Asked by Anonymous
The pH of a 0.010 M aqueous solution of a weak monoprotic acid, HX, is 4.5. What is the value of the acid ionization constant?
Answers
Answered by
DrBob222
pH = -log(H^+) = 4.5
(H^+) = 3.16E-5
.............HX ==> H^+ + X^-
I..........0.10M...0.....0
C............-x.....x.....x
E..........0.10-x...x.....x
x = 3.16E-5. Substitute into Ka expression and solve for Ka.
(H^+) = 3.16E-5
.............HX ==> H^+ + X^-
I..........0.10M...0.....0
C............-x.....x.....x
E..........0.10-x...x.....x
x = 3.16E-5. Substitute into Ka expression and solve for Ka.
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