Question

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.

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.