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#.

1 Answer
Apr 27, 2017

Answer:

The solubility is 0.002 mol/L.

Explanation:

We have the two equilibria:

#"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#

If we add the two equations, we get

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

For this equilibrium,

#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.