pH = 2.62 = -log (H^+); therefore,
(H^+) 0.0024
............HA ==> H^+ + A^-
I........0.0111.....0.....0
C.........-x........x......x
E.......0.0111-x....x......x
x = 0.0024
Ka = (H^+)(A^-)/(HA)
(H^+) = 0.0024 = (A^-)
(HA) = 0.0111-0.0024 = ?
Substitute and solve for Ka.
enough of a monoprotic acid is dissolved in water to produce a 0.0111M solution. The pH of the resulting solution is 2.62. Calculate the Ka for the acid.
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