What is the solubility of silver bromide in a 0.1 mol/L solution of potassium cyanide?

For #"AgBr", K_text(sp) = 7.7 × 10^"-13"#.

For #"Ag(CN)"_2^"-", K_text(f) = 5.6 × 10^8#.

Answer 1

The solubility is 0.002 mol/L.

The two equilibria are here:

#"AgBr(s)" ⇌ "Ag"^"+""(aq)" + "Br"^"-""(aq)"; color(white)(mmml)K_text(sp) = 7.7 × 10^"-13"# #"Ag"^"+""(aq)" + "2CN"^"-""(aq)" → "Ag(CN)"_2^"-""(aq)"; K_text(f) = color(white)(l)5.6 × 10^8#

The result of adding the two equations is

#"AgBr(s)" + "2CN"^"-""(aq)" ⇌ "Ag(CN)"_2^"-""(aq)" + "Br"^"-""(aq)"#

Regarding this balance,

#K = (["Ag(CN)"_2^"-" ]["Br"^"-"])/["CN"^"-"]^2 = K_text(sp)K_text(f) = 7.7 × 10^"-13" × 5.6 × 10^8 = 4.31 × 10^"-4"#
Now, we can set up an ICE table to calculate the solubility of #"AgBr"# in #"KCN"#.
#color(white)(mmmmmm)"AgBr + 2CN"^"-" ⇌ "Ag(CN)"_2^"-" + "Br"^"-"# #"I/mol·L"^"-1":color(white)(mmmmml)0.1color(white)(llmmmm)0color(white)(mmmll)0# #"C/mol·L"^"-1":color(white)(mmmmll)"-2"xcolor(white)(mmmml)"+"xcolor(white)(mmm)"+"x# #"E/mol·L"^"-1":color(white)(mmmml)"0.1-2"xcolor(white)(mmmm)xcolor(white)(mmmll)x#
#K = x^2/(0.1-2x)^2 = 4.31 × 10^"-4"#
#x/(0.1-2x) = 0.0208#
#x = 0.0208(0.1-2x) = "0.002 08" - 0.0415x#
#"1.04 15"x = "0.002 08"#
#x = "0.002 08"/"1.04 15" = 0.002#
∴ The solubility of #"AgBr"# in the #"KCN"# solution is 0.002 mol/L.
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Answer 2

The solubility of silver bromide in a 0.1 mol/L solution of potassium cyanide is increased significantly due to the formation of a soluble complex ion, [Ag(CN)2]-, through the reaction: AgBr(s) + 2KCN(aq) ⇌ KAg(CN)2 + KBr(aq).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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