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You have a beaker containing $\pu{542 mL}$ of a solution of $\pu{0.48 M}$ acetate buffer ($\mathrm{p}K_\mathrm{a} = 4.76$) at $\ce{pH 4.2}$. To this solution, you add $\pu{342 mL}$ of pure water. What is the $\ce{pH}$ of the resulting solution?

The answer is $4.6$. However, I do not know how to arrive at that answer. I know the H-H equation must be used:

$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$

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Adding water means diluting the buffer.

Use the known concentration, the volume of the buffer and the volume of water added to calculate the new concentrations.

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  • $\begingroup$ I went down that path. However, the question just gave the concentration of the buffer. That means ([A-] + [HA] = 0.48M). In order to calculate the new pH, I'd need the concentration of the weak acid and its conjugate base. I was thinking of solving for its ratio using H-H, but I am not sure if I am on the right path. $\endgroup$ – Michael Apr 21 '15 at 23:13
  • $\begingroup$ @Michael That sounds reasonable, since you have the initial pH and the pKa :-) $\endgroup$ – Klaus-Dieter Warzecha Apr 21 '15 at 23:18
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Calculate the number of moles in the original buffer solution - you should get around 0.26 (that is using ur n=cv equation), now add up the new volume to the old one which would give u 0.884L then use your other mole number you got earlier to use in the n=cv equation to get new concentration, it should give around 0.294. No use pH=pKa+log([A-]/[HA]) equation bearing in mind that you just calculated the new concentration of ([A-]/[HA]) so just substitute it with the 0.294 and u should get the new pH to be 4.22 or the 4.2.

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