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

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

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

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Answer 1 · 2017-09-16T17:21:41

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

Explanation:

${Before adding S}^{2-}$ $bb("Before adding S"^"2-")$

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

These will react to form the complex ions ${Cd(CN)}_{4}^{2-}; K_{f} = 6.0 \times 10^{18}$ $"Cd(CN)"_4^"2-"; K_text(f) = 6.0 × 10^18$ and ${Cu(CN)}_{4}^{3-}; K_{f} = 2.0 \times 10^{30}$ $"Cu(CN)"_4^"3-"; K_text(f) = 2.0 × 10^30$ .

We must calculate the concentrations of all species in solution.

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

Thus, we will have 0.020 mol/L ${Cd(CN)}_{4}^{2-}$ $"Cd(CN)"_4^"2-"$ and 0.050 mol/L ${Cu(CN)}_{4}^{3-}$ $"Cu(CN)"_4^"3-"$ .

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

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

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

$K_{f} = \frac{[Cd {(CN)}_{4}^{2-}]}{[{Cd}^{2+}] {[{CN}^{-}]}^{4}} = \frac{0.020}{[{Cd}^{2+}] \times {0.12}^{4}} = 6.0 \times 10^{18}$ $K_text(f) = (["Cd"("CN")_4^"2-"])/(["Cd"^"2+"]["CN"^"-"]^4) = 0.020/(["Cd"^"2+"] × 0.12^4) = 6.0 × 10^18$

$[{Cd}^{2+}] = \frac{0.020}{{0.12}^{4} \times 6.0 \times 10^{18}} = 1.61 \times 10^{-17} l mol/L$ $["Cd"^"2+"] = 0.020/(0.12^4 × 6.0 × 10^18) = 1.61 × 10^"-17" color(white)(l)"mol/L"$

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

$K_{f} = \frac{[Cu {(CN)}_{4}^{3-}]}{[{Cu}^{+}] {[{CN}^{-}]}^{4}} = \frac{0.050}{[{Cu}^{+}] \times {0.12}^{4}} = 2.0 \times 10^{30}$ $K_text(f) = (["Cu"("CN")_4^"3-"])/(["Cu"^"+"]["CN"^"-"]^4) = 0.050/(["Cu"^"+"] × 0.12^4) = 2.0 × 10^30$

$[{Cu}^{2+}] = \frac{0.050}{{0.12}^{4} \times 2.0 \times 10^{30}} = 5.18 \times 10^{-36} l mol/L$ $["Cu"^"2+"] = 0.050/(0.12^4 ×2.0 × 10^30) = 5.18 × 10^"-36" color(white)(l)"mol/L"$

${After adding S}^{2-}$ $bb("After adding S"^"2-")$

Will we get a precipitate of $CdS (K_{sp} = 1 \times 10^{-27})$ $"CdS" (K_text(sp) = 1 × 10^"-27")$ ?

$m m m m m m CdS m ⇌ m {Cd}^{2+} m + m l S^{2-}$ $color(white)(mmmmmm)"CdS"color(white)(m) ⇌color(white)(m)"Cd"^"2+" color(white)(m)+ color(white)(ml)"S"^"2-"$
${I/mol\cdotL}^{-1} : m m m m m l 1.61 \times 10^{-17} m 0.0010$ $"I/mol·L"^"-1":color(white)(mmmmml)1.61 × 10^"-17"color(white)(m)0.0010$

$Q_{sp} = [{Cd}^{2+}] [S^{2-}] = 1.61 \times 10^{-17} \times 0.0010 = 1.61 \times 10^{-20}$ $Q_text(sp) = ["Cd"^"2+"]["S"^"2-"] = 1.61 × 10^"-17" × 0.0010 = 1.61 × 10^"-20"$

$Q_{sp} > K_{sp}$ $Q_text(sp) > K_text(sp)$ , so a precipitate of $CdS$ $"CdS"$ will form.

Will we get a precipitate of ${Cu}_{2} S (K_{sp} = 2.0 \times 10^{-47})$ $"Cu"_2"S" (K_text(sp) = 2.0 × 10^"-47")$ ?

$m m m m m m {Cu}_{2} S m ⇌ m {2Cu}^{+} m + m l S^{2-}$ $color(white)(mmmmmm)"Cu"_2"S"color(white)(m) ⇌color(white)(m)"2Cu"^"+" color(white)(m)+ color(white)(ml)"S"^"2-"$
${I/mol\cdotL}^{-1} : m m m m m l l 5.18 \times 10^{-36} m m 0.0010$ $"I/mol·L"^"-1":color(white)(mmmmmll)5.18 × 10^"-36"color(white)(mm)0.0010$

$Q_{sp} = {[{Cu}^{+}]}^{2} [S^{2-}] = {(5.18 \times 10^{-36})}^{2} \times 0.0010 = 2.67 \times 10^{-74}$ $Q_text(sp) = ["Cu"^"+"]^2["S"^"2-"] = (5.18 × 10^"-36")^2 × 0.0010 = 2.67 × 10^"-74"$

$Q_{sp} < K_{sp}$ $Q_text(sp) < K_text(sp)$ , so a precipitate of ${Cu}_{2} S$ $"Cu"_2"S"$ will not form.

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

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

1 Answer

Explanation:

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Please explain what is happening in the solution before and after adding S2−S^(2-) . I have the calculus in my book, but dont understand what is happening in the solution exactly.

1 Answer

Explanation:

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

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