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

Here’s how you do it.

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

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##.

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