According to wikipedia
the Ostwald's dilution law says that for a weak acid
$K_{d}={\cfrac {[A^{+}][B^{-}]}{[AB]}}={\frac {\alpha ^{2}}{1-\alpha }}\cdot c_{0}$
when I read the derivation of it ,I knew something was wrong
specially this line :
If α is the fraction of dissociated electrolyte, then αc0 is the concentration of each ionic species
here is how i see it :
$AB\rightleftharpoons A^{+}+B^{-}$
each Molecule of AB gets separated into A and B when put into water
so we can make a small table
| | AB | A | B |
| Before dissociation | 1 Mol | 0 Mol | 0 Mol |
| Concentration Before| 1/V Mol/L = C acid Mol/L | 0 Mol/L| 0 Mol/L |
| After dissociation | (1-X) Mol | X Mol | X Mol |
| Concentration After | (1-X)/V Mol/L | X/V Mol/L | X/V Mol/L |
V = Volume of the Acid AND Water,
After dissociation is an equilibrium between AB and (A,B), so we can apply the law of mass action
$$
K_{dilute}=\frac{\frac{\alpha}{v}\cdot \frac{\alpha}{v}}{\frac{\left(1-\alpha\right)}{v}}
$$
and then
$$
K_{dilute}=\frac{x^2}{v\cdot \left(1-x\right)}
$$
and as we referenced the AB Concentration Before dissociation with $c_{acid}$ = 1/V , then
$$
K_{dilute}=\frac{x^2\cdot c_{acid}}{\left(1-x\right)}
$$
and from here the non-sense starts,
the law says that X is a very tiny value so (1-x = 1) making x = 0 but forgetting that this way $x^2$ is even a tinier value and would also would also = 0,then $k_b$ = 0 , but in my opinion the law should be
$$
\alpha ^2+\frac{K_{dilute}}{C_{acid}}\cdot \alpha -\frac{K_{dilute}}{C_{acid}}=0
$$
to get $\alpha$, and to calculate the concentration:
$$
\left[A^+\right]=\left[B^-\right]=\frac{K_{dilute}\ \cdot \left(1-\alpha \right)}{\alpha }
$$
Any explanation would be appreciated
