## Chemistry and Chemical Reactivity (9th Edition)

Since $Q_{sp} \gt K_{sp}$, there will be precipitation of $PbCl_2$.
1. Calculate the molar mass $(NaCl)$: 22.99* 1 + 35.45* 1 = 58.44g/mol 2. Calculate the number of moles $(NaCl)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 1.20}{ 58.44}$ $n(moles) = 0.0205$ 3. Find the concentration in mol/L $(NaCl)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.0205}{ 0.095}$ $C(mol/L) = 0.216$ 4. Write the $K_{sp}$ expression: $PbCl_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2Cl^{-}(aq)$ $K_{sp} = [Pb^{2+}]^ 1[Cl^{-}]^ 2$ 5. Find the $Q_{sp}$ value $Q_{sp} = (1.2 \times 10^{-3})^ 1 \times (0.216)^ 2$ $Q_{sp} = (1.2 \times 10^{-3})^ 1 \times (0.216)^ 2$ $Q_{sp} = 1.2 \times 10^{-3} \times (0.0467)$ $Q_{sp} = 5.60 \times 10^{-5}$ $K_{sp} (PbCl_2) = 1.7 \times 10^{-5}$ Since $Q_{sp} \gt K_{sp}$, there will be precipitation