Calculating equilibrium constant given Ka value? $\ce{NH_3(aq) + H_3O^+(aq)  &lt;-&gt; NH_4^+(aq) + H_2O(l)}$
$\ce{K_a}$ for $\ce{NH_4^+}$ is $5.6\times10^{-10}$.

I am completely lost and I have no idea how to approach this question.
Never mind, I figured it out. All I had to do is reciprocate the value since Ka is the reciprocal of the equilibrium constant for that reaction.