A student prepared 0.10M acetic acid solution and experimentally measured its pH to
be 2.88. Calculate Ka for acetic acid and determine its percent dissociation.
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The pH is 2.88, Hence;
[H3O+] = 10^-2.88= 1.32*10^-3.
Therefore, X = 1.32*10^-3.
CH3CO2H + H2O ⇄ CH3CO2− + H3O+.
Use ice table and the equilibrium concentration is:
0.10 - X X X
Therefore; Ka = [CH3CO2−][H3O+] / [CH3CO2H].
=(X)^2 / (0.10 - 1.32*10^-3)
=1.76*10^-5.Q2
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