Question

Calculate the degree of ionisation and ph of 0.05 M solution of a weak base having the ionization constant (Kb) is1.77 ? 10 raised to power -5 .Also calculate the ionisation constant of the conjugate acid of this base

Answer 1

I hope this may help you

Answer 2

Hey there,

● Answer -
α = 1.044×10^-4
pH = 10.01
Ka = 5.64×10^-10

● Explaination-
# Given-
Kb = 1.77×10^-5
C = 0.05 M

# Solution-
Aa you know,
Ka × Kb = Kw
Ka × 1.77×10^-5 = 10^-14
Ka = 10^-14 / (1.77×10^-5)
Ka = 5.64×10^-10

Degree of ionization is calculated by-
α = √(Ka / C)
α = √(5.64×10^-10 / 0.05)
α = 1.044×10^-4

For the given base -
pOH = -log(α)
pOH = -log(1.044×10^-4)
pOH = 3.981

pH is calculated by-
pH = 14 - pOH
pH = 14 - 3.981
pH = 10.01

Hope this is helpful...