Given is a concentrated solution of a weak electrolyte AxBy of concentration 'c' and dissociation constant 'K'. The degree of dissociation is given by :
- [K cx+y-1xxyy ]x+y
- ( Kcx+y-1xxyy ) 1x+y
- ( cx+y-1xxyyK )x+y
- ( cx+y-1xxyyK ) 1x+y
Answer
(2) ( Kcx+y-1xxyy ) 1x+y
Solution
AxBy xAy+ + yBx-
At equilibrium: [AxBy] = c(1- ), [ Ay+] = xc , [ Bx-] = yc
K = (xc )x(yc )yc(1- )
Weak electrolyte ⇒ α ≪ 1 ⇒ 1 − α ≈ 1
⇒ K = xxyycx+y-1 x+y
⇒ = ( Kcx+y-1xxyy ) 1x+y
Answer: (2)