# "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"?

Aug 8, 2017

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

${s}_{A {g}_{2} S} \left(\text{aqueous soln}\right) \approx 2.51 \times {10}^{- 17}$ $\text{M}$

${s}_{A {g}_{2} S} \left({\text{0.1 M NH}}_{3}\right) \approx 7.57 \times {10}^{- 14}$ $\text{M}$

${s}_{A {g}_{2} S} \left(\text{0.1 M HCl}\right) \approx 1.25 \times {10}^{- 14}$ $\text{M}$

As it turns out, we can conclude that adding $\text{0.1 M}$ ${\text{NH}}_{3}$ makes silver sulfide more soluble than adding $\text{0.1 M}$ $\text{HCl}$ does...

SOLUBILITY IN AQUEOUS SOLUTION

First, we find the solubility of $\text{Ag"_2"S}$ by itself, for perspective and comparison. Unfortunately, we are not given its ${K}_{s p}$, so we would have had to improvise with the given ${\text{pK}}_{s p}$ and ${\text{pK}}_{a}$.

We would have written out the equilibria of $\text{AgCl} \left(s\right)$, $\text{H"_2"S} \left(a q\right)$, and ${\text{HS}}^{-} \left(a q\right)$.

${\text{AgCl"(s) rightleftharpoons "Ag"^(+)(aq) + "Cl}}^{-} \left(a q\right)$, $\text{ } {K}_{s p} = {10}^{- 9.744}$
${\text{H"_2"S"(aq) rightleftharpoons "HS"^(-)(aq) + "H}}^{+} \left(a q\right)$, $\text{ } {K}_{a} = {10}^{- 7.02}$
${\text{HS"^(-)(aq) rightleftharpoons "S"^(2-)(aq) + "H}}^{+} \left(a q\right)$, $\text{ } {K}_{a} = {10}^{- 13.9}$

But what we wanted this whole time was the ${K}_{s p}$ for...

${\text{Ag"_2"S"(s) rightleftharpoons 2"Ag"^(+)(aq) + "S}}^{2 -} \left(a q\right)$

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 ${K}_{s p}$, you would find that it is a brick.

${K}_{s p} = 6.31 \times {10}^{- 50}$

And thus, its "solubility" $s$ would be found as:

$6.31 \times {10}^{- 50} = \left[{\text{Ag"^(+)]^2["S}}^{2 -}\right]$

$\equiv {\left(2 s\right)}^{2} s = 4 {s}^{3}$

$\implies s = {\left(\frac{6.31 \times {10}^{- 50}}{4}\right)}^{1 / 3} = \underline{2.51 \times {10}^{- 17} \text{M}}$

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

SOLUBILITY IN 0.1 M AMMONIA

a)

${\text{Ag}}^{+}$ complexes with ${\text{NH}}_{3}$ to form "Ag"("NH"_3)_2^(+). You've given that its ${\text{pK}}_{f} = - 7.22$, so its formation constant is ${K}_{f} = {10}^{7.22}$. We should thus expect the solubility of $\text{Ag"_2"S}$ to greatly increase, as a LOT of ${\text{Ag}}^{+}$ is consumed.

${\text{Ag"_2"S"(s) rightleftharpoons cancel(2"Ag"^(+)(aq)) + "S}}^{2 -} \left(a q\right)$
$2 \left({\cancel{{\text{Ag"^(+)(aq)) + 2"NH"_3(aq) rightleftharpoons "Ag"("NH}}_{3}}}_{2}^{+} \left(a q\right)\right)$
$\text{--------------------------------------------------------------}$
"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:

${K}_{n e t} = {K}_{s p , A {g}_{2} S} \cdot {\left({K}_{f , A g {\left(N {H}_{3}\right)}_{2}^{+}}\right)}^{2}$

$= 6.31 \times {10}^{- 50} \cdot {\left({10}^{7.22}\right)}^{2} = 1.738 \times {10}^{- 35}$

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)

$\text{I"" "-" "" "" ""0.1 M"" "" "" ""0 M"" "" "" ""0 M}$
$\text{C"" "-" "" "-4s" "" "" "" "+s" "" } + 2 s$
$\text{E"" "-" "" "0.1 - 4s" "" "" "s" "" "" "" } 2 s$

$1.738 \times {10}^{- 35} = \left({\left[{\text{S"^(2-)]["Ag"("NH"_3)_2^(+)]^2)/(["NH}}_{3}\right]}^{4}\right)$

$= \frac{4 {s}^{3}}{0.1 - 4 s} ^ 4$

We can clearly make a small $s$ approximation, so that $4 s \approx 0$ in the denominator. That gives:

$1.738 \times {10}^{- 35} = \frac{4 {s}^{3}}{{0.1}^{4}}$

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

And since ${\text{S}}^{2 -}$ is $1 : 1$ with $\text{Ag"_2"S}$, this is also the solubility of $\text{Ag"_2"S}$. This indeed has increased relative to no ${\text{0.1 M NH}}_{3} \left(a q\right)$, by a factor of about $1000$.

• The formation of "Ag"("NH"_3)_2^(+) decreases $\left[{\text{Ag}}^{+}\right]$, and increases the solubility of $\text{Ag"_2"S}$ from Le Chatelier's Principle.

SOLUBILITY IN 0.1 M HCl

b)

As before, we write the equilibrium that would occur. In this case, we add $\text{0.1 M}$ $\text{HCl}$ to react with the ${\text{Ag}}^{+}$ and the ${\text{S}}^{2 -}$. We assume 100% dissociation of the $\text{HCl}$, and expect that it increases the solubility of $\text{Ag"_2"S}$.

"Ag"_2"S"(s) rightleftharpoons cancel(2"Ag"^(+)(aq)) + cancel("S"^(2-)(aq)) $\text{ } {K}_{s p} = 6.31 \times {10}^{- 50}$
$2 \left(\cancel{\text{Ag"^(+)(aq)) + "Cl"^(-)(aq) rightleftharpoons cancel("AgCl} \left(s\right)}\right)$, $\text{ } {\left({K}_{s p}^{- 1}\right)}^{2} = {10}^{+ 9.744 \cdot 2}$
2(cancel("AgCl"(s)) + "Cl"^(-)(aq) rightleftharpoons "AgCl"_2^(-)(aq)), $\text{ } {K}_{f}^{2} = {10}^{- 4.50 \cdot 2}$
cancel("S"^(2-)(aq)) + "H"^(+)(aq) rightleftharpoons "HS"^(-)(aq), $\text{ } {K}_{b} = {10}^{- 14} / \left({10}^{- 13.9}\right) = 0.7943$
$\text{---------------------------------------------------------------------}$
${\text{Ag"_2"S"(s) + 4"Cl"^(-)(aq) + "H"^(+)(aq) rightleftharpoons 2"AgCl"_2^(-)(aq) + "HS}}^{-} \left(a q\right)$

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

${K}_{n e t} = 6.31 \times {10}^{- 50} \cdot {\left({\left({10}^{- 9.744}\right)}^{- 1}\right)}^{2} \cdot {\left({10}^{- 4.50}\right)}^{2} \cdot 0.7943$

$= 1.542 \times {10}^{- 39}$

$= \left(\left[{\text{AgCl"_2^(-)]^2["HS"^(-)])/(["Cl"^(-)]^4["H}}^{+}\right]\right)$

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

${\text{Ag"_2"S"(s) + 4"Cl"^(-)(aq) + "H"^(+)(aq) rightleftharpoons 2"AgCl"_2^(-)(aq) + "HS}}^{-} \left(a q\right)$

$\text{I"" "-" "" "" ""0.4 M"" "" "" ""0.1 M"" "" ""0 M"" "" "" "" ""0 M}$
$\text{C"" "-" "" "" "-4s" "" "" "-s" "" "+2s" "" "" "" } + s$
$\text{E"" "-" "" "" "0.4 - 4s" "" "0.1 - s" "" "2s" "" "" "" "" } s$

$1.542 \times {10}^{- 39} = \frac{{\left(2 s\right)}^{2} \left(s\right)}{\left(0.4 - 4 s\right) \left(0.1 - s\right)}$

Clearly, we can make the small $s$ approximation here as well. Thus, $4 s \approx s \approx 0$, and:

$1.542 \times {10}^{- 39} = \frac{{\left(2 s\right)}^{2} \left(s\right)}{\left(0.4\right) \left(0.1\right)}$

$= \frac{4 {s}^{3}}{0.04}$

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

This is the solubility of ${\text{HS}}^{-}$, though, and we want it in terms of ${\text{S}}^{2 -}$. So, we briefly go back to the ${\text{HS}}^{-}$ equilibrium:

${\text{HS"^(-)(aq) rightleftharpoons "H"^(+)(aq) + "S}}^{2 -} \left(a q\right)$

${10}^{- 13.9} = \frac{{s}^{2}}{2.5 \times {10}^{- 14} - s}$

Here, we cannot make the small $s$ approximation since ${K}_{a}$ is bigger than the concentration. We solve this in full to get...

$\textcolor{b l u e}{\underline{s = 1.253 \times {10}^{- 14} \text{M}}}$ for ${\text{S}}^{2 -}$,

and thus for $\text{Ag"_2"S}$ by stoichiometry.

• The ${\text{AgCl}}_{2}^{-}$ formation, with ${\beta}_{2} = {K}_{f 1} {K}_{f 2} = {10}^{14.244}$ due to the addition of ${\text{Cl}}^{-}$ from $\text{HCl}$ decreases $\left[{\text{Ag}}^{+}\right]$ and increases the solubility of $\text{Ag"_2"S}$ from Le Chatelier's Principle.
• The ${\text{HS}}^{-}$ dissociation, with ${K}_{a} = {10}^{- 13.9}$, due to the addition of ${\text{H}}^{+}$ from $\text{HCl}$ increases $\left[{\text{S}}^{2 -}\right]$, decreasing the solubility of $\text{Ag"_2"S}$ from Le Chatelier's Principle.

So, it probably makes sense that ${s}_{A {g}_{2} S} \left({\text{0.1 M HCl") < s_(Ag_2S)("0.1 M NH}}_{3}\right)$... the second bullet point competes with the first, and from a cursory look, it is approximately by less than an order of magnitude.

(At the very least, our answers make physical sense, in that they are both larger than the solubility without ${\text{NH}}_{3}$ and $\text{HCl}$.)