Answer
(a)
$[Pb^{2+}] = 1.5 \times 10^{-3}M = 0.0015 M$
$[I^-] = 3.0 \times 10^{-3} M = 0.0030 M$
Work Step by Step
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 = $1.5 \times 10^{-3}$
$[Pb^{2+}] = 1.5 \times 10^{-3}$
$[I^-] = 2(1.5 \times 10^{-3}) = 3.0 \times 10^{-3}$
