A 1.0 L solution of #AgNO_3# and #Pb(NO_3)_2# has a silver ion concentration of 0.020 M and a lead ion concentration of 0.0010 M. A 0.0010 mol sample of #K_2SO_4# (s) is added to the solution. Which precipitate will form?

The #K_(sp)# of #Ag_2SO_4# is #1*10^(-5)# and that of #PbSO_4# is #1*10^(-8)# at 298 K.

a) no precipitate will form
b) just #Ag_2SO_4#
c) just #PbSO_4#
d) both of them will precipitate

1 Answer
Truong-Son N.
May 6, 2018

You'll have to figure out which reaction quotient is bigger than which #K_(sp)#. When #Q > K_(sp)#, the reaction favors the products and Le Chatelier's principle suggests that the reaction shifts towards the solid reactant. Note that you use the concentrations at face value; don't multiply by any coefficients when evaluating #Q#.

I find just #"PbSO"_4# forms.

We know that #["Ag"^(+)] = "0.020 M"# and #["Pb"^(2+)] = "0.0010 M"# from the metal nitrates. We have that #"0.0010 mols K"_2"SO"_4# was added into #"1.0 L"# of solution, so we obtain #"0.0010 M"# #"SO"_4^(2-)#.

Thus, we may generate #"Ag"_2"SO"_4# and/or #"PbSO"_4# precipitate(s). As always, write the reactions and the mass action expressions.

#"Ag"_2"SO"_4(s) rightleftharpoons color(red)2"Ag"^(+)(aq) + "SO"_4^(2-)(aq)#

#Q_(sp)("Ag"_2"SO"_4) = ["Ag"^(2+)]^color(red)(2)["SO"_4^(2-)]#

#= ("0.020 M")^2("0.0010 M")#

#= 4.0 xx 10^(-7)#

Here we have that #color(blue)(Q_(sp)("Ag"_2"SO"_4) "<<" K_(sp)("Ag"_2"SO"_4))#, and #K_(sp) = 1 xx 10^(-5)#. Therefore, no silver sulfate precipitate forms due to the right shift.

#"PbSO"_4(s) rightleftharpoons "Pb"^(2+)(aq) + "SO"_4^(2-)(aq)#

#Q_(sp)("PbSO"_4) = ["Pb"^(2+)]["SO"_4^(2-)]#

#= ("0.0010 M")("0.0010 M")#

#= 1.0 xx 10^(-6)#

Here we have that #color(blue)(Q_(sp)("PbSO"_4) ">>" K_(sp)("PbSO"_4))#, and #K_(sp) = 1 xx 10^(-8)#. Therefore, lead(II) sulfate precipitate forms due to the left shift.

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