Finding equilibrium constant from solubility product constant
Is it a rule that the equilibrium constant of any backwards reaction is equal to the inverse of the equilibrium constant of the forwards reaction?


Yes. The general rule actually is: For a given chemical reaction with equilibrium constant $K$:

$$\ce{aA\+bB-&gt;cC\+dD}$$

If we multiply the chemical reaction by $n$, we get $K_1$ for the following equilibrium:

$$\ce{\mathit{na} A\+\mathit{nb} B-&gt;\mathit{nc} C\+\mathit{nd} D}$$

as $K_1=K^n$.

Reversing the reaction is equivalent to taking $n=-1$, in which case we get: $K_2=\frac{1}{K}$



You must be wondering that:


  Why does the new equilibrium constant $K'=K^n$?


The answer lies in a simple observation on the Law of Mass Action. Recall that for a reaction:  $\ce{A +B —&gt; C + D}$, $K_\mathrm c = \frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}$

Now, notice that $n\ce{A}$ can be written as $\ce{A}+\ce{A}+...+\ce{A}$ ($n$ times) Same applies on $n\ce{B}, n\ce{C}$ and $n\ce{D}$.

So, if we multiply our reaction with $n$, it actually can be rewritten as:

$$(\ce{A}+\cdots+\ce{A})+(\ce{B}+\cdots+\ce{B}) \ce{—&gt;}(\ce{C}+\cdots+\ce{C})+(\ce{D}+\cdots+\ce{D})$$
and, thus, its new $K_\mathrm{c}'$ will be
$$K_\mathrm c' = \frac{([\ce{C}]\cdots[\ce{C}])([\ce{D}]\cdots[\ce{D}])}{([\ce{A}]\cdots[\ce{A}])([\ce{B}]\cdots[\ce{B}])}= \frac{[\ce{C}]^n.[\ce{D}]^n}{[\ce{A}]^n.[\ce{B}]^n}=\left(\frac{[\ce{C}][\ce{D}]}{[\ce{A}][\ce{B}]}\right)^n=K_\mathrm{c}^n$$

Hence, proved.



PS: The previous answer proved the same result, but incorrectly. That proof has now been corrected thanks to the comment by @DavePhD below.