## Chemistry (4th Edition)

(a) $[Pb^{2+}] = 1.5 \times 10^{-3}M = 0.0015 M$ $[I^-] = 3.0 \times 10^{-3} M = 0.0030 M$
Reaction: $$PbI_2(s) \leftrightharpoons Pb^{2+}(aq) + 2I^-(aq)$$ 1. Write the equilibrium constant expression: - The exponent of each concentration is equal to its balance coefficient. $$K_c = \frac{[Products]}{[Reactants]} = \frac{[ Pb^{2+} ][ I^- ]^{ 2 }}{1}$$ 2. At equilibrium, these are the concentrations of each compound: $[ Pb^{2+} ] = x$ $[ I^- ] = 2x$ $$1.4 \times 10^{-8} = (x)(2x)^2$$ x = $1.5 \times 10^{-3}$ $[Pb^{2+}] = 1.5 \times 10^{-3}$ $[I^-] = 2(1.5 \times 10^{-3}) = 3.0 \times 10^{-3}$