Answer
$ [ Cr^{3+} ] = 0.100 \space M$
$ [ Cr^{2+} ] = 1.9 \times 10^{-4} \space M$
$ [ Pb^{2+} ] = 9.3 \times 10^{-5} \space M$
Work Step by Step
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ Cr^{3+} ]& [ Cr^{2+} ]& [ Pb^{2+} ]\\
Initial& 0.100 & 0 & 0 \\
Change& -2 x& 2 x& x\\
Equilibrium& 0.100 -2 x& 2 x& x\\
\end{vmatrix}$$
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ Cr^{2+} ] ^{ 2 }[ Pb^{2+} ] ^{ }}{[ Cr^{3+} ] ^{ 2 }}$$
2. At equilibrium, these are the concentrations of each compound:
$ [ Cr^{3+} ] = 0.100 \space M - 2x$
$ [ Cr^{2+} ] = 0 \space M + 2x$
$ [ Pb^{2+} ] = 0 \space M + x$
$$3.2 \times 10^{-10} = \frac{( 2x)^2( x)}{(0.100-2x)^2}$$
3. Solve for x:
x = $9.3 \times 10^{-5}$
$ [ Cr^{3+} ] = 0.100 \space M - 2(9.3 \times 10^{-5}) = 0.100$
$ [ Cr^{2+} ] = 0 \space M + 2(9.3 \times 10^{-5}) = 1.9 \times 10^{-4}$
$ [ Pb^{2+} ] = 0 \space M + (9.3 \times 10^{-5}) = 9.3 \times 10^{-5}$
