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saikano 2013-4-17 02:31 PM

PH 問題!!

CH3COOH(I)+H2O(I)=H3O)aq)+CH3COO-(aq)

A solution was prepared by dissolving 0.1 mole ethanoic acid into the water
and makes up to 500cm3.
Calculate the pH of the resulting solution.
(Ka of ethanoic acid at 298K is 2.13X10^-5M)


最後答案我計到係pH=2.69

請問我岩唔岩?

↗亢龍無悔↙ 2013-4-17 07:21 PM

Try to list all the concentration for each species at start and at eqm. Set up the eqm law, you will get the answer

chungkin81 2013-4-17 08:06 PM

Initial concentration of CH3COOH = 0.1 / (500/1000) = 0.2 mol dm-3
                          CH3COOH + H2O <==> CH3COO- + H3O+
At start (M):          0.2                                0             0
At eqm (M):          0.2 - x                           x              x

                 Kc = [CH3COO-] [H3O+] / [CH3COOH]
2.13 x 10^-5  = (x) (x) / (0.2 - x)
                  x = 0.00205 or -0.00207 (rejected)

pH = -log[H3O+] = -log(0.00205) = 2.69

Zzlaz 2013-4-18 08:45 PM

P.S.:
if x is ~ 10^-5. we can assume 0.2-x ~= 0.2
it makes the equation simpler (provided that you don't have quadratic equation formula)
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