pH = -log(H^+)
-3.25 = log(H^+)
(H^+) = 5.62E-4
............HA ==> H^+ + A^-
initial.....x.......0.....0
equil.......x...5.62E-4..5.62E-4
Ka = (H^+)(A^-)/(HA)
Substitute (H^+), (A^-) and Ka, and solve for (HA) = x
what is the concentration of a weak acid with a pH of 3.25, if the Ka of the weak acid is 1.6 x 10^-7 ?
2 answers
6.0x10-3