH units $\endgroup$ – DavePhD Sep 16 '17 at 1:29
  • $\begingroup$ So basically, there is a pH change, but the change is negligible... That's what I understand... Is that right? (Also, it's Abhigyan :) ) $\endgroup$ – AbhigyanC Sep 17 '17 at 6:30
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    $\begingroup$ @Abhigyan if you only dilute by a factor of 10, and you consider changes on the order of 0.1 units negligible, then it's negligible. $\endgroup$ – DavePhD Sep 17 '17 at 13:43
  • $\begingroup$ thanks so much... that really clears things up for me, and hope it does the same for many people to come... $\endgroup$ – AbhigyanC Sep 17 '17 at 13:57

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