Question 16.16: Will a precipitate form when 0.150 L of 0.10 M Pb(NO3)2 and ......
Will a precipitate form when 0.150 L of 0.10 M Pb(NO_{3})_{2} and 0.100 L of 0.20 M NaCl are mixed?
STRATEGY
Since ionic compounds are strong electrolytes, Pb(NO_{3})_{2} and NaCl exist in solution as separate cations and anions. Use the solubility guidelines discussed in Section 4.6 to decide which ions might form a precipitate, and calculate their concentrations after mixing. Then calculate the IP for the possible precipitate and compare it with the value of K_{sp}.
| IDENTIFY | |
| Known | Unknown |
| Molarity and volume of reactants (0.150 L of 0.10 M Pb(NO_{3})_{2}) | IP |
| (0.100 L of 0.20 M NaCl) | |
| K_{sp}=1.2\times 10^{-5} (Appendix C) | |
After the two solutions are mixed, the combined solution contains Pb^{2+} , NO_{3} ^{-}, Na^{+}, and Cl^{-} ions and has a volume of 0.150 L + 0.100 L = 0.250 L. Because sodium salts and nitrate salts are soluble in water, the only compound that might precipitate is PbCl_{2}, which has K_{sp}=1.2\times 10^{-5} (Appendix C). To calculate the value of IP for PbCl_{2}, first calculate the number of moles of Pb^{2+} and Cl^{-} in the combined solution:
Moles Pb^{2+} = (0.150 L)(0.10 mol/L) = 1.5 \times 10^{-2} mol
Moles Cl^{-} = (0.100 L)(0.20 mol/L) = 2.0 \times 10^{-2} mol
Then convert moles to molar concentrations:
[Pb^{2+}]=\frac{1.5 \times 10^{-2} mol}{0.250 L}=6.0\times 10^{-2} M[Cl^{-}]=\frac{2.0 \times 10^{-2} mol}{0.250 L}=8.0\times10^{-2} M
The ion product is
IP = [Pb^{2+}]_{t}[Cl^{-}]_{t} ^{2}=(6.0\times 10^{-2})(8.0\times10^{-2})^{2}=3.8\times 10^{-4}
Since K_{sp}=1.2\times 10^{-5}, IP is greater than K_{sp} and PbCl_{2} will precipitate.