Answer to Question #208342 in Chemistry for neil

Question

A 0.045 0 M solution of HA is 0.60% dissociated. Calculate pKa for this acid.

Answer 1

Solution:

Dissociation reaction: HA ⇄ H⁺ + A^-

The K_a expression: K_a = [H⁺][A^-] / [HA]

ICE Table:

___________HA___⇄__H⁺__+__A^-__

Initial:_____0.045_____0______0___

Change:____-x_______+x_____+x___

Equlib:____0.045-x____x______x___

K_a = [H⁺][A^-] / [HA]

K_a = (x)(x) / (0.045 - x)

0.045 - x ≈ 0.045 (assuming that 0.045 >> x)

Therefore,

K_a = x² / 0.045

Since acid is 0.60% dissociated,

0.006 = [H⁺] / [HA] = x / 0.045

x = (0.006)(0.045) = 0.00027

x = 0.00027

Substitute 'x' in K_a = x² / 0.045 to get K_a:

K_a = (0.00027)² / 0.045 = 1.62×10^-6

K_a = 1.62×10^-6

pK_a = -log(K_a) = -log(1.62×10^-6) = 5.79

pK_a = 5.79

Answer: pK_a = 5.79