Calculate the percent ionization of 0.0075 m butanoic acid (ka = 1.5 x 10-5) in a solution containing 0.085 m sodium butanoate.

Part 1: Calculate the percent ionization of 0.0075 m butanoic acid (Ka=1.5x10^-5)

C4H8O2 (aq) + H2O (l) → C4H7O2 (aq) + H3O +

initial 0.0075 0 0

change - X + X + X

final 0.0075-X X X

when Ka is relative smaller to the intial concentration of the acid so we can assume that 0.0075-X≈ 0.0075 by substitution in Ka formula:

Ka = [C4H7O2][H3O+] / [C4H8O2]

1.5x10^-5 = X*X / 0.0075

X^2 = 1.125x10^-7

X = 0.00034 m = 3.4 x 10 ^-4 m

∴ [C4H7O2] = [H3O+] = 3.4X10^-4

∴ percent ionization = [H + equlibrium]/[acid initial] * 100

= 3.4X10^-4/0.0075 * 100

= 4.5 %

part 2) calculate the percent ionization of 0.0075m butanoic acid in a solution containing 0.085m sodium butanoic?

C4H8O2 (aq) + H2O (l) ↔ C4H7O2 (aq) + H3O+

initial 0.0075 0 0.085

change - X + X + X

final 0.0075-X X 0.085+X

we can assume that 0.0075-X≈ 0.0075 & 0.085+X ≈ 0.085

∴Ka = (X * (0.085)) / (0.0075)

(1.5x10^-5) * 0.0075 = 0.085X

∴X = 1.3x 10^-6

∴ percent ionization = (1.3x10^-6) / 0.0075 * 100 = 0.017 %