A chemical's Van't Hoff Factor is an important tool in solution chemistry because it helps you perform stoichiometric calculations concerning solutions. A Van't Hoff Factor is a positive integer that represents the number of smaller components a formula unit/single molecule of a chemical decomposes into when placed into water and dissociated.
Covalent compounds are generally nonelectrolytes and do not decompose further in aqueous solution. The Van't Hoff Factor of a covalently bonded compound is thus usually \(1\), because the result when the chemical is "dissociated" is one molecule, the one that was initially present.
Example 1: The chemical 1-ethanol, with a structural formula \(CH_2OHCH_3\) and molecular formula \(C_2H_6O\), is a nonelectrolyte and does not dissociate in aqueous solution. Thus its Van't Hoff Factor is \(1\).
The one major exception to this rule, however, is acids. An acid will decompose into one or more hydrogen ions and a conjugate base [anion]. The Van't Hoff Factor for these Bronsted-Lowry Acids is equal to the number of ions that result when the original chemical dissociates in aqueous solution.
Example 2: The chemical hydrochloric acid, \(HCl\), dissociates into a single \(H^{+}\) and a single \(Cl^{-}\) when placed in water. Since two ions result, the Van't Hoff Factor is \(2\).
Example 3: The chemical sulfuric acid, \(H_2SO_4\) dissociates into two \(H^{+}\) and a single \(SO_4^{2-}\) when placed in water. There are a total of three ions (not two!) so the Van't Hoff Factor is \(3\).
Finding the Van't Hoff Factor for an ionic compound is basically the same as finding the Van't Hoff Factor for an acid, as described in the previous section. When dissociated, the ionic compound will break down into its component cation and anion. Since two different ions will form from one formula unit of the compound, the Van't Hoff Factor of an ionic compound will always be at least \(2\). However, sometimes more than one cation and/or anion will form from a formula unit, in which case each individual ion is counted separately, even if two or more of the ions are indistinguishable.
Note that all ionic compounds are written in the form \(A_xB_y\), where \(A\) is the cation, \(B\) is the anion, and \(x\) and \(y\) are the respective coefficients. This indicates how many of each ion are used to create the final ionic compound. Once the ionic compound is dissociated, each cation and each anion will be counted when determining the Van't Hoff Factor. Thus, \(x\) cations and \(y\) anions form per formula unit. A total of \(x + y\) ions result, so the Van't Hoff Factor for an ionic compound is \(x + y\).
Example 4: Find the Van't Hoff Factor of \(FrBr\).
Solution: The cation and anion coefficients are both \(1\), so one cation and one anion will be isolated upon dissociation. Thus the Van't Hoff Factor is \(2\).
Example 5: Find the Van't Hoff Factor of Aluminum Dichromate.
Solution: The chemical formula for Aluminum Phosphate is \(Al_2(Cr_2O_7)_3\) (don't be thrown off by the anion containing a metal). The cation coefficient is \(2\), and the anion coefficient is \(3\). A total of five ions are isolated upon dissociation of one formula unit of this chemical, so the Van't Hoff Factor is \(5\).
Finding the Van't Hoff Factor of a compound can be a helpful technique when performing certain types of stoichiometric calculations, generally where we want to find the number of one or more ions constituting a compound.
Example 6: How many moles of ions are in a \(30.0\)-gram sample of sodium bicarbonate?
Solution: The chemical formula for sodium bicarbonate is \(Na_2CO_3\). To convert from grams of sodium bicarbonate to moles of sodium bicarbonate, we need to find the molar mass of sodium bicarbonate:
$$2(23.0) + 12.0 + 3(16.0) = 106.0 \; \frac{g}{mol}$$
Thus we can find the number of moles of Sodium Bicarbonate:
$$30.0 \; g \; Na_2CO_3 \times \frac{1 \; mol \; Na_2CO_3}{106.0 \; g \; Na_2CO_3} = .283 \; mol \; Na_2CO_3$$
Now, we want to find the number of ions, which form from formula units. Thus we must convert from moles to formula units with the use of the Avogadro's Number Conversion:
$$.283 \; mol \; Na_2CO_3 \times \frac{6.022 \times 10^{23} \; formula units \; Na_2CO_3}{1 \; mol \; Na_2CO_3} = 1.70 \times 10^{23} \; formula units \; Na_2CO_3$$
We want to find the number of ions in the sample, however, so we must find the number of ions in one formula unit, the Van't Hoff Factor. In every formula unit, there are two sodium ions and one carbonate ion. Remember that the negative charge comes from a carbonate ion, not a combination of one carbide ion and three oxide ions! Thus, the Van't Hoff Factor is \(3\)--each formula unit dissociates into two sodium cations and one carbonate anion. That makes for a total of three ions per formula unit, and we can convert from formula units to ions:
$$1.70 \times 10^{23} \; formula units \; Na_2CO_3 \times \frac{3 \; ions}{1 \; formula unit \; Na_2CO_3} = 5.11 \times 10^{23} \; ions$$