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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, 2016 at 7:48
  • $\begingroup$ Related-chemistry.stackexchange.com/questions/32176/… $\endgroup$
    – JM97
    Sep 7, 2016 at 11:53

2 Answers 2

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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 $\endgroup$
    – Abhigyan
    Sep 16, 2017 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, 2017 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 :) ) $\endgroup$
    – Abhigyan
    Sep 17, 2017 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, 2017 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... $\endgroup$
    – Abhigyan
    Sep 17, 2017 at 13:57
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I did the experiment. My buffer was a commercial product a simple packet of salts probably phthalate based. I made it up in deionised water to the right volume then measured its pH with a simple all-in-one probe-meter I measured the buffer neat and then again after 1/5 serial dilutions. I rinsed the probe with deionised water between readings.

The pH fell from 4 to 3.45 at dilution number 4 before it climbed again. See the table below. I was not expecting this. The change appears significant. I am inclined to agree that the Ka is dependent on ionic strength and to a greater level than I previously thought.

Relative
Concn
Buffer pH measured

1 4.0

0.2 4.0

0.04 3.7

0.008 3.5

0.0016 3.45

0.00032 3.75

0.000064 4.25

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