What is the maximum #["CO"_3^(2-)]# that can exist in a saturated solution of AgBr?

This problem is why you have to always check before or after you do the small #s# approximation to see if it worked... I got #1.28 xx 10^(-4) "M"# the long way, and #"15.74 M"# the short way.

Clearly, the short way didn't agree.

First, we'll need to find the molar concentration of #"Ag"^(+)# in solution. That gives rise to the common ion effect, which limits how much #"Ag"_2"CO"_3# can dissolve since it also dissolves to give #"Ag"^(+)# in solution.

A saturated #"AgBr"# solution satisfies the solubility product equilibrium:

#"AgBr"(s) rightleftharpoons "Ag"^(+)(aq) + "Br"^(-)(aq)#

#"I"" "-" "" "" "" "0" "" "" "" "0#
#"C"" "-" "" "" "+s" "" "" "+s#
#"E"" "-" "" "" "" "s" "" "" "" "s#

where #s# is defined to be the solubility in pure water for the ion with a stoichiometric coefficient of #1#.

Note that had the stoichiometric coefficient been #nu ne 1#, the concentration would have changed by #nus#. The solubility product constant expression then gives:

#K_(sp) = 5.4 xx 10^(-13) = ["Ag"^(+)]["Br"^(-)] = s^2#

Thus,

#s = sqrt(5.4 xx 10^(-13)) = 7.35 xx 10^(-7) "M" = ["Ag"^(+)]_1#

So, this becomes our initial concentration of #"Ag"^(+)# while dissolving #"Ag"_2"CO"_3# next. (We can do this because concentration is a state function.)

#"Ag"_2"CO"_3(s) " "rightleftharpoons" " 2"Ag"^(+)(aq) " "+" " "CO"_3^(2-)(aq)#

#"I"" "-" "" "" "" "" "" "7.35 xx 10^(-7)" "" "" "" "0#
#"C"" "-" "" "" "" "" "+2s'" "" "" "" "" "" "+s'#
#"E"" "-" "" "" "" "" "7.35 xx 10^(-7) + 2s'" "" "s'#

where #s'# is defined to be the solubility in ion-containing water for the ion with a stoichiometric coefficient of #1#.

In this case, we CANNOT make the small #s# approximation. If we did... we would get a very unreasonable answer.

#K_(sp) = 8.5 xx 10^(-12) = ["Ag"^(+)]^2["CO"_3^(2-)]#

#= (7.35 xx 10^(-7) + 2s')^2(s')#

#stackrel(?" ")(~~) (7.35 xx 10^(-7))^2s'#

That would have given:

#color(red)(s') = ["CO"_3^(2-)] = (8.5 xx 10^(-12))/(7.35 xx 10^(-7))^2#

#=# #color(red)"15.74 M"#

which makes no sense whatsoever! Concentrated nitric acid is about #"16 M"#. Concentrated hydrochloric acid is about #"12 M"#. Concentrated sulfuric acid is around #"18 M"#.

This unreasonable answer probably arose because the two #K_(sp)# values are very similar and thus the two substances are interacting at a non-negligible level.

Let's NOT do that approximation in this case, and solve for the cubic.

#8.5 xx 10^(-12) = (7.35 xx 10^(-7) + 2s')^2(s')#

#= (5.40 xx 10^(-13) + 4cdot7.35 xx 10^(-7)s' + 4s'^2)s'#

#= 5.40 xx 10^(-13)s' + 2.940 xx 10^(-6)s'^2 + 4s'^3#

The final cubic becomes:

#4s'^3 + 2.940 xx 10^(-6)s'^2 + 5.40 xx 10^(-13)s' - 8.5 xx 10^(-12) = 0#

The numerical solution to this cubic can be done in any modern TI calculator to give a physically reasonable answer of:

#color(blue)(s' = ["CO"_3^(2-)] = 1.28 xx 10^(-4) "M")#

And if we check in #K_(sp)#, we should find that this worked...

#8.5 xx 10^(-12) stackrel(?" ")(=) ["Ag"^(+)]^2["CO"_3^(2-)]#

#= (["Ag"^(+)]_1 + 2s')^2(s')#

#= (7.35 xx 10^(-7) + 2 cdot 1.28 xx 10^(-4))^2(1.28 xx 10^(-4))#

#= 8.44 xx 10^(-12) ~~ 8.5 xx 10^(-12)# #color(blue)(sqrt"")#

Yep, it worked!