How to Determine Vant Hoff if You Know Freezing Point

13.nine: Solutions of Electrolytes

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Thus far nosotros have assumed that nosotros could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the bodily concentration of dissolved particles in an electrolyte solution. Nosotros have used this simple model to predict such backdrop as freezing points, melting points, vapor force per unit area, and osmotic pressure. If this model were perfectly correct, nosotros would expect the freezing point low of a 0.ten yard solution of sodium chloride, with 2 mol of ions per mole of \(NaCl\) in solution, to exist 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 non ever the example. 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 human relationship between the actual number of moles of solute added to form a solution and the credible number equally determined by colligative properties is called the van't Hoff factor (\(i\)) and is defined equally follows:Named for Jacobus Hendricus van't Hoff (1852–1911), a Dutch chemical science professor at the University of Amsterdam who won the offset Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions.

\[i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \characterization{13.ix.one}\]

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

The van't Hoff factor is therefore a measure of a difference from ideal behavior. The lower the van 't Hoff factor, the greater the deviation. As the information in Tabular array \(\PageIndex{1}\) show, the van't Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions evidently contain fewer particles than predicted by the number of ions per formula unit. Every bit the concentration of the solute increases, the van't Hoff factor decreases because ionic compounds generally practice not totally dissociate in aqueous solution.

Table \(\PageIndex{1}\): van 't Hoff Factors for 0.0500 M Aqueous Solutions of Selected Compounds at 25°C

Compound	i (measured)	i (ideal)
glucose	one.0	ane.0
sucrose	1.0	one.0
\(NaCl\)	1.9	ii.0
\(HCl\)	one.9	2.0
\(MgCl_2\)	2.7	3.0
\(FeCl_3\)	three.iv	four.0
\(Ca(NO_3)_2\)	2.5	three.0
\(AlCl_3\)	3.two	iv.0
\(MgSO_4\)	1.four	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 \(\PageIndex{1}\)). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as \(Mg^{2+}\), \(Al^{3+}\), \(\ce{SO4^{ii−}}\), and \(\ce{PO4^{three−}}\) have a greater tendency to form ion pairs because of their potent electrostatic interactions. The bodily number of solvated ions present in a solution can be adamant by measuring a colligative belongings at several solute concentrations.

Effigy \(\PageIndex{1}\): Ion Pairs. In full-bodied solutions of electrolytes like \(NaCl\), some of the ions form neutral ion pairs that are not separated past solvent and diffuse as single particles.

Example \(\PageIndex{1}\): Iron Chloride in Water

A 0.0500 M aqueous solution of \(FeCl_3\) has an osmotic pressure level of iv.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 cistron

Strategy:

1. Use Equation 13.9.12 to calculate the expected osmotic force per unit area 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 of measurement, and then apply Equation xiii.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

\[\Pi=MRT=(0.200 \;mol/50) \left[0.0821\;(50⋅atm)/(G⋅mol) \right] (298\; Yard)=4.89\; atm\]

B The observed osmotic pressure is just 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is four.fifteen 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 summate the observed particle concentration from the osmotic pressure of iv.15 atm:

\[4.fifteen\; atm=Thou \left[ 0.0821 \;(L⋅atm)/(K⋅mol)\correct] (298 \;K) \]

\[0.170 mol/L=M\]

The ratio of this value to the expected value of 0.200 Chiliad is 0.170 Grand/0.200 M = 0.850, which over again gives us (0.850)(4) = 3.40 particles per mole of \(FeCl_3\) dissolved. From Equation \ref{thirteen.9.1}, the van't Hoff factor for the solution is

\[i=\dfrac{\text{3.forty particles observed}}{\text{1 formula unit of measurement}\; FeCl_3}=iii.twoscore\]

Exercise \(\PageIndex{one}\): Magnesium Chloride in Water

Calculate the van't Hoff cistron for a 0.050 k aqueous solution of \(MgCl_2\) that has a measured freezing point of −0.25°C.

Answer:

2.7 (versus an ideal value of 3

Cardinal Concepts and Summary

Ionic compounds may non completely dissociate in solution due to activity effects, in which case observed colligative furnishings may be less than predicted.

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Source: https://chem.libretexts.org/Bookshelves/General_Chemistry/Map:_General_Chemistry_%28Petrucci_et_al.%29/13:_Solutions_and_their_Physical_Properties/13.09:_Solutions_of_Electrolytes