What will be the result when 15.0 mL of 0.040 M lead(II) nitrate is mixed with 15.0 mL of 0.040 M sodium chloride? 

${\mathrm{PbCl}}_2\left(s\right)⟶{\mathrm{Pb}}^{2+}\left(aq\right)+2{\mathrm{Cl}}^{-}\left(aq\right)\left[K_{sp}=1.7\times {10}^{-5}\right]$

Solid ${\mathrm{PbCl}}_2$will precipitate and excess ${\mathrm{Pb}}^{2+}$ ions will remain in solution.

Solid ${\mathrm{PbCl}}_2$ will precipitate and excess ${\mathrm{Cl}}^{-}$ ions will remain in solution.

A clear solution with no precipitate will result

Solid  ${\mathrm{PbCl}}_2$ will precipitate and no excess ions will remain in solution.