Chapter 12 Solutions and Their Properties

Solutions

Solution is a two-component system.
• Smaller component = solute
• Larger component = solvent

Energy Changes and the Solution Process

When some solute dissolves into solvent, energetics must become favorable

ΔG = ΔH - TΔS

governs the dissolution process. So there are two components:

Enthalpy (Δ H part)
Entropy (Δ S part)

In all cases, Entropy term is favorable for making a solution. This is because a solution is more disorderly than the separate solute and solvent.

• In order to pull out a solute molecule from its crystaline solid takes energy (ΔHsolute). Always, ΔHsolute > 0
• The solvent also has to make a room for incoming solute molecule. The process also takes energy (ΔHsolvent). Always, ΔHsolute > 0
• The interactin b/w solute molecule and solvent is generally attractive, but varies in strength (ΔHintxn). In this picture, ΔH is larger (more negative) than the magnitude of ΔHsolute + ΔHsolvent. Therefore, overall dissolution process is exothermic. Exothermicity (or endothemicity) really depends on the interaction b/w solute and solvent.

Like dissolves like. Similar molecules have similar groups (substituents), therefore they have similar intermolecular forces so that they can dissolve each other.

Here is the solubility of alcohols in water. Alcohol has OH group, which can hydrogen-bond to water to dissolve. As the chain length gets longer, it becomes more non-polar (fat-like) thus reducing the solubility.

NameFormulaSolubility in water
MethanolCH3OHCompletely
EthanolCH3CH2OHCompletely
PropanolCH3CH2CH2OHCompletely
ButaanolCH3CH2CH2CH2OH74g/100g H2O
PentanolCH3CH2CH2CH2CH2OH27g
HexanolCH3CH2CH2CH2CH2CH2OH6 g
HeptanolCH3CH2CH2CH2CH2CH2CH2OH1.7 g
OctanolCH3CH2CH2CH2CH2CH2CH2CH2OH0 g

Units of Concentration

We know already few units of concentration

Molarity (M)

M = molsolute / 1 Lsol'n

Mole Fraction (Χ)

Χ = molsolute / moltotal

Mass Percent (%mass)

%mass = gsolute / gtotal x 100%
• Parts per million (ppm) = gsolute / gtotal x 106
• Parts per billion (bpm) = gsolute / gtotal x 109

Molality (m)

m = molsolute / 1 kgsolvent

Density (ρ) can also be considered as concentration unit, in this respect.

Conversion between these units is accomplished, none other than, yes, Dimensional Analysis!

Example: Consider an aqueous solution of CO(NH3)2 with its ρ = 1.00 g/mL. Given that m = 2.577 m, what are the %mass of solute, ppm of solute, mole fraction of solvent, and molarity?

μCO(NH3)2 = 60.062 g/mol

$?{g}_{CO{\left(N{H}_{3}\right)}_{2}}=2.577mo{l}_{CO{\left(N{H}_{3}\right)}_{2}}\left(\frac{60.062{g}_{CO{\left(N{H}_{3}\right)}_{2}}}{1mo{l}_{CO{\left(N{H}_{3}\right)}_{2}}}\right)=154.779774g$

Therefore, total mass, mtotal= 1000 g + 154.779774 g = 1154.779774 g, then $%=\left(\frac{154.779774g}{1154.779774g}\right)100%=1.340×{10}^{-5}%$

For ppm, instead of multiplying 100 to the ratio, 106 is mulitiplied to obtain
0.1340 ppm.

For χ, you need the total number of moles, which is the sum of molsolute + molsolvent. The # of moles of solvent is,

$?mo{l}_{{H}_{2}O}=1k{g}_{{H}_{2}O}\left(\frac{{10}^{3}{g}_{{H}_{2}O}}{1k{g}_{{H}_{2}O}}\right)\left(\frac{1mo{l}_{{H}_{2}O}}{18.016{g}_{{H}_{2}O}}\right)=55.506mol$

Therefore, moltotal = 55.560 + 2..577 = 58.10832 mol, so χsolvent = 0.9556.

For molarity, assuming the solution density is 1.00 g/mL, we can convert total mass of solution, which was calculated to be 1154.8 g = 1.1548 L in volume, thus to obtain M = 2.323 M.

Factors affecting Solubility

Effect of Temperature

Generally, higher the temperature more solute can be dissolved.

Solid dissolving in water generally increases, while gas in water decreases.

Except, NaCl doesn't really change its solubility over the range of 0°C to 100°C. In addition, Ce2(SO4)3 actually decrease its solubility in the range of 0°C to about 40°C.

Effect of Pressure

The concentration of gas is proportional to the gas pressure applied.

Therefore, we have

[ ] = k Pgas

where [ ] is the concentration of gas in solution, and k is called Henry's law constant with M/atm unit.

Colligative Properties

Physical changes induced by small amount of solute. They depend only on the concentration, and not the nature and/or type of solute.

4 types

• Vapor pressure lowering - Raoult's Law
• Boiling point elevation
• Freezing point depression
• Osmotic pressure

Raoult's Law: Vapor Pressure Lowering

In comparison to pure liquid, solution has lower vapor pressure. The vapor pressure depends on the mole fraction of solvent, χ1.

For solvent with non-volatile solute

$P 1 = P 1 ∘ χ 1$

where subscripts 1 indicates that it is for solvent, and ° (the circle on P) indicates for the vapor pressure of pure solvent.

You can put the above equation in terms of change in pressure by introducing the solute into pure solvent in a following manner. Since there are only two component, χ is of two parts χ1 and χ2. Furthermore, χ1 + χ2 = 1. So, χ1 = 1 - χ2, and substitute this into P1 = P°1 χ1 yields $P 1 = P 1 ∘ ( 1− χ 2 )= P 1 ∘ − P 1 ∘ χ 2$ Therefore, $P 1 ∘ − P 1 =ΔP= P 1 ∘ χ 2$

Example: How many grams of glucose, C6O6H12 (μ=180.2 g/mol), is required to obtain the vapor pressure change of 1.4 mmHg in 460 mL of water at 30°C? of pure water at 30°C is 31.82 mmHg. Assume density of water at this temperature is 1.00g/mL.

It means that we know ΔP to be 1.4 mmHg. Then, using the last equation, we can arrive at the mole fraction of glucose (χ2).

$χ 2 = ΔP P 1 ∘$ Substituting the numbers in, yields: $χ 2 = 1.4mmHg 31.82mmHg =0.044$ Since χ is a mole fraction, and we know the amount of water we can obtain the number of moles of glucose in the following way: $χ glucose = n glucose n total = n glucose n glucose + n water$ $n glucose = n water χ glucose 1− χ glucose$ So, we need to know the number of moles of water, for which we have 460 mL. $?mo l water =460m L water ( 1g 1mL )( 1mol 180.2g )=2.553mol$ Substitute this into the mole of glucose is, then: $n glucose = ( 2.553mol )0.044 1−0.044 =0.11749mol$ $? g glucose =0.11749mo l glucose ( 180.2g 1mol )=21.172g=21g$

For solvent with volatile solute

The total pressure now contains, not only the solvent vapor, but also contains solute vapor. Therefore, the total pressure is a sum of the two contributions

Ptotal = Psolv + Psolute = P°solv χsolv + P°solute + χsolute

Boiling Point Elevation

Boiling point is higher in solution than that of pure solvent.

ΔT = kb m

where ΔT is the change in boiling point, m is the molality of sol'n, and kb is the boiling point elevation constant. For water as solvent, kb is 0.516 °C kg/mol.

For ionic solute, the equation must be corrected by van't Hoff factor, i

ΔT = i kb m

The van't Hoff factor is close to the number of ions from a solute molecule, but it does depend on the extent of dissociation. Effectively it is the # of ions, for our purpose.

Freezing Point Depression

For freezing point, the effect of having solution is in the opposite of boiling point, and the freezing point measured with sol'n is lower than the pure solvent.

ΔT = kf m

where ΔT is the change in freezing point, m is the molality of sol'n, and kf is the freezing point depression constant. For water as solvent, kf is 1.918 °C kg/mol.

For ionic solute, the equation must be corrected by van't Hoff factor, i

ΔT = i kf m

Example: You are asked to design an antifreeze system of a radiator for automobile use. Let's say that the radiator has a capacity of 10.0 L, and ethylene glycol, C2H6O2, is the mixed with water to make the antifreeze solution. Let's say that it is capable of withstanding the cold temperature of 14°F. How many kg of ethylene glycol does the radiator require to the temperature? Assume that density of water is 1.000 g/mL.

We first need to convert 14°F to Celcius scale. $T( F ∘ )= 5 9 [ T( C ∘ )−32 ]$ $T( C ∘ )= 5 9 ( 14−32 )=−10 C ∘$ Therefore, the ΔT = 10°C. We use this in $\Delta T={k}_{f}m$. We solve for m (molality) with kf = 1.918°C kg/mol. $m= ΔT k f = 10 C ∘ 1.918 C ∘ kg/mol =5.214mol/kg$ Since the capacity of the radiator is 10.0L, therefore, the mass is 10.0 kg, so this can be converted to ethylene glycol mass. $?k g ethylene =10k g water ( 5.214mo l ethylene 1k g water )( 62.07 g ethylene 1mo l ethylene )( 1kg 10 3 g )=3.24kg$

Consider also a question like the following:
Given the area of your yard and the height of snow on your yard, can you calculate the amount of salt (KCl) required to melt all snow, if outside temperature is 25°F and the density of the newly fallen snow is 0.15 g/cm3?

Osmotic Pressure

A bent tube, with a semipermeable membrane where only H2O molecules can pass, contains a solution on the left chamber and pure H2O on the right. In order for the system to reach equilibrium to equalize the concentration, water moves from the right chamber to left. The pressure exerted on the left chamber to stop the flow of water from the right is called osmotic pressure, π. The relationship between π and concentration is as follows,

$\pi =\frac{nRT}{V}=MRT$ where M is the molarity of the solution. It resembles IGL!

Example: Often protein molar mass is obtained from osmotic pressure measurement. If a solution of protein is prepared by dissolving 0.0500 g of the protein in 10.00 mL water, measuring an osmotic pressure of 15.5 mmHg at 20.0°C, what is the molar mass of protein?

Rearrange the equation so that

M = Π/RT

Π = 0.01974 atm,

so, n = [(0.01974 atmi)(0.01 L)]/(0.08206 Latm/molK)(293.15 K) = 8.2046 x 10-6 molprotein

and μ = 0.0500 g/(8.2046 x 10-6 mol) = 6094 g/mol

Fractional Distillation

From vapor pressure lowering for solutions with volatile solute,

$P total = P 1 ∘ χ 1 + P 2 ∘ χ 2$ where subscripts 1 and 2 indicate solvent and solute that consists the solution, respectively.

The diagrams shows the temperature phase diagram as a fxn of mole fraction.

The top part (orange shade) is gas phsae, and the bottom part (pink) is the liquid. The phase boundary in this diagram is separated, due to the volatile nature of the solute. It means that at a certain temperature, the vapor composition (mole fraction) is different from that of the liquid composition (mole fraction).

For example, at 90°C, vapor has 20% toluene, but liquid has 40% toluene. This difference is utilized to concentrate, for example, alcohol from a dilute solution.

Example: Let's say that you are distilling a solution of alcohol with the mole fraction of alcohol to be 0.05. Your distiller operates at 64°C, at which vapor pressure of water is 153 mmHg, and of ethanol is 400 mmHg How many iterations does it take to alchohol content to be higher than 50%mole alchohol?

So, we start with:
χalc = 0.05 : P°alc = 400 mmHg
χwater = 0.95 : P°water = 153 mmHg

Since P = P° χ, vapor pressures are

Palc = (400 mmHg)(0.05) = 20 mmHg
Pwater = (153 mmHg)(0.95) = 145.35 mmHg

Then, Ptotal = 165.35 mmHg So, now we can recompute the mole fraction in vapor by using χ = P/Ptotal

We have:

χalc = 20 mmHg/165.35 mmHg = 0.121
χwater = 145.35 mmHg/165.35 mmHg = 0.879

Now, you see alcohol content went up more than twice as much! If you condense this vapor to a liquid, and do the calculations over, you get:

2nd iteration
Palc = (400 mmHg)(0.121) = 48.4 mmHg
Pwater = (153 mmHg)(0.879) = 134.5 mmHg

Ptotal = 182.9 mmHg

χalc = 48.4 mmHg/182.9 mmHg = 0.265
χwater = 134.5 mmHg/182.9 mmHg = 0.735

3rd iteration
Palc = (400 mmHg)(0.265) = 106.0 mmHg
Pwater = (153 mmHg)(0.735) = 112.4 mmHg

Ptotal = 218.4 mmHg

χalc = 106.0 mmHg/218.4 mmHg = 0.485
χwater = 112.4 mmHg/218.4 mmHg = 0.515

4th iteration
Palc = (400 mmHg)(0.485) = 194.0 mmHg
Pwater = (153 mmHg)(0.515) = 78.8 mmHg

Ptotal = 272.8 mmHg

χalc = 194.0 mmHg/272.8 mmHg = 0.711
χwater = 78.8 mmHg/272.8 mmHg = 0.289

So, after 4 iterations, alcohol content is over 70%.

In actual experiment, the iterations are done naturally by setting up an apparatus such as shown below. You're going to use this set-up in Organic Chemistr next year!

By Original PNG by User:Quantockgoblin, SVG adaptation by User:Slashme - http://en.wikipedia.org/wiki/distillation, Public Domain, Link