Using the voltage from the following Galvanic cell, calculate #K_(sp)# for #Ag_2SO_4# (s)?
#Pb(s)|Pb^(2+)(1.8M)||Ag^(+)("satd "Ag_2SO_4)|Ag(s)#
#E_(cell)=0.83" V"#
1 Answer
#K_(sp) = 1.03 xx 10^(-5)#
compared to the literature value of around
It looks like the point of this is to:
- Find
#E_(cell)^@# using reference values you should be given. - Use
#E_(cell)# to find#Q_c# for these concentrations, assuming that the silver sulfate precipitate is in equilibrium with its ions already. - 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
#"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
#E_(cell) = E_(cell)^@ - ("0.0592 V")/n log Q_c#
Now we solve algebraically for an expression for
#"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
#"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
#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
