Question

Solubility Equilibria Regarding `AgCl(s)` and Complex-Ion Formation `Ag(NH_3)_2^+` Question?

Silver chloride, (`K_(sp) = 1.8*10^-10`), can be dissolved in solutions containing ammonia due to the formation of the soluble complex ion `Ag(NH_3)_2^+` (`K_f = 1.0*10^8`). What is the minimum amount of `NH_3` that would need to be added to dissolve `0.010mol` of `AgCl` in `1.00L` of solution?

I've added the equilibria together into a solubility equation, but can't reason how the answer is `0.095mol`.

Answer 1

If in the end you manage to dissolve `"0.010 M"` of `"Cl"^(-)` in solution for your first equilibrium, and the two sequential reactions occurring are

`"AgCl"(s) rightleftharpoons "Ag"^(+)(aq) + "Cl"^(-)(aq)` `" "bb((1))`

`K_(sp) = ["Ag"^(+)]["Cl"^(-)]`

`"Ag"^(+)(aq) + 2"NH"_3(aq) -> "Ag"("NH"_3)_2^(+)(aq)` `" "bb((2))`

`K_f = (["Ag"("NH"_3)_2^(+)])/(["Ag"^(+)]["NH"_3]^2)`,

then these add to be:

`"AgCl"(s) + 2"NH"_3(aq) -> "Cl"^(-)(aq) + "Ag"("NH"_3)_2^(+)(aq)`

and the net equilibrium constant `beta` would be the product.

`beta = K_(sp)K_f = 1.8 xx 10^(-10) cdot 1.0 xx 10^8 = 1.8 xx 10^(-2)`

`= cancel(["Ag"^(+)])["Cl"^(-)] cdot (["Ag"("NH"_3)_2^(+)])/(cancel(["Ag"^(+)])["NH"_3]^2)`

`= (["Ag"("NH"_3)_2^(+)]["Cl"^(-)])/(["NH"_3]^2)`

The ICE table for the net reaction could then look like:

`"AgCl"(s) + 2"NH"_3(aq) -> "Cl"^(-)(aq) + "Ag"("NH"_3)_2^(+)(aq)`

`"I"" "-" "" "" "["NH"_3]_i" "" ""0 M"" "" "" ""0 M"`
`"C"" "-" "" "-2s" "" "" "+s" "" "" "+s`
`"E"" "-" "["NH"_3]_i - 2s" ""0.010 M"" "" "" "s`

where `["NH"_3]_i` is the initial concentration of ammonia added to cause `"0.010 M"` of `"Cl"^(-)` to dissolve in solution from the initially added `"AgCl"(s)`.

Note that the `"0.010 M"` is after adding `"NH"_3(aq)`, *so it is assigned to the FINAL concentration, and not the initial.*

The initial, whatever it would have been without ammonia, would have been much less than the `"Cl"^(-)` that does dissociate from `"AgCl"(s)` from addition of `"NH"_3(aq)`.

At this point, you know `s`, the solubility, is `"0.010 M"` of `"Cl"^(-)`, so `2s = "0.020 M"` and you can then set up the mass action expression:

`1.8 xx 10^(-2) = ("0.010 M")^2/(["NH"_3]_i - "0.020 M")^2`

This is a perfect square, so:

`0.1342 = ("0.010 M")/(["NH"_3]_i - "0.020 M")`

Flipping the fraction gives:

`7.454 = (["NH"_3]_i - "0.020 M")/("0.010 M")`

And solving for the `"NH"_3` you added gives:

`color(blue)(["NH"_3]_i = "0.0946 M")`

And since this was in `"1.00 L"` solution, you added `"0.095 mols"` of `"NH"_3(aq)` to two sig figs.

Answer 2

`AgCl(s) rightleftharpoons Ag^(+) + Cl^(-)` with `K_(sp) = 1.8*10^-10`
`Ag^(+)(aq) + 2NH_3(aq) rightleftharpoons Ag(NH_3)_2^(+)(aq)` with `K_f = 1.0*10^8`

`AgCl(s) + 2NH_3(aq) rightleftharpoons Ag(NH_3)^+ + Cl^(-)` where,

`K_(sp) * K_f = ([Ag(NH_3)^+][Cl^-])/([NH_3]^2) approx 1.8*10^-2`

`1.8*10^-2 = (0.01x)/(2x)^2`
`0.072x^2-0.01x= 0`
`therefore x = 0.13`

Answer 3

I get 0.095 mol/L.

Explanation:

You want to dissolve 0.01 mol of `"AgCl"`, so you will have 0.010 mol of `"Ag"("NH"_3)_2^"+"` and 0.010 mol of `"Cl"^"-"` at equilibrium. The initial amount of `"NH"_3` is unknown.

The equilibrium reactions and the ICE table will look like this:

`color(white)(mmmmmm)"AgCl(s)"color(white)(mmmmmmll) ⇌ "Ag"^"+""(aq)" +color(white)(m) "Cl"^"-""(aq)"; K_text(sp)`
`ul(color(white)(mmmmmm)"Ag"^"+""(aq)" + "2NH"_3"(aq)" ⇌ "Ag"("NH"_3)_2^"+" + "Cl"^"-""(aq)"; K_text(f))`
`color(blue)(color(white)(mmmmmm)"AgCl(s)" color(white)(l)+ "2NH"_3"(aq)" ⇌ "Ag"("NH"_3)_2^"+" + "Cl"^"-""(aq)"; K_text(oa) = K_text(sp)K_text(f))`

`"I/mol·L"^"-1":color(white)(mmmmmmmml)xcolor(white)(mmmmmml)0color(white)(mmmmm)0`
`"C/mol·L"^"-1":color(white)(mmmmmmll)"-0.020"color(white)(mmmll)"+0.010"color(white)(mm)"+0.010"`
`"E/mol·L"^"-1":color(white)(mmmmmm)x"-0.020"color(white)(mmmml)0.010color(white)(mmm)0.010`

`K_text(oa) = K_text(sp)K_text(f) = 1.8 × 10^"-10" × 1.0 × 10^8 = 0.018`

`K_text(oa) = ([ "Ag"("NH"_3)_2^"+"]["Cl"^"-"])/ (["NH"_3]^2) = 0.010^2/(x"-0.020")^2= 0.018`

`0.010/(x"-0.020") = 0.134`

`0.010 = 0.134(x"-0.020") = 0.134x - 0.00268`

`x = (0.010 + 0.00268)/0.134 = 0.01268/0.134= 0.095`

Check:

`0.010^2/("0.095 - 0.020")^2 = 0.00010/0.075^2 = 0.00010/0.00562 = 0.018`

It checks!