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While studying ionic equilibrium I came across a relation between solubility and solubility product for ternary salt. This is what was written in my book.

Let the solubility of ternary salt $AB_2$ be $s\ mol\ L^{-1}$ $$\ce{AB_2(s)\longrightleftharpoons A^{2+}\ (aq)\ +\ 2B^- (aq) }$$ $$K_{sp}=[A^{2+}][B^-]^2$$ $$K_{sp}=4s^3$$

But this seems to be incorrect. Let's analyse the equation the other way. We can write the equation like this.

$$\ce{AB_2(s)\longrightleftharpoons A^{2+}\ (aq)\ +\ B^- (aq) + B^- (aq) }$$

Now just for an instant let the $B^-$ on the extreme right be considered as $C^-$. I haven't changed the anion, but just named it as another ion so that you understand my point. So the equation looks like this

$$\ce{AB_2(s)\longrightleftharpoons A^{2+}\ (aq)\ +\ B^- (aq) + C^- (aq) }$$

I haven't changed $AB_2$ because $C^- (aq)$ is actually $B^- (aq)$, just written differently. You'll understand later why I did this. Now if the solubility be $s\ mol/L$ then what's the concentration of ions on RHS. $s\ mol/L$ for $A^{2+}$, $s\ mol/L$ for $B^-$ and $s\ mol/L$ for $C^-$. Now use this equation

$$K_{sp}=[A^{2+}][B^-][C^-]$$

$$K_{sp}=(s)(s)(s)$$

$$K_{sp}=s^3$$

If you carefully observe what I have written then you will get to know that concentration of $B+C=2s$ or concentration of $2B=2s$ but concentration of $B=s$ so we should use concentration of B as $s$ in law of mass action and not $2s$. Don't you think I m correct?

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  • $\begingroup$ I think you are deeply confused about the meaning of the equilibrium constant (and its derivation, although writing $[c]^2=[c][c]$ is unorthodox and unnecessary but correct) and its use in deriving the concentration of reagents and products at equilibrium (when a reaction is complete). $\endgroup$ – Buck Thorn Nov 7 '18 at 11:26
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Your wrong... The value of the $K_{sp}$ is $4s^3$, not $s^3$.


Let the solubility of ternary salt $\ce{AB_2}$ be $s\ \text{ mol/L}$

So the chemical reaction is noted correctly as:

$$\ce{AB_2(s) -> A^{2+}\ (aq)\ +\ 2B^- (aq) }$$

You've written the $K_{sp}$ equation correctly as:

$$K_{sp}=\ce{[A^{2+}][B^-]^2}$$

$\ce{[A^{2+}]}$ is the concentration of $\ce{A^{2+}}$, and $\ce{[B-]}$ is the concentration of $\ce{B-}$.

Since $s$ mole/L of $\ce{AB2}$ dissolved, the solution must contain $s$ mole/L of $\ce{A^{2+}}$, and $2s$ mole/L of $\ce{B-}$. Substituting those values into the $K_{sp}$ equation gives:

$$K_{sp}=\ce{[A^{2+}][B^-]^2}= (s)(2s)^2 = 4s^3$$

Note that if I write the $K_{sp}$ equation as:

$$K_{sp}=\ce{[A^{2+}][B^-][B^-]}= (s)(2s)(2s) = 4s^3$$

I get exactly the same answer.

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  • $\begingroup$ you didn't get what I m saying. I have edited my question, I think you'll understand now $\endgroup$ – Loop Back Nov 8 '18 at 5:34

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