Question

From the equilibrium concentrations given, calculate Ka for each of the weak acids and Kb for each of the weak bases.

a) CH3CO2H:

[H3O+]= 1.34 x 10^-3

[CH3CO2H]= 9.866 x 10^-2

[CH3CO2-]= 1.34 x 10^-3

b) ClO-

[OH-]= 4.0 x 10^-4

[HClO]= 2.38 x 10^-5

[ClO-]= 0.273 M

Answer 1

Answer:(a)

For given weak acid i.e Acetic acid (CH₃COOH)

CH₃COOH + H₂O H₃O⁺ + CH₃COO^-

Given equilibrium concentrations are

[H₃O+]= 1.34 x 10^-3

[CH₃CO₂H]= 9.866 x 10^-2

[CH₃CO₂^-]= 1.34 x 10^-3

We have expresion for K_a as,

K_a = [CH₃CO₂^-][H₃O+]/[CH₃CO₂H]

K_a= (1.34x10^-3)(1.34x10^-3)/(9.866x10^-2)

calculating,   

K_a= 1.82x10^-5

(here K_a = Kx[H₂O]; as [H₂O] remains constant)

Answer: (b)

for given weak base,

ClO^- + H₂O OH^- + HClO

Equilibrium concentrations here,

[OH-]= 4.0 x 10^-4

[HClO]= 2.38 x 10^-5

[ClO-]= 0.273

As we have expression for K_b,

K_b= [HClO][OH^-]/[ClO^-]

K_b= (2.38x10^-5)(4.0x10^-4)/(0.273)

calculating,

K_b = 3.48x10^-8

(here K_b = Kx[H₂O]; as [H₂O] remains constant)