How can I determine the van't Hoff factor of a substance from its formula?

1 Answer
Apr 13, 2016

Answer:

Here's how you do it.

Explanation:

The van't Hoff factor, #i#, is the number of particles formed in a solution from one formula unit of solute.

Notice that #i# is a property of the solute. In an ideal solution, #i# does not depend on the concentration of the solution.

For a nonelectrolyte

If the solute is a nonelectrolyte (i.e. it does not separate into ions in solution), #i = 1#

For example, #"sucrose(s) → sucrose (aq)"#.

#i = 1#, because 1 molecule of sucrose forms only one particle in solution.

For a strong electrolyte

If the solute is a strong electrolyte (i.e. it separates into ions in solution), #i > 1#.

Some examples are:

#"NaCl(s)" → "Na"^+("aq") + "Cl"^"-"("aq"); i = 2#

One formula unit of #"NaCl"# will form two particles in solution, an #"Na"^+# ion and a #"Cl"^"-"# ion.

#"CaCl"_2(s) → "Ca"^"2+"("aq") + "2Cl"^"-"("aq"); i = 3#

One formula unit of #"CaCl"_2# will form three particles in solution, a #"Ca"^"2+"# ion and two #"Cl"^"-"# ions.

Here's another example:

#"Fe"_2("SO"_4)_3("s") → "2Fe"^"3+"("aq") + "3SO"_4^"2-"("aq"); i = 5#

For a weak electrolyte

If the solute is a weak electrolyte , it dissociates only to a limited extent.

For example, acetic acid is a weak acid. We often set up an ICE table to calculate the number of particles in a 1 mol/L solution.

#color(white)(mmmmmm)"HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A"^"-"#
#"I/mol·L"^"-1":color(white)(ml) 1color(white)(mmmmmmml) 0color(white)(mmm) 0#
#"C/mol·L"^"-1": color(white)(m)"-"xcolor(white)(mmmmmm) +xcolor(white)(m)+x#
#"E/mol·L"^"-1":color(white)(l) 1-xcolor(white)(mmmmmm)xcolor(white)(mmm) x#

At equilibrium, we have #1-xcolor(white)(l) "mol of HA", xcolor(white)(l) "mol of H"_3"O"^+, and xcolor(white)(l) "mol of A"^"-"#.

#"Total moles" = (1-x + x + x)color(white)(l) "mol" = (1+x)color(white)(l) "mol"#, so #i = 1+x"#.

Usually, #x < 0.05#, so #i < 1.05 ≈ 1#.