# Question 1aa69

Apr 22, 2017

The degree of dissociation α = "0.000 02". The percent dissociation is 0.002 %.

#### Explanation:

The degree of dissociation α is the fraction of the original solute molecules that have dissociated.

If you have a solution of $\text{HCN}$ with a concentration of $c \textcolor{w h i t e}{l} \text{mol/L}$ and degree of dissociation α, then at equilibrium you have

$\textcolor{w h i t e}{m m m m m m l} \text{HCN + H"_2"O" ⇌ "H"_3"O"^"+" + "CN"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m} c \textcolor{w h i t e}{m m m m m m m l l} 0 \textcolor{w h i t e}{m m m l} 0$
"C/mol·L"^"-1":color(white)(m)"-"αc color(white)(mmmmmml) "+"αc color(white)(mll) "+"αc
"E/mol·L"^"-1": color(white)(m) c"-"αc color(white)(mmmmmmll) αc color(white)(mmll) αc

Thus, at equilibrium, the total concentration of all particles is

c "-" αc + αc + αc = c + αc = c(1+α)

The van't Hoff $i$ factor is the number of moles of particles obtained from 1 mol of solute. That is,

i = (color(red)(cancel(color(black)(c)))(1+α))/color(red)(cancel(color(black)(c))) = 1 + α

i = "1.000 02" = 1 + α

α = "1.000 02 - 1" = "0.000 02"

"Percent dissociation" = α × 100 % = "0.000 02" × 100 % = 0.002 %#