Corrected practice final exam in chemistry of solutions (202-SN2-RE)

At equilibrium, the forward rate $k_{1}\left[\mathrm{A}\right]$ and the reverse rate $k_{-1}\left[\mathrm{B}\right]$ exactly balance. This does not make both of them zero, it makes them equal: it is a dynamic equilibrium. From $k_{1}\left[\mathrm{A}\right]=k_{-1}\left[\mathrm{B}\right]$ we get $\frac{\left[\mathrm{B}\right]}{\left[\mathrm{A}\right]}=\frac{k_{1}}{k_{-1}}$, which is by definition $K$. Therefore $K=\frac{4.0\times 10^{-3}}{5.0\times 10^{-4}}=8.0$.

The total amount is conserved: $\left[\mathrm{A}\right]+\left[\mathrm{B}\right]=0.50$. With $\left[\mathrm{B}\right]=8\left[\mathrm{A}\right]$, we get $9\left[\mathrm{A}\right]=0.50$, so $\left[\mathrm{A}\right]\approx 0.0556$ and $\left[\mathrm{B}\right]\approx 0.4444$ mol/L. The ratio is indeed 8.

The approach constant is $k_{1}+k_{-1}=4.5\times 10^{-3}\ \mathrm{s}^{-1}$, so the half-approach time is $\frac{\ln 2}{4.5\times 10^{-3}}\approx 154.0$ seconds, about 2 minutes 34 seconds. Note that it is NOT $\frac{\ln 2}{k_{1}}$: the two reactions run at the same time, and it is their sum that governs how fast the system returns to equilibrium.

A catalyst lowers the activation barrier, which is the same hill seen from both slopes: it therefore multiplies $k_{1}$ AND $k_{-1}$ by the same factor of 100. The ratio $K=\frac{k_{1}}{k_{-1}}$ is unchanged and is still 8, so the final concentrations are exactly those of b). Only the rate changes: $k_{1}+k_{-1}$ is multiplied by 100 and the half-approach time drops to about 1.54 s. This is the sharpest form of the rule from the course: a catalyst acts on the time, never on the destination.

Seen from the products, the barrier is higher than seen from the reactants, by exactly the drop in energy: $E_{a}(\text{reverse})=E_{a}(\text{forward})-\Delta H=85-(-45)=130$ kJ/mol. On the profile, the reactants are on the left, the transition state peaks 85 kJ higher, and the products lie 45 kJ below the reactants, hence 130 kJ below the peak.

$\ln\frac{k_{2}}{k_{1}}=-\frac{85{,}000}{8.314}\times\left(-4.8214\times 10^{-4}\right)\approx 4.929$, so the ratio is $\mathrm{e}^{4.929}\approx 138.3$. Fifty more degrees multiply the forward rate by about 138.

With $E_{a}=130$ kJ: $\ln\frac{k_{2}}{k_{1}}=-\frac{130{,}000}{8.314}\times\left(-4.8214\times 10^{-4}\right)\approx 7.539$, so the ratio is $\mathrm{e}^{7.539}\approx 1879.8$. The reverse reaction is sped up far more, because its barrier is higher, and it is precisely the height of the barrier that makes a reaction sensitive to temperature.

Since $K=\frac{k_{1}}{k_{-1}}$, the factor on $K$ is the quotient of the two factors: $\frac{138.3}{1879.8}\approx 0.0736$. By van 't Hoff: $\ln\frac{K_{2}}{K_{1}}=-\frac{-45{,}000}{8.314}\times\left(-4.8214\times 10^{-4}\right)\approx -2.610$, so $\frac{K_{2}}{K_{1}}\approx 0.0736$. The two routes give exactly the same number, which is no accident: the van 't Hoff equation is just the difference of the two Arrhenius relations, since $\Delta H=E_{a}(\text{forward})-E_{a}(\text{reverse})$.

There is no paradox, because rate and yield answer two different questions. Heating does speed up both directions, so equilibrium is reached much sooner: here about 138 times sooner for the forward direction. But the equilibrium reached is not the same, because the reverse direction was sped up 14 times more than the forward one, which moves the meeting point toward the reactants. This is Le Chatelier's principle again: for an exothermic reaction, raising the temperature disfavours the products. The industrial chemist therefore constantly trades arriving quickly against arriving far, which is exactly the compromise of the Haber process.

$Q=\frac{\left[\mathrm{HI}\right]^{2}}{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}=\frac{0^{2}}{0.100\times 0.100}=0$. Since $Q<K$, the system evolves in the forward direction, toward the formation of hydrogen iodide.

With a concentration change $x$: $\left[\mathrm{H}_{2}\right]=\left[\mathrm{I}_{2}\right]=0.100-x$ and $\left[\mathrm{HI}\right]=2x$. Then $K=\frac{(2x)^{2}}{(0.100-x)^{2}}$, and since both sides are squares of positive quantities we can take the square root: $\frac{2x}{0.100-x}=\sqrt{50.0}\approx 7.0711$. Hence $2x=0.70711-7.0711x$, that is $9.0711x=0.70711$ and $x\approx 0.07795$. The equilibrium concentrations are therefore $\left[\mathrm{H}_{2}\right]=\left[\mathrm{I}_{2}\right]\approx 0.0220$ and $\left[\mathrm{HI}\right]\approx 0.1559$ mol/L. Check: $\frac{0.1559^{2}}{0.0220^{2}}\approx 50.0$.

$Q=\frac{0.300^{2}}{0.0220^{2}}\approx 185.1$. Since $Q>K$, the system now evolves in the reverse direction, consuming hydrogen iodide until the quotient falls back to 50.0.

The statement confuses the direction of evolution with a comparison of rate constants. What $Q>K$ says is that AT THIS INSTANT the reverse rate exceeds the forward rate, because the hydrogen iodide concentration was raised far above its equilibrium value. It says nothing about the constants $k_{1}$ and $k_{-1}$, which have not moved since the temperature has not changed. In fact, as the system returns to equilibrium, the two rates draw closer and end up equal again, without any constant having changed. A rate depends on the concentrations as much as on the constant.

The two successive ionizations are written $K_{a1}=\frac{\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]\left[\mathrm{HS}^{-}\right]}{\left[\mathrm{H}_{2}\mathrm{S}\right]}$ and $K_{a2}=\frac{\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]\left[\mathrm{S}^{2-}\right]}{\left[\mathrm{HS}^{-}\right]}$. Multiplying them side by side, the $\mathrm{HS}^{-}$ ion cancels: $K_{a1}K_{a2}=\frac{\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]^{2}\left[\mathrm{S}^{2-}\right]}{\left[\mathrm{H}_{2}\mathrm{S}\right]}$, hence $\left[\mathrm{S}^{2-}\right]=\frac{K_{a1}K_{a2}\left[\mathrm{H}_{2}\mathrm{S}\right]}{\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]^{2}}$.

$\left[\mathrm{S}^{2-}\right]=\frac{1.0\times 10^{-7}\times 1.3\times 10^{-13}\times 0.10}{(0.30)^{2}}=\frac{1.3\times 10^{-21}}{0.090}\approx 1.44\times 10^{-20}$ mol/L. The sulfide ion concentration, of the order of $10^{-20}$, is therefore driven by the acidity alone, which makes it a remarkably fine adjustment.

For copper sulfide: $Q=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{S}^{2-}\right]=1.0\times 10^{-2}\times 1.44\times 10^{-20}\approx 1.44\times 10^{-22}$, far above $8.0\times 10^{-37}$: it precipitates massively, with a margin of fourteen orders of magnitude. For zinc sulfide, the quotient is the same since both cations are at the same concentration, that is $1.44\times 10^{-22}$, which still exceeds $1.6\times 10^{-24}$ by a factor of 90: it precipitates too.

No, the pH is not suitable, since both sulfides precipitate together and nothing is separated. We must lower $\left[\mathrm{S}^{2-}\right]$, and it varies as the inverse of the square of $\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]$: the solution must therefore be made MORE acidic, that is the pH must be lowered. Since zinc sulfide is the more soluble of the two, by more than twelve orders of magnitude, it is the one that stops precipitating first, and there is a wide window of acidity in which copper is completely precipitated while zinc stays in solution. This is exactly the principle behind the qualitative analysis of cations in groups.

At pH 3.0, $\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{10^{-3}}=1.0\times 10^{-11}$ mol/L. The equilibrium gives $\left[\mathrm{Fe}^{3+}\right]=\frac{K_{sp}}{\left[\mathrm{OH}^{-}\right]^{3}}=\frac{2.8\times 10^{-39}}{\left(1.0\times 10^{-11}\right)^{3}}=\frac{2.8\times 10^{-39}}{1.0\times 10^{-33}}=2.8\times 10^{-6}$ mol/L.

For nickel, the concentration that would be at equilibrium is $\frac{5.5\times 10^{-16}}{\left(1.0\times 10^{-11}\right)^{2}}=5.5\times 10^{6}$ mol/L, an absurd number that no solution can reach. In other words the actual ion product, $1.0\times 10^{-2}\times\left(10^{-11}\right)^{2}=1.0\times 10^{-24}$, is far below $5.5\times 10^{-16}$: the nickel does not precipitate at all.

The residual iron is $\frac{2.8\times 10^{-6}}{1.0\times 10^{-2}}=2.8\times 10^{-4}$, that is $0.028\%$ of the initial iron. More than 99.97% of the iron is precipitated while all of the nickel stays in solution: the separation is excellent.

At pH 8.0, $\left[\mathrm{OH}^{-}\right]=1.0\times 10^{-6}$. The residual iron becomes $\frac{2.8\times 10^{-39}}{10^{-18}}=2.8\times 10^{-21}$ mol/L, a perfectly useless gain since it was already negligible. But the residual nickel falls to $\frac{5.5\times 10^{-16}}{10^{-12}}=5.5\times 10^{-4}$ mol/L, well below $1.0\times 10^{-2}$: the nickel then precipitates to 94.5%. The proposal is therefore bad, even counterproductive: it destroys exactly what we were after, the selectivity. Precipitating more is only progress if you precipitate what you should.

Because the exponent on $\left[\mathrm{OH}^{-}\right]$ differs: it is 3 for a trivalent cation and 2 for a divalent one. The residual iron concentration therefore varies as $\left[\mathrm{OH}^{-}\right]^{-3}$ and that of nickel as $\left[\mathrm{OH}^{-}\right]^{-2}$. Each pH unit gained divides the residual iron by a thousand and the nickel by only a hundred: the gap between the two widens by a factor of ten per pH unit, mechanically. Two cations of the same charge, even with very different solubility products, would separate much less cleanly, because their curves would stay parallel instead of diverging.

At the equivalence point the amounts of substance are equal, and since the concentrations are equal too, the volume added is $25.0$ mL. The total volume is then $50.0$ mL, so the acetate ion formed, $2.50\times 10^{-3}$ mol, is spread over twice the initial volume: its concentration is $0.0500$ mol/L.

At the equivalence point only the acetate ion remains, the conjugate weak base. Its base ionization constant is $K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0\times 10^{-14}}{1.8\times 10^{-5}}\approx 5.56\times 10^{-10}$. Neglecting the autoionization of water, $\left[\mathrm{OH}^{-}\right]=\sqrt{K_{b}C}=\sqrt{5.56\times 10^{-10}\times 0.0500}\approx 5.27\times 10^{-6}$, so $\mathrm{pOH}\approx 5.28$ and $\mathrm{pH}\approx 8.72$. The equivalence point of a weak acid-strong base titration is therefore basic, not neutral.

The colour-change ranges are approximately $[2.7,\ 4.7]$ for methyl orange, $[6.0,\ 8.0]$ for bromothymol blue and $[8.4,\ 10.4]$ for phenolphthalein. Only the last one brackets the equivalence pH of 8.72: that is the one to choose.

The student applies a rule that does not exist: an indicator is not chosen by the nature of the substance being titrated but by the pH at the equivalence point, which depends on the PAIR involved. Methyl orange would change colour around pH 4, well before the equivalence point, in the buffer region where the pH rises very slowly. Since the pH equals $\mathrm{p}K_{a}=4.74$ at the half-equivalence point, a colour change between pH 3.7 and 4.7 happens around the half-equivalence point: the student would stop the titration at about 12 to 13 mL instead of 25, underestimating the concentration of the acid by nearly half. The error is not a matter of shade, it is of the same order as the measurement itself.

For a strong acid-strong base titration. The pH jump there is very large, typically from 3 to 11 within a single drop near the equivalence point, so all three colour-change ranges are crossed within a fraction of a millilitre. The equivalence pH there is exactly 7, since the ions left over are those of a neutral salt. It is the only case where the choice of indicator does not matter, which is precisely what makes it misleading as an introductory example.

The solubility equilibrium requires $\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=K_{sp}$. Since $\left[\mathrm{Cl}^{-}\right]=1.0$, we get directly $\left[\mathrm{Ag}^{+}\right]=1.8\times 10^{-10}$ mol/L.

The Nernst equation for the $\mathrm{Ag}^{+}/\mathrm{Ag}$ couple, with $n=1$, is $E=E^{\circ}+0.0592\log\left[\mathrm{Ag}^{+}\right]=0.80+0.0592\log\left(1.8\times 10^{-10}\right)$. Now $\log\left(1.8\times 10^{-10}\right)\approx -9.745$, so $E\approx 0.80-0.577\approx 0.223$ V.

The couple $\mathrm{AgCl}+\mathrm{e}^{-}\rightleftharpoons\mathrm{Ag}+\mathrm{Cl}^{-}$ is under standard conditions as soon as $\left[\mathrm{Cl}^{-}\right]=1.0$ mol/L, silver chloride and silver being pure solids. The potential calculated in b) is therefore indeed $E^{\circ}\left(\mathrm{AgCl}/\mathrm{Ag}\right)$. Redoing the calculation in letters: $E^{\circ}\left(\mathrm{AgCl}/\mathrm{Ag}\right)=E^{\circ}\left(\mathrm{Ag}^{+}/\mathrm{Ag}\right)+0.0592\log\frac{K_{sp}}{\left[\mathrm{Cl}^{-}\right]}$, which reduces to the requested relation for $\left[\mathrm{Cl}^{-}\right]=1.0$.

The agreement is excellent: $0.223$ V calculated against $0.222$ V tabulated, a one-millivolt difference, entirely due to rounding in $K_{sp}$ and in the factor $0.0592$. This is not a numerical coincidence but a consistency check on the course: the thermodynamics of the redox couple and that of the solubility equilibrium describe the same system and must necessarily agree. Precipitating the silver drops its potential by nearly 0.58 V, simply because the precipitate keeps its concentration more than five billion times lower.

Because its potential is set by a solid in equilibrium, and not by a concentration that would have to be prepared and maintained. As long as solid silver chloride remains and the chloride concentration is known, in practice a saturated potassium chloride solution, the potential restores itself: a little silver that dissolves is immediately compensated by the equilibrium. An ordinary $\mathrm{Ag}^{+}/\mathrm{Ag}$ electrode, on the other hand, drifts as soon as the solution evaporates, dilutes or becomes contaminated. This is the same principle as the buffer in acid-base chemistry: a system with a solid reserve resists disturbances, a simple solution does not.

At 800 K: $\Delta G^{\circ}=180{,}000-800\times 160=180{,}000-128{,}000=+52{,}000$ J/mol, that is $+52$ kJ. The reaction is not spontaneous. At 1400 K: $\Delta G^{\circ}=180{,}000-1400\times 160=180{,}000-224{,}000=-44{,}000$ J, that is $-44$ kJ: it becomes spontaneous.

The crossover occurs when $\Delta G^{\circ}=0$, that is $T=\frac{\Delta H^{\circ}}{\Delta S^{\circ}}=\frac{180{,}000}{160}=1125$ K, which is about $851.9$ °C. Above this temperature the process works, below it does not.

The reaction is endothermic, so the $\Delta H^{\circ}$ term alone opposes spontaneity and does not depend on temperature. The $-T\Delta S^{\circ}$ term, on the other hand, is negative since $\Delta S^{\circ}>0$, and its absolute value grows in proportion to $T$: it therefore necessarily ends up winning. It is the disorder created, here typically by a gas being released from a solid, that pays the energy bill. If $\Delta S^{\circ}$ were negative, both terms would be unfavourable at every temperature and the reaction would be spontaneous at no temperature: heating would only speed up a reaction that does not happen.

One kilogram corresponds to $\frac{1000}{26.98}\approx 37.06$ mol of aluminum. Each mole requires 3 mol of electrons, so $3\times 96{,}485=289{,}455$ C. The total charge is about $37.06\times 289{,}455\approx 1.073\times 10^{7}$ C, or about 2980 ampere-hours.

The energy is $W=QU=1.073\times 10^{7}\times 4.50\approx 4.83\times 10^{7}$ J. In kilowatt-hours, dividing by $3.6\times 10^{6}$, we obtain about $13.4$ kWh per kilogram. That is enormous: producing one tonne of aluminum takes of the order of 13,400 kilowatt-hours, the annual electricity consumption of several households. And this figure is a theoretical minimum, which ignores the faradaic efficiency and thermal losses. This explains two concrete facts: aluminum smelters are built where electricity is abundant and cheap, which is precisely the case in Quebec, and recycling aluminum, which only requires melting it, saves about 95% of this energy.

At the anode, the oxidation: $\mathrm{H}_{2}\rightarrow 2\,\mathrm{H}^{+}+2\,\mathrm{e}^{-}$. At the cathode, the reduction: $\frac{1}{2}\mathrm{O}_{2}+2\,\mathrm{H}^{+}+2\,\mathrm{e}^{-}\rightarrow\mathrm{H}_{2}\mathrm{O}$. So $n=2$ mol of electrons are exchanged per mole of water formed.

From $\Delta G^{\circ}=-nFE^{\circ}$ we get $E^{\circ}=\frac{-\Delta G^{\circ}}{nF}=\frac{237{,}130}{2\times 96{,}485}\approx 1.229$ V. This is the well-known voltage of a hydrogen-oxygen cell, and it is also the reason why the electrolysis of water requires at least this same voltage.

The efficiency is $\frac{-237.13}{-285.83}\approx 0.830$, about 83%. It does not reach 100% because $\Delta G^{\circ}$ and $\Delta H^{\circ}$ differ by $T\Delta S^{\circ}$, and the entropy of this reaction is clearly negative: one and a half moles of gas, one of hydrogen and a half of oxygen, give a single mole of liquid. The system must therefore release heat to the surroundings to compensate for this loss of disorder, and that heat cannot be recovered as electrical work. The missing 17% is not a technical imperfection, it is the $T\Delta S^{\circ}$ imposed by the second law.

In kelvins, $T_{c}=1273$ and $T_{f}=373$, so the Carnot efficiency is $1-\frac{373}{1273}\approx 0.707$, about 70.7%. The cell therefore wins by more than twelve points, and the real gap is much larger still, since a car engine does not reach half of its Carnot efficiency.

It is a reason of principle. The heat engine is limited by Carnot because it first converts all the energy into heat, which destroys the information about the organized form of chemical energy, and then has to climb back up the slope. The cell never goes through heat: it converts chemical energy directly into electrical work, and its only limit is $\Delta G^{\circ}$, precisely the share that can be converted into work. If $\Delta S^{\circ}$ were positive, $\Delta G^{\circ}$ would be more negative than $\Delta H^{\circ}$ and the efficiency would exceed 100%, which is not an absurdity: the cell would absorb heat from its surroundings and convert it into work as well. This is the case for some real cells, and it does not violate the second law since the total entropy increases anyway.

A white precipitate with silver ions that darkens in light is the signature of silver chloride: the anion is therefore the chloride ion. The equation is $\mathrm{Ag}^{+}+\mathrm{Cl}^{-}\rightarrow\mathrm{AgCl}(s)$, and the darkening comes from the photoreduction of silver to metallic silver, the reaction on which film photography was founded.

A gas with a pungent odour that turns damp pH paper blue is a gaseous base: it is ammonia. The cation is therefore the ammonium ion, displaced by a strong base according to $\mathrm{NH}_{4}^{+}+\mathrm{OH}^{-}\rightarrow\mathrm{NH}_{3}(g)+\mathrm{H}_{2}\mathrm{O}$.

It is barely informative. Any soluble salt dissociates into ions and conducts electricity: the measurement only confirms that this is a dissolved ionic compound, which the statement already said. It would have been useful in the opposite case, since weak conduction would point to a weak electrolyte, but here it discriminates nothing.

The chloride ion is the conjugate base of a strong acid, so it is a spectator and does not affect the pH. The ammonium ion, on the other hand, is the conjugate acid of a weak base: it hydrolyzes according to $\mathrm{NH}_{4}^{+}+\mathrm{H}_{2}\mathrm{O}\rightleftharpoons\mathrm{NH}_{3}+\mathrm{H}_{3}\mathrm{O}^{+}$, which acidifies the solution. Check: $\left[\mathrm{H}_{3}\mathrm{O}^{+}\right]=\sqrt{K_{a}C}=\sqrt{5.6\times 10^{-10}\times 0.10}\approx 7.48\times 10^{-6}$, so $\mathrm{pH}\approx 5.13$. The measured value of 5.2 is perfectly consistent, the difference being of the order of the precision of a pH meter.

The salt is ammonium chloride, of formula $\mathrm{NH}_{4}\mathrm{Cl}$. With sodium chloride, both ions would be spectators, the sodium ion because it is the cation of a strong base and the chloride ion as we saw: the pH would therefore be 7.0, that of pure water. This is the clearest demonstration of the four cases of salt hydrolysis: it is not the salt that decides the pH, it is, separately, the strength of the acid and that of the base it comes from.