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 and [ Bx-] = yc , so
K = (xc )x(yc )yc(1 - )
Being a weak electrolyte, α ≪ 1, so 1 − α ≈ 1 and
K = xxyycx + y - 1 x + y
= ( Kcx + y - 1xxyy ) 1x + y
Answer: (2)