Calculate the pH of 0.0070 M butanoic acid, which is a monoprotic acid; Ka=1.52 x 10โˆ’5. Assume that the 5% approximation rule applies. Provide your answer to two places after the decimal?

1 Answer
Apr 9, 2018

color(blue)"pH=3.50"

Explanation:

Butanoic acid (HC_4H_7O_2) is a weak acid .

DISSOCIATION

HC_4H_7O_2 <=>. H^+ + C_4H_7O_2 ^-

The equilibrium constant for this expression is called the acid dissociation constant, Ka.

Ka = 1.52" x"10^-5

The expression for Ka is:

Ka = ([C_4H_7O_2 ^-] [H^+]) / ([HC_4H_7O_2])

AT EQUILIBRIUM
Write each equilibrium concentration in terms of x

x = [C_4H_7O_2 ^-]= [H^+]

C = [HC_4H_7O_2]

So,

K_a = ([x] [x]) / ([C - x])= (x^2) / (C - x)

With the small x approximation, C - x ~~ C so that:

x ~~ sqrt(CK_a) = 3.187xx10^-4 mol/L

Calculating,

pH=-log[H^+]

pH=-log[3.187" x"10^-4]

color(blue)"pH=3.50"

NOTES

Checking the 5% Rule (not used in here)

("3.187 E-4"/0.007)100 = 4.6% < 5%

=>"5% Rule should be considered"

"x-solution negative value was not considered"