Question

A. Calculate the percent ionization of 8.0×10−3 M butanoic acid (Ka=1.5×10−5) B. Calculate the percent ionization of 8.0×10−3 M butanoic acid in a solution containing 8.0×10−2 M sodium butanoate.

Answer 1

A. [Butanoic acid] = 8.0*10−3 M

Ka = 1.5*10−5

Butanoic acid ionizes as:

HA      H+ + A-
Initially 8.0*10−3 0 0
Finally 8.0*10−3 - x x x

Ka = x*x / 8.0*10−3 - x

1.5*10−5 = x*x / 8.0*10−3

1.2*10^-7 = x*x

x = 3.46*10^-4 M

Percent ionization = (3.46*10^-4 / 8.0*10−3) * 100

= 4.325 %

B). [Sodium butanoate] = 8.0*10^-2 M

HA      H+ + A-
Initially 8.0*10−3 0 0
Finally   8.0*10−3 - x x 8.0*10^-2 + x

Ka = x*(8.0*10^-2 + x) / 8.0*10−3 - x

1.5*10−5 = x*8.0*10^-2 + x*x / 8.0*10−3

1.2*10^-7 = x*x + x*8.0*10^-2

x*x + x*8.0*10^-2 - 1.2*10^-7 = 0

x = 1.49*10^-6 M

Percent ionization = (1.49*10^-6 / 8.0*10−3) * 100

= 0.018 %