# 13.9: Solutions of Electrolytes

Thus far we have assumed that we could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the actual concentration of dissolved particles in an electrolyte solution. We have used this simple model to predict such properties as freezing points, melting points, vapor pressure, and osmotic pressure. If this model were perfectly correct, we would expect the freezing point depression of a 0.10 m solution of sodium chloride, with 2 mol of ions per mole of $$NaCl$$ in solution, to be exactly twice that of a 0.10 m solution of glucose, with only 1 mol of molecules per mole of glucose in solution. In reality, this is not always the case. Instead, the observed change in freezing points for 0.10 m aqueous solutions of $$NaCl$$ and KCl are significantly less than expected (−0.348°C and −0.344°C, respectively, rather than −0.372°C), which suggests that fewer particles than we expected are present in solution.

The relationship between the actual number of moles of solute added to form a solution and the apparent number as determined by colligative properties is called the van’t Hoff factor (i) and is defined as follows:Named for Jacobus Hendricus van’t Hoff (1852–1911), a Dutch chemistry professor at the University of Amsterdam who won the first Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions.

$i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \tag{13.9.1}$

Note

As the solute concentration increases the van’t Hoff factor decreases.

The van’t Hoff factor is therefore a measure of a deviation from ideal behavior. The lower the van’t Hoff factor, the greater the deviation. As the data in Table 13.9.1 show, the van’t Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions apparently contain fewer particles than predicted by the number of ions per formula unit. As the concentration of the solute increases, the van’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution.

Table 13.9.1: van’t Hoff Factors for 0.0500 M Aqueous Solutions of Selected Compounds at 25°C
Compound i (measured) i (ideal)
glucose 1.0 1.0
sucrose 1.0 1.0
$$NaCl$$ 1.9 2.0
$$HCl$$ 1.9 2.0
$$MgCl_2$$ 2.7 3.0
$$FeCl_3$$ 3.4 4.0
$$Ca(NO_3)_2$$ 2.5 3.0
$$AlCl_3$$ 3.2 4.0
$$MgSO_4$$ 1.4 2.0

Instead, some of the ions exist as ion pairs, a cation and an anion that for a brief time are associated with each other without an intervening shell of water molecules (Figure 13.9.1). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as $$Mg^{2+}$$, $$Al^{3+}$$, $$SO_4^{2−}$$, and PO_4^{3−}\) have a greater tendency to form ion pairs because of their strong electrostatic interactions. The actual number of solvated ions present in a solution can be determined by measuring a colligative property at several solute concentrations.

Figure 13.9.1: Ion Pairs. In concentrated solutions of electrolytes like $$NaCl$$, some of the ions form neutral ion pairs that are not separated by solvent and diffuse as single particles.

Example 13.9.1: Iron Chloride in Water

A 0.0500 M aqueous solution of $$FeCl_3$$ has an osmotic pressure of 4.15 atm at 25°C. Calculate the van’t Hoff factor $$i$$ for the solution.

Given: solute concentration, osmotic pressure, and temperature

Asked for: van’t Hoff factor

Strategy:

1. Use Equation 13.9.12 to calculate the expected osmotic pressure of the solution based on the effective concentration of dissolved particles in the solvent.
2. Calculate the ratio of the observed osmotic pressure to the expected value. Multiply this number by the number of ions of solute per formula unit, and then use Equation 13.9.1 to calculate the van’t Hoff factor.

Solution:

A If $$FeCl_3$$ dissociated completely in aqueous solution, it would produce four ions per formula unit [Fe3+(aq) plus 3Cl−(aq)] for an effective concentration of dissolved particles of 4 × 0.0500 M = 0.200 M. The osmotic pressure would be

$\Pi=MRT=(0.200 \;mol/L) \left[0.0821\;(L⋅atm)/(K⋅mol) \right] (298\; K)=4.89\; atm$

B The observed osmotic pressure is only 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is 4.15 atm/4.89 atm = 0.849, which indicates that the solution contains (0.849)(4) = 3.40 particles per mole of $$FeCl_3$$ dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm:

$4.15\; atm=M \left[ 0.0821 \;(L⋅atm)/(K⋅mol)\right] (298 \;K)$

$0.170 mol/L=M$

The ratio of this value to the expected value of 0.200 M is 0.170 M/0.200 M = 0.850, which again gives us (0.850)(4) = 3.40 particles per mole of $$FeCl_3$$ dissolved. From Equation 13.9.1, the van’t Hoff factor for the solution is

$i=\dfrac{\text{3.40 particles observed}}{\text{1 formula unit}\; FeCl_3}=3.40$

Exercise 13.9.1: Magnesium Chloride in Water

Calculate the van’t Hoff factor for a 0.050 m aqueous solution of $$MgCl_2$$ that has a measured freezing point of −0.25°C.

Answer: 2.7 (versus an ideal value of 3

### Key Concepts and Summary

Ionic compounds may not completely dissociate in solution due to activity effects, in which case observed colligative effects may be less than predicted.