Question

1 liter of solution contains 0.020mol `Cd^(2+)` , 0.050mol `Cu^(+)` and 0.40mol KCN. Will CdS and `Cu_2S` precipitate if we add 0.0010mol `S^(2- )` to the solution?

Please explain what is happening in the solution before and after adding `S^(2-)` .
I have the calculus in my book, but dont understand what is happening in the solution exactly.

Answer 1

WARNING! Long answer! A precipitate of `"CdS"` will form.

Explanation:

`bb("Before adding S"^"2-")`

The initial solution contains `"Cd"^"2", "Cu"^"+"`, and `"CN"^"-"`.

These will react to form the complex ions `"Cd(CN)"_4^"2-"; K_text(f) = 6.0 × 10^18` and `"Cu(CN)"_4^"3-"; K_text(f) = 2.0 × 10^30`.

We must calculate the concentrations of all species in solution.

Because the formation constants are so large, essentially all the `"Cd"^"2+"` and `"Cu"^"+"` will be converted to their complex ions.

Thus, we will have 0.020 mol/L `"Cd(CN)"_4^"2-"` and 0.050 mol/L `"Cu(CN)"_4^"3-"`.

`["CN"^"-"]` will decrease by 0.080 mol/L in forming the `"Cd"` complex and by 0.10 mol/L in forming the `"Cu"` complex.

At equilibrium, `["CN"^"-"] = "0.40 mol/L - 0.28 mol/L" = "0.12 mol/L"`

`color(white)(mmmmmmm)"Cd"^"2+" + "4CN"^"-" ⇌ "Cd(CN)"_4^"2-"`
`"I/mol·L"^"-1":color(white)(mll)0.020color(white)(mll)0.40color(white)(mmmml)0`
`"C/mol·L"^"-1":color(white)(m)"-0.020"color(white)(ml)"-0.28"color(white)(mml)"+0.020"`
`"E/mol·L"^"-1":color(white)(mml)xcolor(white)(mmll)0.12color(white)(mmml)0.020`

`K_text(f) = (["Cd"("CN")_4^"2-"])/(["Cd"^"2+"]["CN"^"-"]^4) = 0.020/(["Cd"^"2+"] × 0.12^4) = 6.0 × 10^18`

`["Cd"^"2+"] = 0.020/(0.12^4 × 6.0 × 10^18) = 1.61 × 10^"-17" color(white)(l)"mol/L"`

`color(white)(mmmmmmml)"Cu"^"+" + "4CN"^"-" ⇌ "Cu(CN)"_4^"3-"`
`"I/mol·L"^"-1":color(white)(mll)0.050color(white)(mll)0.40color(white)(mmmml)0`
`"C/mol·L"^"-1":color(white)(m)"-0.050"color(white)(ml)"-0.28"color(white)(mml)"+0.050"`
`"E/mol·L"^"-1":color(white)(mml)xcolor(white)(mmll)0.12color(white)(mmml)0.050`

`K_text(f) = (["Cu"("CN")_4^"3-"])/(["Cu"^"+"]["CN"^"-"]^4) = 0.050/(["Cu"^"+"] × 0.12^4) = 2.0 × 10^30`

`["Cu"^"2+"] = 0.050/(0.12^4 ×2.0 × 10^30) = 5.18 × 10^"-36" color(white)(l)"mol/L"`

`bb("After adding S"^"2-")`

Will we get a precipitate of `"CdS" (K_text(sp) = 1 × 10^"-27")`?

`color(white)(mmmmmm)"CdS"color(white)(m) ⇌color(white)(m)"Cd"^"2+" color(white)(m)+ color(white)(ml)"S"^"2-"`
`"I/mol·L"^"-1":color(white)(mmmmml)1.61 × 10^"-17"color(white)(m)0.0010`

`Q_text(sp) = ["Cd"^"2+"]["S"^"2-"] = 1.61 × 10^"-17" × 0.0010 = 1.61 × 10^"-20"`

`Q_text(sp) > K_text(sp)`, so a precipitate of `"CdS"` will form.

Will we get a precipitate of `"Cu"_2"S" (K_text(sp) = 2.0 × 10^"-47")`?

`color(white)(mmmmmm)"Cu"_2"S"color(white)(m) ⇌color(white)(m)"2Cu"^"+" color(white)(m)+ color(white)(ml)"S"^"2-"`
`"I/mol·L"^"-1":color(white)(mmmmmll)5.18 × 10^"-36"color(white)(mm)0.0010`

`Q_text(sp) = ["Cu"^"+"]^2["S"^"2-"] = (5.18 × 10^"-36")^2 × 0.0010 = 2.67 × 10^"-74"`

`Q_text(sp) < K_text(sp)`, so a precipitate of `"Cu"_2"S"` will not form.