Acids & Bases Chemistry Math Question 8

Question

Calculate the percentage of acetic acid that exists in ionized form when dissolved in water at 0.3 M total concentration.

K_a(HOAc) = 1.76 x 10^-5.

Answer 1

Let acetic acid be represented by HOAc

HOAc ==> H⁺ + OAc^-

Ka = [H⁺][OAc^-]/[HAc]

1.76x10^-5 = (x)(x)/0.3-x and if we assume that x is small, we can neglect it in the denominator...

1.76x10^-5 = x²/0.3

x² = 5.28x10^-6

x = 2.3x10^-3 (note: this is small (<1%) compared to 0.3M so assumption above was valid)

[H⁺] = [OAc^-] = 2.3x10^-3 M

Percent ionized = 2.3x10^-3 M/0.3 M (x100%) = 0.8 % (to 1 significant figure)