Chemistry: The Molecular Science (5th Edition)

$Q_{sp} < K_{sp}$ Therefore, there is no precipitation.
1. Write the $K_{sp}$ expression: $AgCl(s) \lt -- \gt 1Ag^{+}(aq) + 1Cl^{-}(aq)$ $K_{sp} = [Ag^{+}]^ 1[Cl^{-}]^ 1$ 2. Calculate the $Q_{sp}$: $Q_{sp} = (1.0 \times 10^{-5})^ 1 \times (1.0 \times 10^{-5})^ 1$ $Q_{sp} = (1.0 \times 10^{-5}) \times (1.0 \times 10^{-5})$ $Q_{sp} = (1.0 \times 10^{-10})$ 3. Compare this value to the $K_{sp}$: $1.0 \times 10^{-10} < 1.8 \times 10^{-10}$ $Q_{sp} < K_{sp}$ Therefore, there is no precipitation.