[tashkent]

Greetings!

I am having problem solving #82 of the GRE Chemistry Practice Book:

82) A weak acid, HA, has a Ka of 1 x 10-5.  If 0.100 mole of this acid is dissolved in one liter of water, the percentage of acid dissociated at equilibrium is closest to:

a)   0.100%
b)   1.00%
c)   99.0%
d)   99.9%
e)   100%

Initially, HA should be 0.1 M.  Let H+ and A- be x.  Multiply Ka with [HA] to get 1 x 10-6.  Use this value in the “Change” part.  During the change, HA is deducted by 1 x 10-6, while H+ and A- are increased by 1 x 10-6.  At equilibrium, HA = 0.1 – 1 x 10-6, while H+ and A- have the value 1 x 10-6.

To get percent ionization, [H]+equil/[HA]initial x 100.  Therefore, 1 x 10-6/0.1 x 100 = 0.001%.  The problem is, I can’t find my answer in any of the choices above.  Where did I go wrong?  I will appreciate any help from you guys.  The answer should be letter B.

Hope to hear from you.  Thanks!

Sincerely,
Tashkent

[Donaldson Tan]

HA <-> H⁺ + A^-

let k be the degree of dissociation

before dissociation,
concentration of HX = 0.100mol/L

after dissociation,
concentration of HA = 0.100(1-k)mol/L
concentration of H⁺ = concentration of A^- = 0.100k

K_a = [ H⁺ ][ A^- ] / [HA]
1x10^-5 = (0.100k)² / 0.100(1-k)

since HA is a weak acid, k is very small.
1-k ~ 1

1x10^-5 ~ 0.100k²

k = 0.01 = 1.00%

answer is (b)