Asked by Michael

calculate the pH and the propionate ion concentration [c3h502-] of a solution that is .06M in potassium propionate [kc3h502] and .085M in propionic acid [hc3h502]
Answered by DrBob222
propionic acid = HPr = 0.085M
KPr = potassium priopionate = 0.06

........HPr ==> H^+ + Pr^-
I.......0.085....0.....0
C........-x.......x....x
E......0.085-x....x....x

Ka = (H^+)(Pr^-)/(HPr)
Substitute Ka.
(H^+) = x
(Pr^-) = x + 0.06
(HPr) = 0.085
Solve for H^+ and convert to pH. You may need to use the quadratic formula.
Answered by Sarah
pH=2.98
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