Ignoring the difference of concentrations of reactants. Equilibrium
Let's look at your example and set up the ICE (intial-change-equilibrium) problem. Then I'll give you more inclusive of thumb. 

Let's say reaction equation is

$\ce{A + B &lt;=> 2C}$$

The law of mass action is:

$$K = \dfrac{[\ce{C}]^2}{[\ce{A}][\ce{B}]}=10^{-8}$$

Let's set up the ICE matrix. The change in concentration $x$, will be subtracted from $[\ce{A}]$ and $[\ce{B}]$ and added to $[\ce{C}]$.

$$\begin{array}{|c|c|c|c|}\hline
\ &amp; [\ce{A}]\ (\mathrm{M}) &amp; [\ce{B}]\ (\mathrm{M}) &amp; [\ce{C}]\ (\mathrm{M}) \\ \hline
I &amp; 10^{-20} &amp; 10^{-20} &amp; 0 \\
C &amp; -x &amp; -x &amp; +2x \\
E &amp; 10^{-20} - x &amp; 10^{-20} -x &amp; 2x \\ \hline
\end{array}$$

Now we pug in to the law of mass action:

$$10^{-8} = \dfrac{x^2}{\left(10^{-20} -x \right)^2}$$

Since all numbers have to be positive, let's take the square root of both sides:

$$10^{-4} = \dfrac{x}{10^{-20} -x }$$

Now we have a ratio that says that $x$ is four orders of magnitude smaller than $[\ce{a}]_i -x$. I could have easily used the initial concentration of $\ce{B}$ Put another way:

$$x \approx  K\cdot [\ce{A}]_i$$

If this value is much less than $[\ce{A}]_i$, then we can simplify.