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


1 Answer 1


You made $K_b$ = 0. That is where you were wrong. Taking approximation of Neglecting $\alpha$ with respect to 1 only gives us advantage of not solving the quadratic equation.

From there you can write $\alpha = √(K_b/C)$ here you can see $\alpha$ comes inversely proportional to $√C_o$ or we can say directly proportional to $√V$. From here you can deduce if $V$ approaches infinity $\alpha$ approaches 1. And that's the law.

  • $\begingroup$ my problem with this is that this is not mathematically accurate, setting a value to 0 means deleting a value, if 1 - x = 1 because x was too small with respect to 1,then ka = 0 because x^2 is too small with respect to ka $\endgroup$ Dec 23, 2015 at 3:26
  • $\begingroup$ That's why its just an approximation for small values of $\alpha$. The value of $\alpha$ is that much small that we neglect it in front of 1 since it is much much less than 1. But we cannot say value of K equal to zero. It could be of order of -5 or less than that but it is not zero. That's How this approximation works. If you are still not satisfied yet take a weak acid pH calculation problem and do by two methods. First by neglecting $\alpha$ with respect to one and other by solving quadratic. You will see the results are almost same in both cases $\endgroup$
    – Vaibhav
    Dec 23, 2015 at 6:32
  • $\begingroup$ I'd appreciate it if you don't add stuff like "hi", "thanks", "hope this helped" etc. to your posts. They don't add anything to the answer or the question. $\endgroup$
    – M.A.R.
    Dec 23, 2015 at 19:58

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