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Topic: Electrochemistry and Buffer  (Read 2381 times)

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Offline Abhinab

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Electrochemistry and Buffer
« on: June 24, 2014, 08:44:05 PM »
The question is:
Calculate the pH in the titration of 25 ml of 0.1M acetic acid by sodium hydroxide after the addition of 35 ml of 0.1M NaOH . The solution becomes either acidic, basic or neutral when 25 ml of 0.1M NaOH is added, why??

in the first part, we can write the chemical reaction as,

CH3COOH + NaOH  :rarrow: CH3COONa + H2O

but we know that CH3COOH is not completely ionised and also NaOH is added in excess amount. So, how can we calculate number of moles of NaOH remaining in the solution and how much CH3COOH is ionised??

for the second part, I don't have any idea..
please help..  :( :(

Offline Borek

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Re: Electrochemistry and Buffer
« Reply #1 on: June 25, 2014, 03:04:45 AM »
At the equivalence point solution is not different from the one prepared by dissolving produced salt. Do you know how to calculate pH of the sodium acetate solution?

After the equivalence the only thing that matters is the excess NaOH. CH3COO- is a weak base and effect of its presence is negligible. But if you want, you can try to incorporate it in the calculation, just construct ICE table for the CH3COO- hydrolysis with a concentration of OH- from the excess NaOH as initial value.
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