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

## For $\text{AgBr", K_text(sp) = 7.7 × 10^"-13}$. For "Ag(CN)"_2^"-", K_text(f) = 5.6 × 10^8.

Apr 27, 2017

The solubility is 0.002 mol/L.

#### Explanation:

We have the two equilibria:

$\text{AgBr(s)" ⇌ "Ag"^"+""(aq)" + "Br"^"-""(aq)"; color(white)(mmml)K_text(sp) = 7.7 × 10^"-13}$
"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

$\text{AgBr(s)" + "2CN"^"-""(aq)" ⇌ "Ag(CN)"_2^"-""(aq)" + "Br"^"-""(aq)}$

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 $\text{AgBr}$ in $\text{KCN}$.

$\textcolor{w h i t e}{m m m m m m} \text{AgBr + 2CN"^"-" ⇌ "Ag(CN)"_2^"-" + "Br"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m m m m l} 0.1 \textcolor{w h i t e}{l l m m m m} 0 \textcolor{w h i t e}{m m m l l} 0$
$\text{C/mol·L"^"-1":color(white)(mmmmll)"-2"xcolor(white)(mmmml)"+"xcolor(white)(mmm)"+} x$
$\text{E/mol·L"^"-1":color(white)(mmmml)"0.1-2} x \textcolor{w h i t e}{m m m m} x \textcolor{w h i t e}{m m m l l} x$

K = x^2/(0.1-2x)^2 = 4.31 × 10^"-4"

$\frac{x}{0.1 - 2 x} = 0.0208$

$x = 0.0208 \left(0.1 - 2 x\right) = \text{0.002 08} - 0.0415 x$

$\text{1.04 15"x = "0.002 08}$

$x = \text{0.002 08"/"1.04 15} = 0.002$

∴ The solubility of $\text{AgBr}$ in the $\text{KCN}$ solution is 0.002 mol/L.