Asked by Victoria
In a 1.070 M aqueous solution of a monoprotic acid, 4.81% of the acid is ionized. What is the value of its Ka?
Answers
Answered by
DrBob222
........HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
This tells you that the 1.070 HA ionized 4.81% which means
(H^) = 1.070 x 0.0481 = ?
(A^-) = same as (H^+).
(HA) = [1.070 - (H^+)]
Substitute these values into the Ka expression and solve for Ka.
Ka = (H^+)(A^-)/(HA)
This tells you that the 1.070 HA ionized 4.81% which means
(H^) = 1.070 x 0.0481 = ?
(A^-) = same as (H^+).
(HA) = [1.070 - (H^+)]
Substitute these values into the Ka expression and solve for Ka.
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