# 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.

Sep 16, 2017

WARNING! Long answer! A precipitate of $\text{CdS}$ will form.

#### Explanation:

$\boldsymbol{\text{Before adding S"^"2-}}$

The initial solution contains $\text{Cd"^"2", "Cu"^"+}$, and $\text{CN"^"-}$.

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 $\text{Cd"^"2+}$ and $\text{Cu"^"+}$ will be converted to their complex ions.

Thus, we will have 0.020 mol/L $\text{Cd(CN)"_4^"2-}$ and 0.050 mol/L $\text{Cu(CN)"_4^"3-}$.

$\left[\text{CN"^"-}\right]$ will decrease by 0.080 mol/L in forming the $\text{Cd}$ complex and by 0.10 mol/L in forming the $\text{Cu}$ complex.

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

$\textcolor{w h i t e}{m m m m m m m} \text{Cd"^"2+" + "4CN"^"-" ⇌ "Cd(CN)"_4^"2-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l l} 0.020 \textcolor{w h i t e}{m l l} 0.40 \textcolor{w h i t e}{m m m m l} 0$
$\text{C/mol·L"^"-1":color(white)(m)"-0.020"color(white)(ml)"-0.28"color(white)(mml)"+0.020}$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m m l} x \textcolor{w h i t e}{m m l l} 0.12 \textcolor{w h i t e}{m m m l} 0.020$

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"

$\textcolor{w h i t e}{m m m m m m m l} \text{Cu"^"+" + "4CN"^"-" ⇌ "Cu(CN)"_4^"3-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l l} 0.050 \textcolor{w h i t e}{m l l} 0.40 \textcolor{w h i t e}{m m m m l} 0$
$\text{C/mol·L"^"-1":color(white)(m)"-0.050"color(white)(ml)"-0.28"color(white)(mml)"+0.050}$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m m l} x \textcolor{w h i t e}{m m l l} 0.12 \textcolor{w h i t e}{m m m l} 0.050$

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"

$\boldsymbol{\text{After adding S"^"2-}}$

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

$\textcolor{w h i t e}{m m m m m m} \text{CdS"color(white)(m) ⇌color(white)(m)"Cd"^"2+" color(white)(m)+ color(white)(ml)"S"^"2-}$
$\text{I/mol·L"^"-1":color(white)(mmmmml)1.61 × 10^"-17} \textcolor{w h i t e}{m} 0.0010$

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

${Q}_{\textrm{s p}} > {K}_{\textrm{s p}}$, so a precipitate of $\text{CdS}$ will form.

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

$\textcolor{w h i t e}{m m m m m m} \text{Cu"_2"S"color(white)(m) ⇌color(white)(m)"2Cu"^"+" color(white)(m)+ color(white)(ml)"S"^"2-}$
$\text{I/mol·L"^"-1":color(white)(mmmmmll)5.18 × 10^"-36} \textcolor{w h i t e}{m m} 0.0010$

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

${Q}_{\textrm{s p}} < {K}_{\textrm{s p}}$, so a precipitate of $\text{Cu"_2"S}$ will not form.