-log[H⁺] = 5.42
... [H⁺] = 10^-5.42 = 3.80E-6
Ka = [H⁺][A⁻] / [HA]
... = 3.80E-6 * 3.80E-6 /
... (.095 - 3.80E-6)
... = ?
The pH of a 0.095 M solution of an unknown monoprotic acid is 5.42. Calculate the Ka of the acid.
A) 3.8 x 10-6 B) 3.6 x 10-7 C) 1.5 x 10-10 D) 2.6 x 10-9 E) 2.8 x 10-8
2 answers
........HA ==> H^+ + Ac^-
I....0.095.....0.....0
C......-x......x.....x
E...0.095-x....x.....x
The problem tells you that the pH is 5.42 which means you can calculate (H^+) from pH = -log(H^+) and that is x in the above. Then (Ac^-) is the same and
you can calculate 0.095-x. Substitute each of those into the Ka expression and solve for Ka.
I....0.095.....0.....0
C......-x......x.....x
E...0.095-x....x.....x
The problem tells you that the pH is 5.42 which means you can calculate (H^+) from pH = -log(H^+) and that is x in the above. Then (Ac^-) is the same and
you can calculate 0.095-x. Substitute each of those into the Ka expression and solve for Ka.
2 answers