NCERT Class XI Chemistry Equilibrium Solutions

Question : 59

Total: 73

The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01 M in HCl also?

Solution:

Ka = 1.32 × 10−5 ; C = 0.05 M ; α = √

or α = √

1.32×10−5

5×10−2

= 1.62 × 10−2
[H+] = Cα = 0.05 × 1.62 × 10–2 = 8.1 × 10–4
pH = – log(8.1 × 10–4) = 3.09
In 0.01 M HCl, [H+] = 0.01 M
C2H5COOH ⇌ C2H5COO−+H+

Initial‌molar‌conc.	C	0	0
Eqm.molar‌conc.	C(1−α)	Cα	Cα

Applying law of chemical eqm.
Ka =

[C2H5COO−][H+]

[C2H5COOH]

Cα×0.01

C(1−α)

; Ka =

Cα×0.01

1.32 × 10−5 = 0.01α ⇒ α = 1.32 × 10−3

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