Equilibrium

An equilibrium mixture of ${\text{N}}_2$, ${\text{H}}_2$ and ${\text{NH}}_3$ is sitting in a vessel. You inject argon, which takes no part in the reaction. Does the equilibrium shift?

It depends entirely on what is held constant.

At constant volume, the partial pressures of the three reacting gases are unchanged — argon simply occupies the same space alongside them. $Q$ is untouched and nothing happens.

At constant pressure, the vessel must expand to accommodate the argon. Every reacting gas is now spread through a larger volume, so every partial pressure falls. The system responds by shifting towards the side with more gaseous moles, which here means dissociating ammonia.

Almost every hard question in this chapter is a distinction of that kind. Which quantity is fixed, whether an approximation is still valid, which of several simultaneous equilibria dominates, and what actually appears in the expression.

1. The several forms of $K$, and what changes it

$K_p=K_c{\left(RT\right)}^{\Delta n_g},K_p=K_x{P}^{\Delta n_g}$

where $\Delta n_g$ counts gaseous moles only, and $K_x$ uses mole fractions. When $\Delta n_g=0$ all three are numerically equal.

Only temperature changes $K$. Concentration, pressure and catalysts change the position of equilibrium or the speed of arrival, never the constant itself. Pure solids and pure liquids are omitted from the expression, since their activities are one.

$\ln \frac{K_2}{K_1}=-\frac{\Delta H^{\circ }}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Illustration 1

For ${\text{PCl}}_5\rightleftharpoons {\text{PCl}}_3+{\text{Cl}}_2$ at $500$ K, $K_c=0.04$. Find $K_p$ and $K_x$ at a total pressure of $2$ atm.

$\Delta n_g=2-1=+1$

$K_p=0.04\times {\left(0.0821\times 500\right)}^1=0.04\times 41.05=1.64$ atm

$K_x=K_p{P}^{-\Delta n_g}=\frac{1.64}{2}=0.82$

The three differ whenever $\Delta n_g\ne 0$, and a question that supplies one while asking for another is testing precisely that conversion.

2. Degree of dissociation from vapour density

When a gas dissociates into more moles, the average molar mass falls and so does the measured vapour density. For $\text{A}\to n\text{B}$,

$\alpha =\frac{D-d}{\left(n-1\right)d}$

where $D$ is the theoretical vapour density of the undissociated gas and $d$ the observed value. The total number of moles rises by the factor $1+\left(n-1\right)\alpha$, and since $Md$ is constant, the density falls by the same factor.

Illustration 2

Phosphorus pentachloride has a theoretical vapour density of $104.25$. At a certain temperature the observed value is $62$. Find the degree of dissociation.

${\text{PCl}}_5\to {\text{PCl}}_3+{\text{Cl}}_2$ gives $n=2$.

$\alpha =\frac{104.25-62}{\left(2-1\right)\times 62}=\frac{42.25}{62}=0.68$

Sixty-eight per cent dissociated. If the observed density had been half the theoretical value, $\alpha$ would have been exactly $1$ — a useful limiting check on any answer of this type.

3. Le Chatelier, applied carefully

The principle is qualitative, but its applications have precise conditions.

The temperature row is the only one that changes $K$ itself. All the others change $Q$ and let the system return to the same $K$.

Illustration 3

For $2{\text{SO}}_2+{\text{O}}_2\rightleftharpoons 2{\text{SO}}_3$, $\Delta H=-198$ kJ mol${}^{-1}$. Predict the effect of raising the temperature, compressing the mixture, and adding helium at constant volume.

Raising temperature: shifts backwards towards the endothermic direction, and $K$ itself falls.

Compressing: three moles of gas become two, so it shifts forward.

Helium at constant volume: partial pressures unchanged, so no effect.

This is why the contact process runs at only about $450^{\circ }$C. Higher temperatures would speed the reaction but destroy the yield, so the temperature chosen is a compromise, not an optimum for either.

4. Weak acids and bases, and when the approximation breaks

For a weak acid of concentration $C$,

$K_a=\frac{C{\alpha }^2}{1-\alpha }\Rightarrow \alpha \approx \sqrt{\frac{K_a}{C}}\text{(Ostwald)}$

$\left[{\text{H}}^{+}\right]=\sqrt{K_{a}C},\text{pH}=\frac{1}{2}\left(p{K}_a-\log C\right)$

The approximation drops the $1-\alpha$ term, and it is valid while $\alpha <5\%$, that is while $C/K_a>400$. Below that the full quadratic is needed.

Dilution raises $\alpha$ but lowers $\left[{\text{H}}^{+}\right]$: diluting a hundredfold multiplies $\alpha$ by ten while dividing the hydrogen ion concentration by ten. For a polyprotic acid, the second ionisation constant is typically $10^5$ times smaller than the first, so only the first contributes measurably to the pH.

Illustration 4

Find the pH and degree of dissociation of $0.01$ M acetic acid, with $K_a=1.8\times 10^{-5}$, and check the approximation.

$C/K_a=\frac{0.01}{1.8\times 10^{-5}}=556>400$, so the approximation holds.

$\left[{\text{H}}^{+}\right]=\sqrt{1.8\times 10^{-5}\times 0.01}=4.24\times 10^{-4}$ M, giving pH $=3.37$

$\alpha =\frac{4.24\times 10^{-4}}{0.01}=4.24\%$

Just inside the limit. At $0.001$ M the ratio falls to $56$ and the quadratic becomes necessary, which is exactly the kind of check Advanced expects you to make unprompted.

5. Buffers

A buffer is a weak acid with its conjugate base in comparable amounts. Its pH follows the Henderson-Hasselbalch relation:

$\text{pH}=p{K}_a+\log \frac{\left[\text{salt}\right]}{\left[\text{acid}\right]}$

Two consequences follow immediately. At half neutralisation the salt and acid concentrations are equal, so $\text{pH}=p{K}_a$ — which is how $p{K}_a$ is measured. And the useful buffer range is $p{K}_a\pm 1$, since beyond a ten-to-one ratio the buffer is nearly exhausted.

Because the ratio appears in the expression, dilution does not change a buffer's pH — both concentrations fall by the same factor.

Illustration 5

A buffer contains $0.2$ M acetic acid and $0.3$ M sodium acetate. Find its pH, and the pH after adding $0.05$ mol of sodium hydroxide to one litre. Take $p{K}_a=4.74$.

Initially: $\text{pH}=4.74+\log \frac{0.3}{0.2}=4.74+0.176=4.92$

The base converts $0.05$ mol of acid to salt: acid becomes $0.15$ M, salt becomes $0.35$ M.

$\text{pH}=4.74+\log \frac{0.35}{0.15}=4.74+0.368=5.11$

A rise of only $0.19$, where the same base added to pure water would take the pH from $7$ to nearly $12.7$. That resistance is the whole purpose of a buffer.

6. Salt hydrolysis: four cases

A salt's solution is neutral only if both parent acid and base were strong.

with $K_h=\frac{K_w}{K_a}$ for the second case. The fourth is the interesting one: it contains no concentration term at all, so diluting ammonium acetate does not change its pH.

Illustration 6

Find the pH of $0.1$ M sodium acetate, with $p{K}_a=4.74$.

$\text{pH}=7+\frac{1}{2}\left(4.74\right)+\frac{1}{2}\log \left(0.1\right)$

$=7+2.37-0.5=8.87$

Basic, as expected, because acetate is the conjugate base of a weak acid and takes a proton back from water. The stronger the parent acid, the weaker its conjugate base and the closer the salt is to neutral.

7. Solubility product: three ways to change a solubility

For ${\text{A}}_x{\text{B}}_y\rightleftharpoons x{\text{A}}^{y+}+y{\text{B}}^{x-}$,

$K_{sp}={\left(xs\right)}^x{\left(ys\right)}^y$

so $s=\sqrt{K_{sp}}$ for a $1:1$ salt but $s={\left(K_{sp}/4\right)}^{1/3}$ for a $1:2$ salt — which is why comparing solubilities across different stoichiometries by $K_{sp}$ alone is meaningless.

Three ways to change solubility, and Advanced uses all three:

Common ion lowers it, by raising the concentration of one product. Lowering the pH raises the solubility of any salt of a weak acid, because hydrogen ions remove the anion. Complex formation raises it, which is why silver chloride dissolves in ammonia, and why excess chloride eventually redissolves it as ${\text{AgCl}}_2^{-}$.

Illustration 7

The $K_{sp}$ of silver chloride is $1.8\times 10^{-10}$. Find its solubility in pure water and in $0.01$ M sodium chloride.

Pure water: $s=\sqrt{1.8\times 10^{-10}}=1.34\times 10^{-5}$ M

In $0.01$ M chloride: $s\times 0.01=1.8\times 10^{-10}$, so $s=1.8\times 10^{-8}$ M

Suppressed by a factor of about $750$. The added chloride so far exceeds what the salt itself supplies that the salt's own contribution can be neglected entirely.

Illustration 8

Compare the solubilities of $\text{AgCl}$ ($K_{sp}=1.8\times 10^{-10}$) and ${\text{Ag}}_2{\text{CrO}}_4$ ($K_{sp}=1.1\times 10^{-12}$).

$\text{AgCl}$: $s=1.34\times 10^{-5}$ M

${\text{Ag}}_2{\text{CrO}}_4$: $K_{sp}=4s^3$, so $s={\left(\frac{1.1\times 10^{-12}}{4}\right)}^{1/3}=6.5\times 10^{-5}$ M

The chromate has the smaller $K_{sp}$ yet is nearly five times more soluble, because its expression contains a cube. Comparing solubility products across different stoichiometries is the classic trap in this topic.

8. Selective precipitation and simultaneous equilibria

When one reagent can precipitate two different ions, the salt with the smaller solubility product forms first, and the separation is practical if the first is essentially complete before the second begins.

The concentration of precipitant at which each salt starts to appear is found by setting $Q=K_{sp}$ for each in turn. Separation is considered clean if the first ion has fallen below about $10^{-6}$ M by the time the second starts.

Hydrogen sulphide is the classic controlled reagent, because its sulphide concentration is governed by pH:

${\text{H}}_2\text{S}\rightleftharpoons 2{\text{H}}^{+}+{\text{S}}^{2-},\left[{\text{S}}^{2-}\right]=\frac{K_{a1}{K}_{a2}\left[{\text{H}}_2\text{S}\right]}{{\left[{\text{H}}^{+}\right]}^2}$

The inverse square dependence is what makes the method work. In acidic solution the sulphide concentration is driven down to around $10^{-21}$ M, enough only for the most insoluble sulphides; in ammoniacal solution it rises by many orders of magnitude and the rest follow.

Illustration 9

A solution is $0.01$ M in both ${\text{Cu}}^{2+}$ and ${\text{Zn}}^{2+}$. Given $K_{sp}$ values of $8\times 10^{-37}$ for $\text{CuS}$ and $3\times 10^{-22}$ for $\text{ZnS}$, find the range of sulphide concentration that separates them.

$\text{CuS}$ begins to precipitate when $\left[{\text{S}}^{2-}\right]=\frac{8\times 10^{-37}}{0.01}=8\times 10^{-35}$ M.

$\text{ZnS}$ begins when $\left[{\text{S}}^{2-}\right]=\frac{3\times 10^{-22}}{0.01}=3\times 10^{-20}$ M.

Any sulphide concentration between these two values precipitates copper completely and leaves zinc entirely in solution.

A window fifteen orders of magnitude wide. Controlling it by pH is trivially easy, which is why the second and fourth analytical groups are separated exactly this way.

Illustration 10

Silver nitrate is added slowly to a solution $0.01$ M in both chloride and chromate. Which precipitates first, and how much of it has gone before the second appears? Take $K_{sp}$ as $1.8\times 10^{-10}$ for $\text{AgCl}$ and $1.1\times 10^{-12}$ for ${\text{Ag}}_2{\text{CrO}}_4$.

$\text{AgCl}$ needs $\left[{\text{Ag}}^{+}\right]=\frac{1.8\times 10^{-10}}{0.01}=1.8\times 10^{-8}$ M.

${\text{Ag}}_2{\text{CrO}}_4$ needs $\left[{\text{Ag}}^{+}\right]=\sqrt{\frac{1.1\times 10^{-12}}{0.01}}=1.05\times 10^{-5}$ M.

Chloride goes first. When the red chromate finally appears, $\left[{\text{Cl}}^{-}\right]=\frac{1.8\times 10^{-10}}{1.05\times 10^{-5}}=1.7\times 10^{-5}$ M — over $99.8\%$ precipitated.

This is Mohr's method, where the appearance of red silver chromate signals that the chloride titration is complete.

9. Titration curves and choosing an indicator

An indicator is itself a weak acid, changing colour over roughly

$\text{pH}=p{K}_{In}\pm 1$

so a usable indicator must have its range inside the steep portion of the titration curve.

Hence a strong acid against a strong base has a vertical stretch spanning pH $3$ to $11$ and either indicator works. A weak acid against a strong base has its equivalence point above $7$ and a much shorter steep section, so only phenolphthalein is usable. A weak base against a strong acid is the mirror image and needs methyl orange.

Illustration 11

Explain why methyl orange cannot be used for the titration of acetic acid against sodium hydroxide.

The equivalence point lies at about pH $8.9$, since the product is sodium acetate, which hydrolyses.

Methyl orange changes between pH $3.1$ and $4.4$, which is reached long before the equivalence point.

The colour would change while a substantial fraction of the acid remains untitrated, giving a large systematic error.

Match the indicator's range to the equivalence pH, not to the acid. The strength of the acid matters only through where it puts that equivalence point.

Illustration 12

A weak base of $p{K}_b=4.75$ is titrated with hydrochloric acid. Find the equivalence pH for a final salt concentration of $0.05$ M, and choose an indicator.

$\text{pH}=7-\frac{1}{2}\left(4.75\right)-\frac{1}{2}\log \left(0.05\right)$

$=7-2.375+0.65=5.28$

Methyl orange, changing between $3.1$ and $4.4$, is too early; methyl red, changing between $4.4$ and $6.2$, brackets this value well.

Choosing an indicator is a calculation, not a recollection. Compute the equivalence pH first and then look for a range containing it.

Summary

• Inert gas at constant $V$ does nothing; at constant $P$ it shifts towards more gaseous moles.
• $K_p=K_c{\left(RT\right)}^{\Delta n_g}=K_x{P}^{\Delta n_g}$; all three coincide when $\Delta n_g=0$.
• Only temperature changes $K$. Pure solids and liquids never appear in the expression.
• $\alpha =\frac{D-d}{\left(n-1\right)d}$ from vapour density; halving the density means complete dissociation for $n=2$.
• Le Chatelier: heating shifts endothermically; a catalyst shifts nothing.
• Ostwald: $\alpha =\sqrt{K_a/C}$ and $\text{pH}=\frac{1}{2}\left(p{K}_a-\log C\right)$, valid while $C/K_a>400$.
• Diluting a weak acid raises $\alpha$ but lowers $\left[{\text{H}}^{+}\right]$; only the first ionisation of a polyprotic acid matters.
• Henderson: $\text{pH}=p{K}_a+\log \frac{\left[\text{salt}\right]}{\left[\text{acid}\right]}$; at half neutralisation $\text{pH}=p{K}_a$.
• Buffer range is $p{K}_a\pm 1$, and dilution does not change a buffer's pH.
• Hydrolysis: $\text{pH}=7+\frac{1}{2}p{K}_a+\frac{1}{2}\log C$ for a weak acid salt; the weak-weak case has no concentration term.
• $K_{sp}$ gives $s=\sqrt{K_{sp}}$ for $1:1$ but ${\left(K_{sp}/4\right)}^{1/3}$ for $1:2$ — never compare $K_{sp}$ across stoichiometries.
• Solubility falls with a common ion, rises at low pH for salts of weak acids, and rises again through complex formation.
• Selective precipitation: the smaller $K_{sp}$ goes first; separation is clean if the first ion falls below $10^{-6}$ M before the second starts.
• $\left[{\text{S}}^{2-}\right]\propto 1/{\left[{\text{H}}^{+}\right]}^2$, which is why pH control separates the analytical sulphide groups.
• An indicator works only if $p{K}_{In}\pm 1$ lies inside the steep section; weak acid against strong base needs phenolphthalein, weak base against strong acid needs methyl orange or methyl red.