TBR Gen-Chem Section V, passage 5, #33

[AMSOHN]

Hello,

I am just not following the explanation to this problem. Would anyone be kind enough to clarify? Much appreciated.

Thanks,
AMSOHN

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[BerkReviewTeach]

This question is asking you to consider what you have in solution at those two points. At point (a), you have only the weak acid in solution, as you have yet to add any titrant. At point (d), you have completely neutralized the weak base and thus converted it into its conjugate weak acid. So, both point (a) and point (d) represent solutions of weak acid in water. The pH should be about the same for both points, although the molarity is different making the pHs slightly different. It's a log scale, so the two pHs should be less than 1 different (maybe 0.3 to 0.5 different).

Hope this helps.

The takehome message is that the pH at the start of a weak acid titration is about the same as the equivalence pH for the titration of its weak conjugate base by a strong acid.

 

[AMSOHN]

Hi BerkReviewTeach,

Thank you for your helpful reply. I guess the part I was confused about, and conceptually this may not matter in solving the problem, was the final concentration at point (d), 0.05M. How was that concentration found? Once again, thank you very much.

-AMSOHN

 

[BerkReviewTeach]

The solution starts at 0.10 M before you add anything in. At equivalence, you have doubled the solution volume while keeping the same number of moles for the compound (although its now protonated), so the molarity is cut in half. That makes it 0.05 M at equivalence.

 

[AMSOHN]

Ah, apologies for the ignorant question. In any case, thank you for the clarification.