Question #1aa69

1 Answer
Apr 22, 2017

Answer:

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 #"HCN"# with a concentration of #c color(white)(l)"mol/L"# and degree of dissociation #α#, then at equilibrium you have

#color(white)(mmmmmml)"HCN + H"_2"O" ⇌ "H"_3"O"^"+" + "CN"^"-"#
#"I/mol·L"^"-1": color(white)(mm)c color(white)(mmmmmmmll)0color(white)(mmml)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 %#