Answer
Since $Q_{sp} \gt K_{sp}$, there will be precipitation of $PbCl_2$.
Work Step by Step
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
