Titration Question & reverse titration?

  • Thread starter srizen
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  • #1
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Homework Statement



. A 50 mL acetate solution is titrated with a 0.2342 M HCl solution. The equivalence point occurs at
28.52 mL. Acetate is the conjugate base of acetic acid, for which Ka = 1.8 × 10-5
.

b) Write down the titration reaction.
c) Find the pH of the acetate solution at the start of the titration.
d) What is the pH after 14.26 mL of HCl have been added?

Homework Equations



Ka = [prod]/[react] using the ICE table method

moles = CV

KaKb=Kw = 1.0*10^-14

ph= -logH+

The Attempt at a Solution



for part b, my assumption is:
CH3COO-(aq) +HCl(aq) ---> CH3COOH(aq) + Cl(aq)

For part C i was thinking of converting Ka to Kb then using the Ice table...but i don't think that's going to work anymore. I've just been stuck on this for almost two days now. Any help would be greatly appreciated .
 
Last edited:

Answers and Replies

  • #2
Borek
Mentor
28,476
2,873
At the start you have just a solution of a weak base CH3COO-. Finding pH is not different from finding pH of a - say - ammonia solution.

Does it help?
 

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