What is the pH of 0.51M Ca(CH3COO)2 solution
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What is the pH of 0.51M Ca(CH3COO)2 solution

[From: ] [author: ] [Date: 12-05-13] [Hit: ]
02M.Kb = [HC2H2O2] [OH-] / [C2H3O2^-] = Kw / Ka = 1x10^-14 / 1.8x10^-5 = 5.[OH-] = 2.pOH = -log[OH-] = 4.pH = 14 - pOH = 9.......
thanks! pls show the solution :D

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Calcium acetate will dissocate completely to form Ca2+ and two acetate ions, giving an acetate ion concentration of 1.02M.

Acetate ion hydrolyzes (reacts with water) this way:
C2H3O2^- + HOH <==> HC2H3O2 + OH-

Kb = [HC2H2O2] [OH-] / [C2H3O2^-] = Kw / Ka = 1x10^-14 / 1.8x10^-5 = 5.6×10^-10
[OH-] = 2.4×10^-5
pOH = -log[OH-] = 4.62
pH = 14 - pOH = 9.38
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