# #"AgCl"# has #"pK"_(sp) = 9.744#, #"H"_2"S"# has #"pK"_a = 7.02#, #"HS"^(-)# has #"pK"_a = 13.9#, and #"Ag"("NH"_3)_2^(2+)# has #"pK"_f = -7.22#. Find the solubility of #"Ag"_2"S"# in water, #"0.1 M NH"_3#, and #"0.1 M HCl"#?

##### 1 Answer

Well... here's my best attempt at this. I got:

#s_(Ag_2S)("aqueous soln") ~~ 2.51 xx 10^(-17)# #"M"#

#s_(Ag_2S)("0.1 M NH"_3) ~~ 7.57 xx 10^(-14)# #"M"#

#s_(Ag_2S)("0.1 M HCl") ~~ 1.25 xx 10^(-14)# #"M"#

As it turns out, we can conclude that adding

**DISCLAIMER:** *LONG ANSWER!*

**SOLUBILITY IN AQUEOUS SOLUTION**

First, we find the solubility of

We would have written out the equilibria of

#"AgCl"(s) rightleftharpoons "Ag"^(+)(aq) + "Cl"^(-)(aq)# ,#" "K_(sp) = 10^(-9.744)#

#"H"_2"S"(aq) rightleftharpoons "HS"^(-)(aq) + "H"^(+)(aq)# ,#" "K_a = 10^(-7.02)#

#"HS"^(-)(aq) rightleftharpoons "S"^(2-)(aq) + "H"^(+)(aq)# ,#" "K_a = 10^(-13.9)#

But what we wanted this whole time was the

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

This would have required a hassle of Hess's law and equilibrium constant properties. But to save us the agony of doing this, for we are lazy and can clearly look this up in this modern day and age...

If you look up the

#K_(sp) = 6.31 xx 10^(-50)#

And thus, its "solubility"

#6.31 xx 10^(-50) = ["Ag"^(+)]^2["S"^(2-)]#

#-= (2s)^2s = 4s^3#

#=> s = ((6.31 xx 10^(-50))/4)^(1//3) = ul(2.51 xx 10^(-17) "M")#

That becomes our reference point for reality checks. That aside...

**SOLUBILITY IN 0.1 M AMMONIA**

#"Ag"_2"S"(s) rightleftharpoons cancel(2"Ag"^(+)(aq)) + "S"^(2-)(aq)#

#2(cancel("Ag"^(+)(aq)) + 2"NH"_3(aq) rightleftharpoons "Ag"("NH"_3)_2^(+)(aq))#

#"--------------------------------------------------------------"#

#"Ag"_2"S"(s) + 4"NH"_3(aq) rightleftharpoons "S"^(2-)(aq) + 2"Ag"("NH"_3)_2^(+)(aq)#

The resultant equilibrium has the **net equilibrium constant**:

#K_(n et) = K_(sp,Ag_2S) cdot (K_(f,Ag(NH_3)_2^(+)))^2#

#= 6.31xx10^(-50) cdot (10^(7.22))^2 = 1.738 xx 10^(-35)#

And this should convince us to use the following ICE table to not get lost:

#"Ag"_2"S"(s) + 4"NH"_3(aq) rightleftharpoons "S"^(2-)(aq) + 2"Ag"("NH"_3)_2^(+)(aq)#

#"I"" "-" "" "" ""0.1 M"" "" "" ""0 M"" "" "" ""0 M"#

#"C"" "-" "" "-4s" "" "" "" "+s" "" "+2s#

#"E"" "-" "" "0.1 - 4s" "" "" "s" "" "" "" "2s#

#1.738 xx 10^(-35) = (["S"^(2-)]["Ag"("NH"_3)_2^(+)]^2)/(["NH"_3]^4)#

#= (4s^3)/(0.1 - 4s)^4#

We can clearly make a small

#1.738 xx 10^(-35) = (4s^3)/(0.1^4)#

#=> color(blue)(ul(s = 7.57 xx 10^(-14) "M")) -= ["S"^(2-)]#

And since

- The formation of
#"Ag"("NH"_3)_2^(+)# *decreases*#["Ag"^(+)]# , and*increases*the solubility of#"Ag"_2"S"# from Le Chatelier's Principle.

**SOLUBILITY IN 0.1 M HCl**

As before, we write the equilibrium that would occur. In this case, we add

#"Ag"_2"S"(s) rightleftharpoons cancel(2"Ag"^(+)(aq)) + cancel("S"^(2-)(aq))# #" "K_(sp) = 6.31 xx 10^(-50)#

#2(cancel("Ag"^(+)(aq)) + "Cl"^(-)(aq) rightleftharpoons cancel("AgCl"(s)))# ,#" "(K_(sp)^(-1))^2 = 10^(+9.744 cdot 2)#

#2(cancel("AgCl"(s)) + "Cl"^(-)(aq) rightleftharpoons "AgCl"_2^(-)(aq))# ,#" "K_(f)^2 = 10^(-4.50 cdot 2)#

#cancel("S"^(2-)(aq)) + "H"^(+)(aq) rightleftharpoons "HS"^(-)(aq)# ,#" "K_b = 10^(-14)/(10^(-13.9)) = 0.7943#

#"---------------------------------------------------------------------"#

#"Ag"_2"S"(s) + 4"Cl"^(-)(aq) + "H"^(+)(aq) rightleftharpoons 2"AgCl"_2^(-)(aq) + "HS"^(-)(aq)#

Combine the first with twice the second and third reactions, together with the fourth, to get:

#K_(n et) = 6.31 xx 10^(-50) cdot ((10^(-9.744))^(-1))^2 cdot (10^(-4.50))^2 cdot 0.7943#

#= 1.542 xx 10^(-39)#

#= (["AgCl"_2^(-)]^2["HS"^(-)])/(["Cl"^(-)]^4["H"^(+)])#

The ICE table that goes with this would be...

#"Ag"_2"S"(s) + 4"Cl"^(-)(aq) + "H"^(+)(aq) rightleftharpoons 2"AgCl"_2^(-)(aq) + "HS"^(-)(aq)#

#"I"" "-" "" "" ""0.4 M"" "" "" ""0.1 M"" "" ""0 M"" "" "" "" ""0 M"#

#"C"" "-" "" "" "-4s" "" "" "-s" "" "+2s" "" "" "" "+s#

#"E"" "-" "" "" "0.4 - 4s" "" "0.1 - s" "" "2s" "" "" "" "" "s#

#1.542 xx 10^(-39) = ((2s)^2(s))/((0.4 - 4s)(0.1 - s))#

Clearly, we can make the small

#1.542 xx 10^(-39) = ((2s)^2(s))/((0.4)(0.1))#

#= (4s^3)/(0.04)#

#=> ul(s = 2.5 xx 10^(-14) "M"#

This is the solubility of

#"HS"^(-)(aq) rightleftharpoons "H"^(+)(aq) + "S"^(2-)(aq)#

#10^(-13.9) = (s^2)/(2.5 xx 10^(-14) - s)#

Here, we cannot make the small

#color(blue)(ul(s = 1.253 xx 10^(-14) "M"))# for#"S"^(2-)# ,and thus for

#"Ag"_2"S"# by stoichiometry.

- The
#"AgCl"_2^(-)# formation, with#beta_2 = K_(f1)K_(f2) = 10^(14.244)# due to the addition of#"Cl"^(-)# from#"HCl"# *decreases*#["Ag"^(+)]# and*increases*the solubility of#"Ag"_2"S"# from Le Chatelier's Principle. - The
#"HS"^(-)# dissociation, with#K_a = 10^(-13.9)# , due to the addition of#"H"^(+)# from#"HCl"# *increases*#["S"^(2-)]# ,*decreasing*the solubility of#"Ag"_2"S"# from Le Chatelier's Principle.

So, it probably makes sense that

(At the very least, our answers make physical sense, in that they are both larger than the solubility without