Corrected exercises: solutions and colligative properties (202-SN2-RE)

$\Delta H_{sol}=+649+(-623)=+26$ kJ/mol. The lattice term wins by a small margin: the dissolution ABSORBS heat, and the solution cools. This is the principle of the instant cold pack sold in pharmacies.

$\Delta H_{sol}=+2258+(-2340)=-82$ kJ/mol. This time hydration wins, and the dissolution RELEASES heat. Both terms are three to four times larger than for ammonium nitrate, because the calcium ion carries two charges and there are two chloride ions per formula unit: the lattice energy grows with the product of the charges.

Amount: $n=\dfrac{20.0}{80.04}=0.2499$ mol. Heat absorbed: $Q=26\times 0.2499=6.50$ kJ, that is $6.50\times 10^{3}$ J. This heat is taken from the whole contents, of mass $100+20=120$ g: $\Delta T=-\dfrac{6500}{120\times 4.18}=-13.0$ °C. The final temperature is therefore $25.0-13.0=12.0$ °C. The MINUS sign comes from the chemical system taking the heat from the solution; writing it the wrong way round gives $38$ °C and a hot pack, which should have raised a flag.

Consequence: in practice $\Delta H_{sol}$ is NEVER calculated from this cycle, it is MEASURED in a calorimeter. An uncertainty of $1\%$ on the lattice energy of calcium chloride, that is $23$ kJ/mol, is by itself more than a quarter of the quantity being sought. This is the textbook case of a trap met everywhere in the laboratory: a small difference between two large quantities cannot be calculated, it has to be measured. The cycle is there to UNDERSTAND the sign, not to predict it to the nearest kilojoule.

The lithium ion is much SMALLER than the potassium ion, $76$ pm against $138$ pm, for the same $+1$ charge. Its charge density, and therefore the electric field at its surface, is much greater: it organizes more water molecules around itself and its enthalpy of hydration is much more negative. The lattice energy also increases as the ion shrinks, but more slowly, so the balance tips toward hydration: $\Delta H_{sol}$ is $-37$ kJ/mol for lithium chloride against $+17$ kJ/mol for potassium chloride. The rule to remember: SMALL AND HIGHLY CHARGED means strongly hydrated.

$\Delta G_{sol}=26\,000-298\times 108=26\,000-32\,184=-6184$ J/mol, that is $-6.2$ kJ/mol. The dissolution is therefore SPONTANEOUS even though it cools the solution: entropy pays the bill. Watch out for the unit trap, which costs a mark on every exam: the enthalpy is in kilojoules, the entropy in joules per kelvin. Without converting, you get $26-32\,184=-32\,158$ kJ/mol, an absurd value that no order-of-magnitude check would let through.

Because the two effects do not weigh the same. Destroying a crystal, where every ion sits in a fixed position, and spreading the ions through the whole volume of the solvent creates an enormous amount of disorder. The water of hydration does organize itself around each ion, which costs entropy, but that only involves a shell of a few molecules per ion. The balance remains strongly positive. There are counterexamples: for a very small, highly charged ion such as $Al^{3+}$, the order imposed on the water wins and $\Delta S_{sol}$ becomes NEGATIVE.

$\Delta G_{sol}=-12\,500-298\times(-95)=-12\,500+28\,310=+15\,810$ J/mol, that is $+15.8$ kJ/mol. The POSITIVE sign says that the dissolution of oxygen is only slightly favourable: its equilibrium constant is small, and water in equilibrium with air holds only eight or nine milligrams of it per litre. Entropy works against the dissolution here, which is normal: a gas going from the dispersed state to the dissolved state LOSES disorder.

The dissolution stops being spontaneous when $\Delta G_{sol}$ becomes positive, that is when $T<\dfrac{\Delta H_{sol}}{\Delta S_{sol}}=\dfrac{26\,000}{108}=241$ K, about $-32$ °C. Above this temperature the $-T\Delta S$ term wins; below it, it is no longer enough. This is exactly the crossover temperature seen in thermodynamics, applied to a dissolution.

For a SOLID whose dissolution is endothermic, heating supplies what the process consumes: the solubility INCREASES with temperature, as anyone can see with sugar. For a GAS, whose dissolution is exothermic, heating shifts the equilibrium the other way: the solubility DECREASES, which makes a warm soft drink go flat and suffocates the fish of a warmed river. In both cases it is Le Chatelier's principle applied to the dissolution equilibrium, treating heat as a reactant or a product. The rule is therefore a single one, and it is the signs of $\Delta H_{sol}$ that differ, not the law.

We apply van 't Hoff between $T_{1}=298.15$ K and $T_{2}=308.15$ K: $\ln\dfrac{K_{2}}{K_{1}}=-\dfrac{-12\,500}{8.314}\left(\dfrac{1}{308.15}-\dfrac{1}{298.15}\right)=1503.5\times(-1.089\times 10^{-4})=-0.1637$. So $K_{2}=1.3\times 10^{-3}\times e^{-0.1637}=1.10\times 10^{-3}$ mol/(L·atm). Temperatures are in KELVINS, and the double minus sign is where mistakes happen: with a negative $\Delta H$, the factor $-\dfrac{\Delta H}{R}$ is POSITIVE.

At $25$ °C: $[O_{2}]=k_{H}P_{O_{2}}=1.3\times 10^{-3}\times 0.21=2.73\times 10^{-4}$ mol/L, that is $2.73\times 10^{-4}\times 32.0\times 1000=8.74$ mg/L. At $35$ °C: $1.10\times 10^{-3}\times 0.21=2.32\times 10^{-4}$ mol/L, that is $7.42$ mg/L. We use the PARTIAL pressure of oxygen, not the total atmospheric pressure: this is the most common error with Henry's law, and it multiplies the result by five.

$\dfrac{7.42-8.74}{8.74}=-0.151$, a drop of $15.1\%$ for ten degrees. That is considerable for such a modest change in temperature.

At $18$ °C, that is $291.15$ K: $\ln\dfrac{K}{K_{25}}=1503.5\times 8.07\times 10^{-5}=0.1213$, so $K=1.47\times 10^{-3}$ and $[O_{2}]=9.86$ mg/L. At $28$ °C, that is $301.15$ K: $\ln\dfrac{K}{K_{25}}=-0.0502$, so $K=1.24\times 10^{-3}$ and $[O_{2}]=8.31$ mg/L. The drop is $1.55$ mg/L, that is $15.8\%$, and the $6.0$ mg/L threshold is not crossed.

The direction can be read directly: the dissolution equation is written $O_{2}(g)\rightleftharpoons O_{2}(aq)+\text{heat}$ since $\Delta H_{sol}<0$. Heating means adding a product, so the equilibrium shifts to the LEFT: the gas leaves. The van 't Hoff calculation only puts numbers on what Le Chatelier announces. As for the real severity: the $8.31$ mg/L figure is only half of the problem, because a fish's metabolism, like any biochemical reaction, SPEEDS UP with temperature, roughly by a factor of two for ten degrees according to Arrhenius. Warm water therefore offers less oxygen to animals that need more of it, and the two effects multiply instead of adding. On top of that, in summer the river is already warm and low: it is the combination that kills, not the ten degrees taken in isolation.

$i=\dfrac{\Delta T_{f}}{K_{f}\,m}=\dfrac{0.1885}{1.86\times 0.100}=\dfrac{0.1885}{0.186}=1.0134$. The value is only just above $1$: acetic acid behaves almost like a nonelectrolyte, which is precisely the definition of a WEAK acid. A student who announced $i=2$ because the formula releases one $H^{+}$ would be off by a factor of about seventy-five on $\alpha$.

$i=1+\alpha$, so $\alpha=i-1=0.0134$, that is $1.34\%$. In other words, out of a thousand acetic acid molecules, only thirteen have given up their proton.

$K_{a}=\dfrac{0.100\times(0.0134)^{2}}{1-0.0134}=\dfrac{0.100\times 1.806\times 10^{-4}}{0.9866}=1.83\times 10^{-5}$. The agreement with the tabulated value, $1.8\times 10^{-5}$, is excellent. The denominator $1-\alpha$ changes almost nothing here, $1.3\%$, but it must be written: it is what distinguishes the exact formula from the approximation, and it becomes decisive as soon as $\alpha$ exceeds $5\%$.

The uncertainty propagates first onto $i$: $\Delta i=\dfrac{0.002}{0.186}=0.0108$. Now $\alpha=i-1$ is a DIFFERENCE, and the absolute uncertainty survives intact: $\alpha=0.0134\pm 0.0108$, a relative uncertainty of $\dfrac{0.0108}{0.0134}=80\%$. The degree of dissociation is therefore known only between $0.3$ and $2.4\%$, and $K_{a}$, which depends on $\alpha^{2}$, between $1\times 10^{-6}$ and $6\times 10^{-5}$: almost two orders of magnitude. This is the classic trap of the small difference between two nearby quantities, here $i$ and $1$. The method is sound and made history, but it is metrologically disastrous for a weak acid.

A pH measurement. The true degree of dissociation gives $[H^{+}]=c\alpha=0.100\times 0.0134=1.34\times 10^{-3}$ mol/L, so $pH=-\log(1.34\times 10^{-3})=2.87$. The decisive advantage is that pH measures $[H^{+}]$ DIRECTLY, without going through a difference: an ordinary pH meter reads to the hundredth of a unit, which gives $[H^{+}]$ to within $2\%$ and $K_{a}$ to within $5\%$, sixteen times better than freezing-point depression. As a general rule, the same one as in the physics lab: when the quantity sought is a small difference, you change to a method that measures it directly rather than improving the thermometer.

Sodium chloride gives $0.010$ mol/L of $Na^{+}$ with charge $+1$ and $0.010$ mol/L of $Cl^{-}$ with charge $-1$: $I=\dfrac{1}{2}\left(0.010\times 1^{2}+0.010\times 1^{2}\right)=0.010$ mol/L. For a $1{:}1$ electrolyte, the ionic strength is EQUAL to the concentration: this is the only case where the two can be confused.

Magnesium sulfate gives $0.010$ mol/L of $Mg^{2+}$ and $0.010$ mol/L of $SO_{4}^{2-}$, both with charge $\pm 2$: $I=\dfrac{1}{2}\left(0.010\times 4+0.010\times 4\right)=0.040$ mol/L, that is FOUR TIMES more for the same concentration. The charge enters SQUARED: it is the charge that rules, not the number of ions.

Sodium chloride: $\log\gamma_{\pm}=-0.509\times 1\times\sqrt{0.010}=-0.0509$, so $\gamma_{\pm}=10^{-0.0509}=0.889$. Magnesium sulfate: $\log\gamma_{\pm}=-0.509\times 4\times\sqrt{0.040}=-0.407$, so $\gamma_{\pm}=0.392$. The second salt suffers both penalties, a charge product four times larger AND an ionic strength four times larger: its activity falls to less than $40\%$ of its concentration.

Yes, perfectly. The deviation from ideality is measured by $2-i$: it is $0.1$ for sodium chloride and $0.7$ for magnesium sulfate, SEVEN times more, of the same order as the ratio of the deviations of $\gamma$ from $1$, $0.111$ against $0.608$, five times more. Physically the cause is the same: doubly charged ions attract each other enough to travel as PAIRS part of the time, with $Mg^{2+}SO_{4}^{2-}$ then counting as ONE particle instead of two. Colligative properties count particles; they therefore see this pairing directly.

The student will observe that the salt becomes MORE SOLUBLE, even though no common ion was added and the temperature has not moved. This is the salt effect, and it follows from everything above: the tabulated $K_{sp}$ is a constant of ACTIVITIES, $K_{sp}=\gamma_{+}[M^{+}]\gamma_{-}[X^{-}]$. Adding potassium nitrate raises the ionic strength to $0.10$ mol/L, so the $\gamma$ values fall; for the product to stay constant, the CONCENTRATIONS must rise. The increase is commonly $20$ to $30\%$ for a $1{:}1$ salt, more for a $2{:}2$ salt. The moral of the chapter: a tabulated constant is thermodynamic, so it is written in activities; treating it as a constant of concentrations is legitimate only in very dilute solution.

The LIQUID-VAPOR curve moves down, because the solute occupies part of the surface and dilutes the water molecules: fewer of them escape, so the vapor pressure of the liquid falls, exactly what Raoult's law says. The SOLID-VAPOR curve stays put, because the solid that crystallizes is PURE ice: the solute stays in the liquid and has no effect on sublimation. The solid-liquid curve, which starts from the intersection of the first two, therefore shifts to the left as a consequence. Everything else, freezing-point depression and boiling-point elevation, can be read off the figure.

Glucose is a nonelectrolyte, so $i=1$. Freezing: $\Delta T_{f}=K_{f}m=1.86\times 1.00=1.86$ °C, and the temperature goes DOWN: $-1.86$ °C. Boiling: $\Delta T_{b}=K_{b}m=0.512\times 1.00=0.512$ °C, and the temperature goes UP: $100.512$ °C. The two constants are not of the same order: the effect on freezing is nearly FOUR times larger, which explains why we salt roads and do not make a pot of water boil faster by salting it.

Sodium chloride releases two ions, so $i=2$: $\Delta T_{f}=2\times 1.86=3.72$ °C and $\Delta T_{b}=2\times 0.512=1.02$ °C. The solution freezes at $-3.72$ °C and boils at $101.02$ °C. What matters for a colligative property is the number of PARTICLES, not their nature: one mole of salt is worth two moles of sugar.

The triple point is the INTERSECTION of the three curves. Since the liquid-vapor curve moves down and the solid-vapor curve stays put, their intersection slides along the sublimation curve, to the LEFT and DOWN: lower temperature, lower pressure. This is consistent with b): the triple point is very close to the freezing point.

No, the ice is NOT sweet: it is made of pure water, and all the sugar has stayed in the liquid, which has therefore become more concentrated. The process is called FREEZE CONCENTRATION, and it is used to concentrate fruit juices, cider vinegar or beer without heating them, and therefore without destroying the volatile aromas or the vitamins. The same physics gives ice wine, the purification of metals by zone melting, and sea ice, which is almost fresh although the water it comes from is salty. In practice, a little concentrated liquid always stays trapped between the crystals, so rapid freezing purifies less well than slow freezing: this is why the industrial process takes its time.

$P=x_{eth}P_{eth}^{\circ}+x_{water}P_{water}^{\circ}=0.90\times 59+0.10\times 23.8=53.1+2.38=55.5$ mmHg. Since both liquids are VOLATILE, both terms count: Raoult's law in the form $P=x_{solvent}P^{\circ}$ holds only for a nonvolatile solute, and applying it here would lose $2.4$ mmHg.

$\dfrac{60-55.5}{55.5}=0.081$, that is $+8.1\%$. The measured pressure EXCEEDS the ideal pressure: this is a POSITIVE deviation.

$\Delta H_{mix}>0$, mixing is endothermic, and the two facts always go together. Pure water is a dense network of hydrogen bonds, where each molecule forms close to four; ethanol, whose single hydroxyl group is hindered by a hydrophobic chain, can offer only one. Mixing therefore amounts to breaking a very cohesive network and replacing it with fewer bonds: this costs energy, and the molecules, less strongly held, escape more readily. A positive deviation, an endothermic mixing and an increased pressure are three readings of the same fact.

This point is called an AZEOTROPE, and here it is a MINIMUM in boiling temperature, since $78.2$ °C is below both pure-substance temperatures. At this composition, the vapor has exactly the SAME composition as the liquid: distillation no longer enriches anything and the mixture boils like a pure substance. This is why distilled alcohol levels off at $95.6\%$, however many plates the column has. To go beyond, we stop distilling: we pass the mixture over a molecular sieve, which holds back the water by its size, or we add a third component as an entrainer. The general lesson: a sufficiently strong positive deviation always creates a minimum-boiling azeotrope.

Acetone carries an electron-rich carbonyl oxygen, and chloroform a hydrogen made acidic by the three chlorines: together they form a HYDROGEN BOND that neither could form alone. The molecules are therefore held more strongly in the mixture than in the pure liquids: the vapor pressure falls below the ideal value, the deviation is NEGATIVE, and the mixture warms up, $\Delta H_{mix}<0$. In distillation this gives a MAXIMUM-boiling azeotrope, $64.5$ °C at about $20\%$ acetone, which boils HIGHER than both of its components. The practical consequence is the opposite of the previous case: it is no longer the distillate that runs into the azeotrope, it is the column RESIDUE that accumulates it.

FALSE. The criterion for spontaneity is $\Delta G=\Delta H-T\Delta S$, not $\Delta H$ alone. For ammonium nitrate, $\Delta H_{sol}=+26$ kJ/mol but $\Delta S_{sol}=+108$ J/(mol·K), so $\Delta G_{sol}=-6.2$ kJ/mol at $25$ °C: the dissolution is spontaneous AND cools the solution. Correct version: “an endothermic dissolution is spontaneous as soon as the entropy gain compensates, which is the case for most salts above a certain temperature”.

FALSE. The number of particles actually present cannot be read off the formula, it is MEASURED. Acetic acid is only $1.34\%$ dissociated at $0.100$ mol/kg, so $i=1+\alpha=1.013$, almost $1$. Only STRONG electrolytes approach their theoretical $i$, and even then only optimistically, since ion pairing moves them away from it. Correct version: “$i=1+\alpha$ for a weak monoprotic acid, and $\alpha$ is obtained from $K_{a}$ or from a measurement”.

FALSE, and it is the opposite. The dissolution of a gas is EXOTHERMIC, $\Delta H_{sol}\approx-12.5$ kJ/mol for oxygen, because there is no lattice to destroy, only solvation to gain. Heating therefore means adding a product to the dissolution equilibrium: the gas escapes. A warm soft drink fizzes less, and a river warmed by ten degrees loses $16\%$ of its dissolved oxygen. Correct version: “the solubility of a gas DECREASES as temperature rises, by Le Chatelier”.

FALSE. This is precisely what makes distillation possible. In an ideal mixture, each component contributes a partial pressure $x_{i}P_{i}^{\circ}$, so its fraction in the VAPOR is $y_{i}=\dfrac{x_{i}P_{i}^{\circ}}{P_{total}}$: the component with the larger $P^{\circ}$ is overrepresented. An equimolar mixture of ethanol and water at $25$ °C gives a vapor that is $71\%$ ethanol. Correct version: “the vapor is always enriched in the more volatile component, EXCEPT at the azeotropic point, where the two compositions coincide”.

FALSE. The tabulated $K_{sp}$ is a constant of ACTIVITIES, $K_{sp}=\gamma_{+}[M^{+}]\,\gamma_{-}[X^{-}]$. Adding an inert salt raises the ionic strength, and so lowers the activity coefficients; for the product to stay constant, the concentrations must RISE. The salt becomes more soluble, commonly by $20$ to $30\%$: this is the salt effect, not to be confused with the common-ion effect, which goes the other way. Correct version: “$K_{sp}$ depends only on temperature, but it is written with activities; as ionic strength increases, the solubility in concentration increases”.

Because a solution does NOT freeze all at once. At $-5$ °C, part of the water has crystallized as pure ice and the rest forms a liquid that has become more concentrated, and so whose freezing point has dropped even further: freezing-point depression chasing its own tail. This residual liquid disappears only at the EUTECTIC temperature, $-21.1$ °C for a water-sodium chloride system, where everything finally solidifies. Now it is precisely this concentrated liquid that would damage the product and that, under vacuum, would start to boil and spatter the sample. The product is therefore frozen well below it, typically at $-40$ °C.

The pressure must stay BELOW the triple point, otherwise the ice would melt instead of sublimating: the pressure must therefore be below $611$ Pa. In millimetres of mercury, $611\times\dfrac{760}{101\,325}=4.58$ mmHg. In practice we work much lower, around $10$ to $50$ Pa, so that sublimation is fast enough.

$E=m\,\Delta H_{sub}=1.00\times 2835=2835$ kJ. Over $20$ h, that is $20\times 3600=72\,000$ s: $P=\dfrac{2.835\times 10^{6}}{72\,000}=39.4$ W. That is barely the power of a light bulb: freeze-drying is not expensive in POWER, it is expensive in TIME, and it is the vacuum to be maintained for twenty hours that weighs on the bill.

Freeze-drying first requires freezing, so removing $334$ kJ, then sublimating, so supplying $2835$ kJ: in all, $334+2835=3169$ kJ of energy to move around. Drying by evaporation at $100$ °C costs only $2260$ kJ. The extra cost is $\dfrac{3169-2260}{2260}=0.40$, that is $+40\%$, not counting the energy of the cooling unit and the vacuum pump, which easily multiplies the real bill by three.

Because the liquid route would destroy the product, in at least two ways. First CONCENTRATION: as it evaporates, the water leaves behind salts and sugars that become more and more concentrated, and whose ionic strength denatures proteins, exactly what exercise 5 describes. Second HEAT: at $100$ °C, a vaccine, an enzyme or a volatile aroma will not survive. A third, mechanical effect comes on top: the surface tension of the receding liquid collapses the pores of the product, which becomes hard and no longer rehydrates. When ice sublimates, it leaves in its place a void of the same shape: the product keeps its porous structure, and a grain of freeze-dried coffee redissolves in a few seconds. We pay $40\%$ in energy to never meet a liquid phase.

$f=\dfrac{100}{100+8.0\times 100}=\dfrac{100}{900}=0.111$. So $0.111$ g remain in the water, and $0.889$ g have been extracted, a yield of $88.9\%$. Note that $K=8$ does NOT mean that $8/9$ is extracted by luck: that is exactly what the calculation says, because the two volumes are equal.

Each extraction leaves $f_{1}=\dfrac{100}{100+8.0\times 50}=\dfrac{100}{500}=0.200$. Since the two are independent, the fractions MULTIPLY: $f=0.200^{2}=0.0400$. $0.040$ g remain, and the yield reaches $96.0\%$. The same ether, split in two, brings the loss down from $11.1$ to $4.0\%$.

$f_{1}=\dfrac{100}{100+8.0\times 25}=\dfrac{100}{300}=0.3333$, so $f=0.3333^{4}=0.01235$. $0.0123$ g remain and the yield rises to $98.8\%$. The loss has been divided by more than three again, still with the same $100$ mL.

In general, $f=\left(\dfrac{V_{aq}}{V_{aq}+\dfrac{KV}{n}}\right)^{n}=\left(\dfrac{1}{1+\dfrac{KV}{nV_{aq}}}\right)^{n}$. Setting $u=\dfrac{KV}{V_{aq}}=8.0$, this is written $\left(1+\dfrac{u}{n}\right)^{-n}$, whose limit as $n\to\infty$ is $e^{-u}=e^{-8}=3.4\times 10^{-4}$: only $0.34$ mg would remain, a yield of $99.97\%$. We recognize the limit that defines the number $e$. In practice we stop at about three or four extractions: the gain becomes marginal while each manipulation costs time and loses product in the glassware.

An amine is a BASE. In acidic medium, it picks up a proton and becomes an $RNH_{3}^{+}$ cation, a charged and therefore very hydrophilic species: its partition coefficient collapses and it stays in the water. In basic medium, it becomes neutral again and goes back into the ether. It can thus be moved back and forth at will by adjusting the pH, and this is the very principle of the acid-base extraction that separates an amine from a carboxylic acid and a neutral compound in the same mixture. To increase $K$ without touching the pH, we SALT the aqueous phase, dissolving sodium chloride in it up to saturation: the ions tie up the water molecules, which become less available to solvate the organic product, and it is driven toward the ether. This is salting out, yet another ionic-strength effect, and it is the most profitable move in the whole organic laboratory.