# How to correctly apply Law of mass action

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?

• 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). – Buck Thorn Nov 7 '18 at 11:26

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.

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