Using the voltage from the following Galvanic cell, calculate #K_(sp)# for #Ag_2SO_4# (s)?

#K_(sp) = 1.03 xx 10^(-5)#

compared to the literature value of around #1.20 xx 10^(-5)#.

It looks like the point of this is to:

1. Find #E_(cell)^@# using reference values you should be given.
2. Use #E_(cell)# to find #Q_c# for these concentrations, assuming that the silver sulfate precipitate is in equilibrium with its ions already.
3. Find #K_(sp)#, knowing that saturated solutions satisfy the condition given in #(2)#.

You should be given:

#E_(red)^@("Pb"^(2+)->"Pb") = -"0.13 V"#
#E_(red)^@("Ag"^(+)->"Ag") = "0.80 V"#

Thus,

#E_(cell)^@ = "0.80 V" - (-"0.13 V") = "0.93 V"#.

The next thing is that we can set up #Q_c#:

#"Pb"(s) + 2"Ag"^(+)(aq) -> "Pb"^(2+)(aq) + 2"Ag"(s)#

#Q_c = (["Pb"^(2+)])/(["Ag"^(+)]^2)#

#= ("1.8 M")/(["Ag"^(+)]^2)#

And now let's put that off to the side, and solve for it last. At #25^@ "C"#, we have the Nernst equation again:

#E_(cell) = E_(cell)^@ - ("0.0592 V")/n log Q_c#

Now we solve algebraically for an expression for #Q_c#.

#"0.0592 V"/nlog Q_c = E_(cell)^@ - E_(cell)#

For now,

#=> log Q_c = n/"0.0592 V"(E_(cell)^@ - E_(cell))#

#" "" "" "" "= ("2 mol e"^(-)/"1 mol atoms")/("0.0592 V") cdot ("0.93 V" - "0.83 V")#

#" "" "" "" "= 3.38#

Therefore,

#Q_c = 10^(3.38) = 2390#

Lastly, we can solve for #["Ag"^(+)]# and therefore #K_(sp)#.

#"2390 M"^(-1) = ("1.8 M")/(["Ag"^(+)]^2)#

#["Ag"^(+)] = sqrt("1.8 M"/("2390 M"^(-1))) = "0.0274 M"#

Two silver cations and one sulfate anion are in the equilibrium:

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

Due to the coefficients above, it must be recognized that:

#ul(["Ag"^(2+)] = 2["SO"_4^(2-)])#

From this, we find the #K_(sp)#:

#color(blue)(K_(sp)) = ["Ag"^(+)]^2["SO"_4^(2-)]#

#= ("0.0274 M")^2("0.0274 M"/2)#

#= color(blue)(1.03 xx 10^(-5))#

This is not that far off from the actual one, yay! The actual value should have been #1.20 xx 10^(-5)#.