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

Apr 13, 2016

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, $\text{sucrose(s) → sucrose (aq)}$.

$i = 1$, 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 $\text{NaCl}$ will form two particles in solution, an ${\text{Na}}^{+}$ ion and a $\text{Cl"^"-}$ ion.

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

One formula unit of ${\text{CaCl}}_{2}$ will form three particles in solution, a $\text{Ca"^"2+}$ ion and two $\text{Cl"^"-}$ 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.

$\textcolor{w h i t e}{m m m m m m} \text{HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{m m m m m m m l} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1": color(white)(m)"-} x \textcolor{w h i t e}{m m m m m m} + x \textcolor{w h i t e}{m} + x$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{l} 1 - x \textcolor{w h i t e}{m m m m m m} x \textcolor{w h i t e}{m m m} x$

At equilibrium, we have $1 - x \textcolor{w h i t e}{l} \text{mol of HA", xcolor(white)(l) "mol of H"_3"O"^+, and xcolor(white)(l) "mol of A"^"-}$.

$\text{Total moles" = (1-x + x + x)color(white)(l) "mol" = (1+x)color(white)(l) "mol}$, so i = 1+x".

Usually, $x < 0.05$, so i < 1.05 ≈ 1.