Chapter 15 - Priciples of Chemical Equilibrium - Exercises - Direction and Extent of Chemical Change - Page 691: 44

$[ Cr^{3+} ] = 0.100 \space M$ $[ Cr^{2+} ] = 1.9 \times 10^{-4} \space M$ $[ Pb^{2+} ] = 9.3 \times 10^{-5} \space M$

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 = $9.3 \times 10^{-5}$ $[ Cr^{3+} ] = 0.100 \space M - 2(9.3 \times 10^{-5}) = 0.100$ $[ Cr^{2+} ] = 0 \space M + 2(9.3 \times 10^{-5}) = 1.9 \times 10^{-4}$ $[ Pb^{2+} ] = 0 \space M + (9.3 \times 10^{-5}) = 9.3 \times 10^{-5}$

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