Question

The degree of dissociation of a 0.1 M monobasic
acid is 0.4%. Its dissociation constant is
(a) 0.4 x 10-4
(b) 4.0 x 10-4
(c) 1.6 x 10-6
(d) 0.8 x 10-5​

Answer 1

Given:  

The degree of dissociation of a 0.1 M monobasic acid is 0.4%.  

To find:

Its dissociation constant is

Solution:

From given, we have,

The degree of dissociation of a 0.1 M monobasic acid is 0.4%.  

⇒ α = 0.4 % and C = 0.1 M

α = 0.4 % = 0.4/100 = 0.004

we use the formula,

The dissociation constant, kd = (αC × αC) / C(1 - α)

⇒ kd = α²C/(1 - α)

substituting the values in the above equation, we get,

kd = (0.004² × 0.1)/(1 - 0.004)

kd = 1.6 × 10^{-6}/0.996

∴ kd = 1.6 × 10^{-6}

The dissociation constant is 1.6 × 10^{-6}. ∴ Option (c) is correct.