Question

300 mL of 0.200 M Pb(NO3)2 is added to 200 mL of 0.0500 M NaCl(aq) at 25°C. Will a precipitate of PbCl2 form at 25 °C? Tip: You have to calculate appropriate Qsp for this system. Ks- 1.7x10 for PbCl2 in water at 25 °C. (You MUST show your work to justify your answer.)

Answer 1

Explanation:

It is given that volume of $Pb(NO_{3})_{2}$ is 300 mL and molarity is 0.200 M.

Volume of NaCl is 200 mL and molarity is 0.050 M.

The chemical reaction will be as follows.

            $PbCl_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^{-}(aq)$

$K_{sp}$ for $PbCl_{2}$ is given as $1.7 \times 10^{-5}$.

As, molarity is number of moles present in liter of solution.

Hence, moles of $Pb^{2+}$(aq) will be calculated as follows.

         moles of $Pb^{2+}$(aq) = $0.300 L \times 0.200 M$

                                                = 0.06 mol

                      $[Pb^{2+}]$ = $\frac{0.06 mol}{0.5 L}$

                                            = 0.120 M

Mole of $Cl^{-}(aq)$ = $0.2 L \times 0.05 M$  

                                  = 0.010 M

Now,   Q = $[Pb^{2+}][Cl^{-}]^{2}$

                  = $0.120 \times (0.010)^{2}$

                  = $1.2 \times 10^{-5}$    

As, Q < $K_{sp}$ hence, there will be no formation of $PbCl_{2}$ precipitate.