Answer
d) $$K_{sp} = 3.4 \times 10^{-6}$$
Work Step by Step
For the reaction: $$A_2X(s) \leftrightharpoons 2A^+(aq)+X^{2-}(aq)$$
1. Write the equilibrium constant expression:
- The exponent of each concentration is equal to its balance coefficient.
$$K_c = \frac{[Products]}{[Reactants]} = \frac{[ A^+ ]^{ 2 }[ X^{2-} ]}{1}$$
2. At equilibrium, these are the concentrations of each compound:
$[ A^+ ] = 2x$
$[ X^{2-} ] = x$
$0.019 = 2x \longrightarrow x = 0.0095$
$[X^{2-}] = 0.0095$
$$K_{sp} = (0.019)^2(0.0095) = 3.4 \times 10^{-6}$$
