# If Na_2S is added to a solution containing AgCl, a black precipitate of Ag_2S forms. Why does this happen using Le Chateliers's principle?

Jul 22, 2018

$\text{Ag"_2 "S}$ is less soluble in water than $\text{AgCl}$. The production of $\text{Ag"_2 "S}$ consumes ${\text{Ag}}^{+}$ and drives the disassociation of $\text{AgCl}$ into ions.

#### Explanation:

Silver chloride $\text{AgCl}$ is barely soluble in water. It's aqueous disassociation to its component ions would thus be a reversible process.

${\text{AgCl" (s) rightleftharpoons "Ag"^(+)(aq) + "Cl}}^{-} \left(a q\right)$
${K}_{s p} = 1.77 \times {10}^{- 10}$ at room temperature. (Wikipedia)

Sodium sulfide $\text{Na"_2 "S}$ ionize completely when dissolved in water to form significant quantities of ${\text{Na}}^{+}$ and ${\text{S}}^{2 -}$:

${\text{Na"_2 "S" (aq) to 2 color(white)(l)"Na"^(+) (aq) + "S}}^{2 -} \left(a q\right)$

The silver ion would combine readily with the sulfide ${\text{S}}^{2 -}$ ion to produce silver sulfide, a black precipitate.

$2 \textcolor{w h i t e}{l} \text{Ag"^(+) (aq) + "S"^(2-) (aq) to "Ag"2 "S} \left(s\right)$
${K}_{s p} = \boldsymbol{6.31 \times {10}^{- 50}}$ in water. (Wikipedia)

The production of $\text{Ag"_2"S}$ removes ${\text{Ag}}^{+}$ from the solution system; by the Le Chatlier's principle, the disassociation equilibrium of $\text{AgCl}$ would keep proceeding to the right allowing for the formation of even more $\text{Ag"_2 "S}$. On the other hand, the solubility of $\text{Ag"_2 "S}$- a compound with strong covalent characteristics- is way lower than that of $\text{AgCl}$ meaning that $\text{Ag"_2 "S}$ is weaker an electrolyte. The aqueous system would thus favor the production of $\text{Ag"_2 "S}$ over that of $\text{AgCl}$.

Jul 22, 2018

Consider the solubility expression that governs the solubility of the silver salts....

#### Explanation:

$A g C l \left(s\right) \stackrel{{H}_{2} O}{r} i g h t \le f t h a r p \infty n s A {g}^{+} + C {l}^{-}$

And this is certainly a measurable equilibrium, the which we express as ${K}_{\text{sp}} = \left[A {g}^{+}\right] \left[C {l}^{-}\right]$. This site quotes ${K}_{\text{sp}} \left(A g C l\right) = 1.77 \times {10}^{-} 10$..

And so we know that the ion product is ${K}_{\text{sp}} = \left[A {g}^{+}\right] \left[C {l}^{-}\right] = 1.77 \times {10}^{-} 10$. And if we were so minded, certainly we could quote a solubility in $g \cdot {L}^{-} 1$ for the silver salt under these conditions.

On the other hand for $A {g}_{2} S$, ${K}_{\text{sp}} = {\left[A {g}^{+}\right]}^{2} \left[{S}^{2 -}\right] = 8 \times {10}^{-} 51$; it is hard to believe that this value is legit. Now while silver chloride is insoluble, silver sulfide is even LESS soluble.....

$A {g}_{2} S \left(s\right) \stackrel{{H}_{2} O}{\rightarrow} 2 A {g}^{+} + {S}^{2 -}$

....the equilibrium quantity of silver ion in solution would precipitate out until it satisfied the solubility expression for silver sulfide.

If you are in second year you should work out the gram solubility of each silver salt under standard conditions....

Anyway back to old Le Chatelier's principle...you gots…

$A g C l \left(s\right) \stackrel{{H}_{2} O}{r} i g h t \le f t h a r p \infty n s A {g}^{+} + C {l}^{-}$

But because silver ion is REMOVED from the equilibrium, AS MIGHTILY insoluble silver sulfide, the equilibrium should move from right to left as we face the page. And so MORE silver chloride should dissolve, and silver ion reprecipitates as silver sulfide.