1
$\begingroup$

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:

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

|improve this question
$\endgroup$
1
$\begingroup$

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.

|improve this answer
$\endgroup$
  • $\begingroup$ 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
-3
$\begingroup$

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.

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.