Answer
$[A] = 0.80 \space M$
$[B] = 0.20 \space M$
Work Step by Step
1. Write the $K_c$ expression for a = 1 and b = 1:
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[B]}{[A]}$$
2. Draw the equilibrium table:
\begin{matrix}
& [A] & [B] \\
Initial & 1.0 \space M & 0 \\
Change & -x & + x \\
Equilibrium & 1.0 \space M - x & x
\end{matrix}
3. Substitute the concentrations into the expression and solve for x:
$$K_c = 4.0 = \frac{1.0 - x}{x}$$ $$4.0x = 1.0 -x$$ $$5.0x = 1.0$$ $$x = 0.20$$
Therefore:
$[A] = 1.0 \space M - x = (1.0 - 0.20) \space M = 0.80 \space M$
$[B] = x = 0.20 \space M$
