HA -----> A- + H+ , Ka = 5.5*10^-5
At equilibrium:
[H+] = [A-] = x (let)
Ka = [A-] * [H+] / [HA]
Ka = x * x / 5x
x = 5 * Ka = 5 * 5.5 * 10^-5 = 27.5*10^-5
[H+] = x = 27.5*10^-5 M
pH = -log [H+] = - log(27.5*10^-5) = 3.56 ...Answer
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