Enough of a monoprotic acid is dissolved in water to produce a 1.141.14 M solution. The pH of the resulting solution is 2.712.71. Calculate the Ka for the acid.
pH = 2.71
[H+] = 10^-pH
= 10^-2.71
= 1.95 x 10^-3 M
HA ---------> H+ + A-
1.14 0 0 ----------> I
-x +x +x ---------> C
1.14-x x x ---------->C
Ka = [H+][A-]/[HA]
Ka = x^2 / 1.14-x
Ka = (1.95 x 10^-3)^2 / (1.14 - 1.95 x 10^-3)
Ka = 3.80 x 10^-6 / (1.14 - 1.95 x 10^-3)
Ka = 3.34 x 10^-6
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