The degree of dissociation (α) of a weak electrolyte AxBy is related to van’t Hoff factor (i) by the expression

Question

The degree of dissociation (α) of a weak electrolyte AxBy is related to van’t Hoff factor (i) by the expression (A) α=(i-1/x+y-1) (B) α=(x+y-1/i-1) (C) α=(x+y+1/i-1) (D) α=(i-1/x+y+1)

Tardigrade Admin · Accepted Answer

AxBy arrow xAy+ + yBx- α = (i-1/n-1) where n is the number of ions produced. α = (i -1/(x+y-1))

Q. The degree of dissociation $(α)$ of a weak electrolyte $A_{x} B_{y}$ is related to van’t Hoff factor $(i)$ by the expression

$α = \frac{i - 1}{x + y - 1}$

$α = \frac{x + y - 1}{i - 1}$

$α = \frac{x + y + 1}{i - 1}$

$α = \frac{i - 1}{x + y + 1}$

$A_{x} B_{y} \to x A^{y +} + y B^{x -}$
$α = \frac{i - 1}{n - 1}$
where $n$ is the number of ions produced.
$α = \frac{i - 1}{( x + y - 1 )}$

Q. The degree of dissociation (α) of a weak electrolyte Ax​By​ is related to van’t Hoff factor (i) by the expression

α=x+y−1i−1​

α=i−1x+y−1​

α=i−1x+y+1​

α=x+y+1i−1​

Ax​By​→xAy++yBx− α=n−1i−1​ where n is the number of ions produced. α=(x+y−1)i−1​