Answer
$Q_{sp} < K_{sp}$
Therefore, there is no precipitation.
Work Step by Step
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.
