Chemistry of Solutions 202-SN2-RE • CEGEP, Montreal
Revision sheet: solutions and colligative properties (202-SN2-RE)
Solutions come back in the second chemistry course with tools the first one did not have: enthalpy, entropy, free energy, the equilibrium constant and Ka. The chapter then changes in nature: we no longer just calculate a freezing-point depression, we explain why a salt dissolves while cooling, why a gas leaves water when it is heated, and why a tabulated constant never applies as is to a concentrated solution.
This sheet gives what is really lost: the rule applied to the wrong type of electrolyte, the unit forgotten between kilojoules and joules, partial pressure confused with total pressure, the conclusion about precision that nobody draws. The corrected exercises on the same chapter make you practise each of these moves. It is written for CEGEP students in Chemistry of Solutions, 202-SN2-RE (formerly 202-NYB).
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The thread of the chapter
Dissolving is an EQUILIBRIUM, not a given. Everything the second course can do with an equilibrium, writing a constant, differentiating it with respect to temperature, correcting it with an activity, therefore applies to dissolution. And a colligative property is not a result in itself: it is a measuring instrument, whose precision must be put in numbers before it is trusted.
Dissolution as an equilibrium: cycle, criterion and constant
•ΔHsol=ΔHlattice+ΔHhydration, the first term always POSITIVE and the second always NEGATIVE. Their sum is a small difference between two large numbers: it is measured in a calorimeter, it is not calculated.
•Criterion for spontaneity: ΔGsol=ΔHsol−TΔSsol. An endothermic dissolution is spontaneous as soon as entropy compensates, and it stops being so below T=ΔSsolΔHsol.
•Henry's law: [X]=kHPX, with the PARTIAL pressure of the gas. The constant kH is nothing other than the constant of the equilibrium X(g)⇌X(aq).
•It therefore obeys van 't Hoff: lnK1K2=−RΔH(T21−T11), with temperatures in KELVINS.
•A GAS has ΔHsol<0, so it becomes LESS soluble when hot. A SOLID most often has ΔHsol>0, so it becomes more soluble. In both cases it is Le Chatelier, with heat counting as a reactant or as a product.
The Hess cycle of ammonium nitrate, drawn to scale: two terms of several hundred kilojoules whose difference is only 26 kJ/mol.
The unit trap in the ΔG calculation costs a mark on every exam: the enthalpy is tabulated in kilojoules per mole, the entropy in joules per mole per kelvin. Convert BEFORE subtracting.
Colligative properties, and the three ways the real world departs from the ideal
•ΔTf=iKfm and ΔTb=iKbm, with Kf=1.86 and Kb=0.512 °C·kg/mol for water. What matters is the number of PARTICLES, not their nature.
•WEAK electrolyte: i=1+α for a monoprotic acid, and Ka=1−αcα2. For acetic acid at 0.1 mol/kg, α is 1.3% and i is 1.01, not 2.
•STRONG electrolyte: the measured i stays below its theoretical value, because the ions travel in pairs part of the time. The gap is put in numbers by the ionic strength I=21∑cizi2 and by Debye-Hückel, logγ±=−0.509∣z+z−∣I.
•Every tabulated constant, Ksp as well as Ka, is written with ACTIVITIES a=γc. Adding an inert salt lowers the γ values, and so INCREASES the equilibrium concentrations: this is the salt effect, not to be confused with the common-ion effect.
•Raoult, PA=xAPA∘: POSITIVE deviation when mixing is endothermic, NEGATIVE when it is exothermic. A strong deviation creates an azeotrope, a minimum-boiling one in the first case and a maximum-boiling one in the second.
The liquid-vapor curve of the solution, dashed, passes below that of pure water; the solid-vapor curve stays put. On the 1 atm isobar, freezing goes down and boiling goes up.
On the phase diagram, only the LIQUID-vapor curve moves down, because the solid that crystallizes is PURE solvent. This is the source of freezing-point depression, boiling-point elevation and freeze concentration alike.
The mistakes that cost marks
These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.
1.Judging spontaneity from the enthalpy alone
3 marks, and a conclusion that is the opposite of reality
What not to write
“ΔHsol=+26 kJ/mol, so ammonium nitrate does not dissolve spontaneously.”
What to write
“ΔGsol=26000−298×108=−6.2 kJ/mol: the dissolution is spontaneous and cools the solution.”
ΔH is positive, but −TΔS exceeds it: ΔG is negative and the dissolution is spontaneous. The sign of ΔH alone settles nothing.
Why: The criterion is ΔG, not ΔH. Spreading the ions of a crystal through the whole volume of the solvent creates a lot of disorder, and the −TΔS term wins easily at room temperature. The pharmacy instant cold pack is exactly this case: it works because the dissolution is both spontaneous and endothermic.
2.Mixing kilojoules and joules in the ΔG calculation
1 to 2 marks, and a result five thousand times too large
What not to write
“ΔG=26−298×108=−32158 kJ/mol.”
What to write
“ΔG=26000−298×108=−6184 J/mol, that is −6.2 kJ/mol.”
Why: Tables give the enthalpy in kJ/mol and the entropy in J/(mol·K). This is the most mechanical mistake of the chapter, and the check that catches it is free: no dissolution has a free energy of thirty-two thousand kilojoules per mole, a number far beyond any chemical bond.
3.Giving a weak electrolyte the factor i of its formula
3 marks, and a depression announced as twice too large
What not to write
“Acetic acid releases one H+, so i=2 and ΔTf=2×1.86×0.100=0.372 °C.”
What to write
“α=1.34%, so i=1.013 and ΔTf=0.1885 °C.”
A weak electrolyte releases almost nothing: i=1.013 gives a depression barely above that of a molecular solute. The red bar, i=2, is double.
Why: The van 't Hoff factor counts the particles ACTUALLY present, not those the formula allows. For a weak acid, i=1+α with α obtained from Ka, and α is here a little over one percent. The right reflex: before writing i, ask whether the electrolyte is strong or weak.
In air at 1 atm, oxygen contributes only 0.21 atm: this partial pressure, in green, is what goes into Henry's law, not the total pressure.
Why: Henry's law involves the PARTIAL pressure of the gas in question, the only quantity that measures its tendency to go into solution. The order-of-magnitude check is immediate: water in equilibrium with air holds eight or nine milligrams of oxygen per litre, never forty.
5.Forgetting the square of the charges in the ionic strength
2 marks, then a completely wrong activity coefficient
What not to write
“Magnesium sulfate at 0.010 mol/L: I=21(0.010+0.010)=0.010 mol/L.”
Each ion weighs cz2: at the same concentration, the ions of MgSO4 weigh four times those of NaCl. Without the square, we get the red line, four times too low.
Why: Ionic strength weights ions by the SQUARE of their charge, because that is how electrostatic energy depends on them. A 2:2 salt therefore gives four times the ionic strength of a 1:1 salt of the same concentration, and its activity coefficient falls from 0.89 to 0.39. Ionic strength equals the concentration only for a 1:1 electrolyte.
6.Treating a tabulated constant as a constant of concentrations
2 marks, and a misreading of the salt effect
What not to write
“Adding potassium nitrate, which has no common ion, does not change the solubility.”
What to write
“Ksp is written with activities: the ionic strength goes up, the γ values go down, the concentrations go up.”
A salt with no common ion INCREASES solubility: the activity coefficients fall, and s gains 30% at 0.05 mol/L of potassium nitrate. The red line is the wrong idea.
Why: A thermodynamic constant is written in activities, a=γc. As long as the solution is very dilute, γ is almost 1 and the two can be confused; once the ionic strength reaches a few hundredths of a mole per litre, the real solubility exceeds by 20 to 30% what the tabulated Ksp predicts. The salt effect INCREASES solubility, the common-ion effect decreases it: these are two opposite phenomena that must not be confused.
7.Applying Raoult's law with a single term when both liquids are volatile
1 to 2 marks, and a deviation put in numbers wrongly afterwards
What not to write
“P=xethanolPethanol∘=0.90×59=53.1 mmHg.”
What to write
“P=0.90×59+0.10×23.8=55.5 mmHg.”
Two volatile liquids, two partial pressures dashed: the total pressure, in green, is their sum. The ethanol-only term, in red, forgets the water vapor.
Why: The form P=xsolventP∘ holds only for a NONVOLATILE solute, a sugar or a salt. As soon as both components evaporate, each contributes its partial pressure and the total adds them. This is also what makes distillation possible: the vapor is enriched in the more volatile component, 71% ethanol for an equimolar mixture.
8.Pouring in all the extraction solvent at once
2 marks, and ten times more product lost in the laboratory
What not to write
“100 mL of ether in one go: it is the same volume, so the same yield.”
What to write
“Once, 88.9%; four times 25 mL, 98.8%: the remaining fractions MULTIPLY.”
The same volume of ether, poured in four portions, leaves 1.2% of the product in the water instead of 11.1%: the remaining fractions multiply from one extraction to the next.
Why: Each extraction leaves the fraction Vaq+KVoVaq, and successive extractions multiply these fractions instead of adding them. Splitting the same total volume therefore makes the loss drop geometrically, down to the limit e−KV/Vaq. In practice we stop at about three or four portions, the gain becoming marginal.
Which method to choose
Which relation to apply according to what the problem gives
The quantity sought and the nature of the solute
If a nonvolatile solid, and we want a temperature of change of state → direct colligative relation: ΔTf=iKfm or ΔTb=iKbm
Example: glucose at 1.00 mol/kg: −1.86 °C and 100.512 °C
If an electrolyte, and we do not know how many particles it gives → strong, i is the number of ions, corrected downward; weak, i=1+α with α obtained from Ka
Example: acetic acid at 0.100 mol/kg: α=1.34%, i=1.013
If a dissolved GAS, at a fixed temperature → Henry's law with the PARTIAL pressure
Example: [O2]=1.3×10−3×0.21=2.7×10−4 mol/L
If the same solubility at ANOTHER temperature → van 't Hoff on kH, temperatures in kelvins
Example: from 25 to 35 °C, oxygen loses 15%
If two VOLATILE liquids → Raoult with both terms, then compare with the measured pressure
Example: 0.90×59+0.10×23.8=55.5 mmHg against 60 measured
If a concentrated solution, or a tabulated constant to be used → calculate the ionic strength and correct with the activity coefficients
Example: MgSO4 at 0.010 mol/L: I=0.040 and γ±=0.39
Two questions to ask before any formula, and in this order: is the solute VOLATILE, and is it an ELECTROLYTE? The first decides between one-term and two-term Raoult; the second decides the value of i, and therefore all the colligative results.
How the answer is expected to be written
A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.
Presenting a solution calculation on the exam
When to use it: The problem gives thermodynamic data or a colligative measurement and asks for a conclusion: spontaneity, solubility, concentration or constant.
1Identify the nature of the solute, volatile or not, strong, weak or nonelectrolyte, and deduce i BEFORE writing any formula.
2Write the relation used in literal form, then substitute, watching the units: kelvins for van 't Hoff, joules for ΔG, partial pressure for Henry.
3Give the result with a number of digits consistent with the data, two or three most often, and its unit.
4Conclude with a sentence that answers the question asked, and not only the calculation requested: spontaneous or not, more soluble or less, compatible or not.
Concluding sentence
“ΔGsol=−6.2 kJ/mol at 25 °C: the dissolution is spontaneous although endothermic, the entropy of dispersal of the ions compensating the cost of the lattice. It would stop being so below 241 K.”
The trap: Stopping at the number without commenting on it. A negative ΔG, an i of 1.013 or a γ of 0.39 earn their marks only if the sentence that follows says what they mean for the solution under study.
Marking: Typically 1 mark for identifying the solute and i, 2 for the relation and its application, 1 for the units and significant figures, 1 for the conclusion.
Check before you hand in
Five minutes of checking recover more marks than one more problem started in a hurry.
The factor i before everything else
Before applying a colligative formula, say aloud whether the solute is a nonelectrolyte, a strong electrolyte or a weak electrolyte. The three answers give i=1, i equal to the number of ions, and i barely above 1.
Acetic acid: weak, so i=1.01 and not 2.
The order of magnitude of a dissolved gas
Water in equilibrium with air holds about 8 to 9 mg/L of oxygen, and a fizzy drink a few grams of carbon dioxide per litre. A result ten times larger almost always signals total pressure put in place of partial pressure.
2.7×10−4 mol/L, that is 8.7 mg/L: consistent.
The sign of ΔH and the direction of the temperature effect
A solid that dissolves by absorbing heat: more soluble when hot. A gas, whose dissolution releases heat: less soluble when hot. If the van 't Hoff calculation gives the opposite, the double minus sign has been lost.
ΔHsol<0 for oxygen, so kH DECREASES as T rises.
Ionic strength is not concentration
It equals concentration only for a 1:1 electrolyte. As soon as an ion carries two charges, the ionic strength is at least three times the concentration.
MgSO4 at 0.010 mol/L gives I=0.040 mol/L.
The kilojoule check
No free energy of dissolution exceeds a few tens of kilojoules per mole. A result in the tens of thousands signals an entropy left in joules against an enthalpy in kilojoules.
The oxygen of a warmed river, from van 't Hoff to the conclusion
A power plant discharges its cooling water and warms a river from 18 °C to 28 °C. Henry's law constant for oxygen is 1.3×10−3 mol/(L·atm) at 25 °C and ΔHsol=−12.5 kJ/mol.
Air contains 21% oxygen, of molar mass 32.0 g/mol, and trout need at least 6.0 mg/L.
The situation: the same river, under the same air, before and after the power plant. We want to know what the warmed water loses in dissolved oxygen, and whether the trout threshold holds.
Step 1
At 18 °C: lnK25K=8.31412500(291.151−298.151)=+0.121, so K=1.47×10−3.
Why
Temperatures are in KELVINS, and the factor −RΔH is POSITIVE since ΔH is negative. The double minus sign is the only delicate step of the calculation.
Step 2
At 28 °C: lnK25K=−0.0502, so K=1.24×10−3 mol/(L·atm).
Why
The constant DECREASES as temperature rises, as an exothermic dissolution requires. If the calculation gave the opposite, a sign has been lost.
Step 3
Concentrations: 1.47×10−3×0.21=3.08×10−4 mol/L, that is 9.86 mg/L; then 8.31 mg/L.
Why
We use the PARTIAL pressure, 0.21 atm, not atmospheric pressure. Then we multiply by the molar mass and by a thousand to get milligrams per litre.
Step 4
Drop: 9.86−8.31=1.55 mg/L, that is 15.8%.
Why
Put the change in numbers in absolute value AND as a percentage: the percentage is what compares from one situation to another, the absolute value is what compares with the biological threshold.
Step 5
Comparison with the threshold: 8.31>6.0 mg/L, the threshold is not crossed.
Why
The conclusion must answer the question asked, not just give the number. And it must name what the calculation does not cover.
The conclusion, written out
“Warming takes the dissolved oxygen from 9.86 to 8.31 mg/L, a drop of 15.8%. The 6.0 mg/L threshold is not crossed, but the margin above the threshold falls from 3.86 to 2.31 mg/L, forty percent less; and fish metabolism increases with temperature, so that demand grows while supply shrinks.”
The classic mistake on this problem: Taking the total atmospheric pressure instead of the partial pressure, which multiplies everything by five, or losing the sign of ΔH and concluding that warm water holds more oxygen.
Learn by heart
•Lattice plus hydration: a small difference between two large numbers.
•The criterion is ΔG, never ΔH alone.
•Enthalpy in kilojoules, entropy in joules: convert before subtracting.
•Henry takes the PARTIAL pressure, and kH is an equilibrium constant.
•A gas is less soluble when hot, a solid is more soluble: it is Le Chatelier.
•Weak electrolyte: i=1+α, never the number of ions in the formula.
•Ionic strength: the charge counts SQUARED.
•Tabulated constants are written with activities, not concentrations.
•Two volatile liquids: Raoult has TWO terms, and the vapor is enriched.
•Extraction: split the solvent, the remaining fractions multiply.
Frequently asked questions
Why does a salt that cools the water dissolve anyway?
Because the criterion for spontaneity is free energy, not enthalpy. Dissolving ammonium nitrate absorbs twenty-six kilojoules per mole, but spreading its ions through the whole volume of the solvent gains one hundred and eight joules per mole per kelvin of entropy. At twenty-five degrees, the entropy term is worth thirty-two kilojoules and wins: the free energy is minus six kilojoules per mole, the dissolution is spontaneous and the solution cools. This is exactly the principle of the instant cold pack sold in pharmacies. Below two hundred and forty-one kelvins, entropy would no longer be enough.
Why do gases dissolve less well in hot water?
Because their dissolution releases heat, unlike that of most solids. There is no crystal lattice to destroy, only solvation to gain, so the enthalpy of solution is negative, about minus twelve kilojoules per mole for oxygen. Heating therefore amounts to adding a product to the dissolution equilibrium, and Le Chatelier's principle shifts it toward the gas. This is why a warm soft drink fizzes less and why a river warmed by ten degrees loses nearly sixteen percent of its dissolved oxygen.
How do you find the van 't Hoff factor of a weak acid?
You do not read it off the formula, you calculate it from the degree of dissociation. For a monoprotic acid, each dissociated molecule gives two particles instead of one, so the factor is one plus alpha. And alpha is obtained from the acid dissociation constant: for acetic acid at one tenth of a mole per kilogram, it is about one point three percent, which gives a factor of one point zero one three, and not two. A weak electrolyte therefore behaves almost like a sugar from the colligative point of view.
What is activity, and why not stick to concentrations?
Because an ion in solution is never alone: it is surrounded by an atmosphere of oppositely charged ions that screens it, so it acts as if it were less concentrated than it is. Activity is the product of the concentration and a coefficient below one, and all tabulated constants, solubility product as well as acid dissociation constant, are written with activities. As long as the solution is very dilute the coefficient is almost one and the two can be confused; once the ionic strength reaches a few hundredths, the real solubility exceeds by twenty to thirty percent what the constant predicted. This is the salt effect.
Why does the distillation of alcohol level off at ninety-five percent?
Because of an azeotrope. Ethanol and water form a mixture that departs strongly upward from Raoult's law, because mixing breaks the hydrogen-bond network of water without rebuilding as much. This deviation is large enough for the mixture at ninety-five point six percent to boil lower than pure ethanol. At this composition, the vapor has exactly the same composition as the liquid: distillation no longer enriches anything, however many plates the column has. To go beyond, the process is changed: molecular sieve or entrainer.
Practise it
Corrected exercises: Solutions and colligative properties, 202-SN2-RE
A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.
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Get in touch for a first session. We work on dissolution equilibria and colligative properties together with the write-up expected on the exam, so that the marks for the method stop being lost.
