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A solution is 0.15 M in both Pb2+ and Ag+. If Cl- is added to this solution, what is [Ag+] when PbCl2 begins to precipitate?

Short Answer

Expert verified

When PbCl2starts to precipitate, the concentration of Ag+ is 1.55.10-8.

Step by step solution

01

Define precipitate

The precipitation of a compound may occur when its concentration exceeds its . This can be due to temperature changes, solvent evaporation, or by mixing solvents. Precipitation occurs more rapidly from a strongly solution.

02

Reaction of the solution

PbCl2(²õ)⇌P²ú2+(aq) + 2Cl- (aq)

AgCl (s)⇌Ag+ (aq) + Cl- (aq)

03

Determining solubility

\(PbC{l_2}:{K_{sp}} = 1.6 \times 1{0^{ - 5}}\)

\({\rm{AgCl}}:{{\rm{K}}_{np}} = 1.6 \cdot {10^{ - 10}}\)

\(\left( {{\rm{A}}{{\rm{g}}^ + }} \right) = 0.15{\rm{M\;and\;}}\left( {{\rm{P}}{{\rm{b}}^{2 + }}} \right) = 0.15{\rm{M}}\)

04

Calculating the concentration of Cl- ions

\begin{aligned}{{K_{sp}}=\left[{{\rm{P}}{{\rm{b}}^{2+}}}\right]{{\left[{{\rm{C}}{{\rm{l}}^-}}\right]}^2}}\\{1.6\cdot{{10}^{-5}}=0.15\cdot{{\left[{{\rm{C}}{{\rm{l}}^-}}\right]}^2}}\\{\left[{{\rm{C}}{{\rm{l}}^-}}\right]=\sqrt{\frac{{1.6\cdot{{10}^{-5}}}}{{0.15}}}}\\{\left[{{\rm{C}}{{\rm{l}}^-}}\right]=1.03\cdot{{10}^{-2}}}\end{aligned}

05

Calculating the concentration of Ag+ ions

\begin{aligned}{{{\rm{K}}_{sp}}=\left[{{\rm{A}}{{\rm{g}}^+}}\right]\left[{{\rm{C}}{{\rm{l}}^-}}\right]}\\{1.6\cdot{{10}^{-10}}=\left[{{\rm{A}}{{\rm{g}}^+}}\right]\cdot1.03\cdot{{10}^{-2}}}\\{\left[{{\rm{A}}{{\rm{g}}^+}}\right]=\frac{{1.6\cdot{{10}^{-10}}}}{{1.03\cdot{{10}^{-2}}}}}\\{\left[{{\rm{A}}{{\rm{g}}^+}}\right]=1.55\cdot {{10}^{-8}}}\end{aligned}

When PbCl2starts to precipitate, the concentration of Ag+ is 1.55.10-8.

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Precipitation and Dissolution

1. Complete the changes in concentrations for each of the following reactions:

\(\begin{array}{l}(a)AgI(s) \to nA{g^ + }(aq) + {I^ - }(aq)\\ x \_ \\(b)CaC{O_3}(s) \to C{a^{2 + }}(aq) + C{O_3}^{2 - }(aq)\\ \_\quad x\\(c)Mg{(OH)_2}(s) \to nM{g^{2 + }}(aq) + 2O{H^ - }(aq)\\ x \quad \_\_\\(d)M{g_3}{\left( {P{O_4}} \right)_2}(s) \to n3M{g^{2 + }}(aq) + 2P{O_4}^{3 - }(aq)\\ x\_\\(e)C{a_5}{\left( {P{O_4}} \right)_3}OH(s) \to n5C{a^{2 + }}(aq) + 3P{O_4}^{3 - }(aq) + O{H^ - }(aq)\\ \_ \_ x\end{array}\)

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