At a certain temperature, the equilibrium constant $K_c$ for the reaction $H_2(g) + I_2(g) \rightleftharpoons 2HI(g)$ is 49. If the equilibrium concentrations of $H_2$ and $I_2$ are both $0.10\,mol\,dm^{-3}$, what is the equilibrium concentration of HI?

Using $K_c = [HI]^2 / ([H_2][I_2])$, we get $49 = [HI]^2 / (0.10 \times 0.10) = [HI]^2 / 0.01$. Therefore $[HI]^2 = 49 \times 0.01 = 0.49$, giving $[HI] = \sqrt{0.49} = 0.70\,mol\,dm^{-3}$. Option A results from forgetting to take the square root, and option D from dividing instead of multiplying.