Question

The pH of a 0.50M HCOOK solution is 8.72. Calculate Ka for HCOOH. (10)

Answer 1

HCOOK is a strongelectrolyte, will dissociate in 100% as

HCOO- + K+

note that HCOO- is a weak conjugate base, which will form hydrolysis with water to form its weak acid:

HCOO- + H2O = HCOOH + OH-

note that this is basic, as shown above, pH = 8.72

then

Kb = [HCOOH][OH-]/[HCOO-]

initially

[HCOOH]= 0

[OH-] = 0

[HCOO-] = 0.50

in equilibrium

[HCOOH]= 0 + x

[OH-] = 0 + x

[HCOO-] = 0.50 - x

and we know that

pH = 8.72; so we can get pOH:

pOH = 14-pH = 14-8.72 = 5.28

[OH-] = 10^-pOH = 10^-5.28 = 5.25*10^-6 M

since x = [OH-] = 5.25*10^-6

then x = 5.25*10^-6

substitute

[HCOOH]= 0 + x = 5.25*10^-6

[OH-] = 0 + x = 5.25*10^-6

[HCOO-] = 0.50 - x = 0.50-5.25*10^-6 = 0.49999475 M

substitute in Kb

Kb = ( 5.25*10^-6)( 5.25*10^-6)/0.49999475

Kb = 5.5125*10^-11

we need Ka, so

Kw = Ka*Kb, by definition for water at 25 C

Kw = 10^-14 constant

so

Ka = Kw/Kb = (10^-14)/(5.5125*10^-11) =0.0001814

Ka = 1.81*10^-4, this is for HCOOH acid