If the Ka of a monoprotic weak acid is 8.8 × 10-6, what is the pH of a 0.36 M solution of this acid?
The corresponding equation will be:
HA + H2O <--> A^-1 + H3O^+1
Ka = [A-][H3O+] / [HA] = 8.8x10^-6
At equilibrium, Let X = [H30+] and [A-] = X and [HA] = 0.36 - X
X^2 / (0.36 - X) = 8.8x10^-6
Since it is a weak acid, so assume X << 0.36 therefore
0.36-X = 0.36
X^2/0.36 = 8.8x10^-6
X^2 = 3.168 x10^-6
X = 1.78 x10^-3 = [H+]
pH = -log[H+]
pH = - log (1.78 x 10^-3)
pH = 2.74
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