"Equilibrium" — Practice

Question 1

For $	ext{N}_2	ext{O}_4ightleftharpoons2	ext{NO}_2$ at $400$ K, $K_p=48$ atm. Find $K_c$ and state what happens to the degree of dissociation when the total pressure is doubled at constant temperature.

Accepted Answer

$\Delta n_g=+1$, so $K_p=K_c\left(RTight)$.

$K_c=\dfrac{48}{0.0821	imes400}=\dfrac{48}{32.84}=1.46$ mol L$^{-1}$

Doubling the pressure shifts the equilibrium towards **fewer** gaseous moles, that is towards $	ext{N}_2	ext{O}_4$, so the degree of dissociation **falls**.

$K_p$ itself is unchanged, since only temperature affects it.

Question 2

The observed vapour density of a sample of dinitrogen tetroxide at a certain temperature is $30.67$. Find the degree of dissociation, given a molar mass of $92$.

Accepted Answer

$D=\dfrac{92}{2}=46$; $n=2$ for $	ext{N}_2	ext{O}_4ightarrow2	ext{NO}_2$.

$\alpha=\dfrac{46-30.67}{\left(2-1ight)	imes30.67}=\dfrac{15.33}{30.67}=0.50$

Fifty per cent dissociated.

As a check, at complete dissociation the observed density would be $23$, exactly half the theoretical value.

Question 3

For the Haber process, $\Delta H=-92$ kJ mol$^{-1}$. State and justify the effect on the yield of ammonia of raising the temperature, raising the pressure, adding a catalyst, and adding argon at constant volume.

Accepted Answer

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

**Raising pressure:** four moles of gas become two, so the shift is forward. Yield increases.

**Catalyst:** no shift whatever, only faster arrival at the same equilibrium.

**Argon at constant volume:** partial pressures unchanged, so no effect at all.

The industrial compromise of about $450^{\circ}$C and $200$ atm reflects exactly these four facts.

Question 4

Find the pH of $0.001$ M acetic acid ($K_a=1.8	imes10^{-5}$), checking whether the usual approximation applies.

Accepted Answer

$C/K_a=\dfrac{0.001}{1.8	imes10^{-5}}=55.6$, well below $400$, so the approximation is **not** valid.

Solving $\dfrac{x^{2}}{0.001-x}=1.8	imes10^{-5}$:

$x^{2}+1.8	imes10^{-5}x-1.8	imes10^{-8}=0$, giving $x=1.25	imes10^{-4}$

$	ext{pH}=3.90$

The approximate route would have given $3.87$, and $\alpha$ here is $12.5\%$ — well outside the five per cent limit.

Question 5

How many grams of sodium acetate must be added to $500$ mL of $0.2$ M acetic acid to give a buffer of pH $5.0$? Take $pK_a=4.74$ and the molar mass of sodium acetate as $82$.

Accepted Answer

$5.0=4.74+\log\dfrac{\left[	ext{salt}ight]}{0.2}$

$\log\dfrac{\left[	ext{salt}ight]}{0.2}=0.26\qquad\Rightarrow\qquad \dfrac{\left[	ext{salt}ight]}{0.2}=1.82$

$\left[	ext{salt}ight]=0.364$ M, so in $500$ mL that is $0.182$ mol.

Mass $=0.182	imes82=14.9$ g

Question 6

A buffer is $0.1$ M in both ammonia and ammonium chloride, with $pK_b=4.75$. Find its pH, and the pH after adding $0.02$ mol of hydrochloric acid to one litre.

Accepted Answer

$	ext{pOH}=pK_b+\log\dfrac{\left[	ext{salt}ight]}{\left[	ext{base}ight]}=4.75+\log1=4.75$, so $	ext{pH}=9.25$.

The acid converts $0.02$ mol of ammonia to ammonium: base becomes $0.08$ M, salt becomes $0.12$ M.

$	ext{pOH}=4.75+\log\dfrac{0.12}{0.08}=4.75+0.176=4.93$

$	ext{pH}=9.07$, a fall of only $0.18$ despite adding a strong acid.

Question 7

Find the pH of $0.1$ M ammonium chloride, and of $0.1$ M ammonium acetate. Take $pK_a=4.74$ for acetic acid and $pK_b=4.75$ for ammonia.

Accepted Answer

Ammonium chloride is a strong acid with a weak base:

$	ext{pH}=7-	frac12\left(4.75ight)-	frac12\log\left(0.1ight)=7-2.375+0.5=5.13$

Ammonium acetate is a weak acid with a weak base:

$	ext{pH}=7+	frac12\left(4.74ight)-	frac12\left(4.75ight)=7-0.005=6.995$

Essentially neutral, and **independent of concentration** — diluting it changes nothing.

Question 8

$0.1$ M formic acid ($pK_a=3.75$) is titrated with $0.1$ M sodium hydroxide. Find the pH at the equivalence point and choose a suitable indicator.

Accepted Answer

Equal volumes are mixed, so the salt concentration is $0.05$ M.

$	ext{pH}=7+	frac12\left(3.75ight)+	frac12\log\left(0.05ight)=7+1.875-0.65=8.23$

Phenolphthalein, changing between $8.3$ and $10.0$, brackets this closely and is the correct choice.

Methyl orange would change near pH $4$, long before the equivalence point, and would give a large systematic error.

Question 9

Explain why the titration of a weak acid with a weak base has no satisfactory indicator.

Accepted Answer

Both the acid and the base resist pH change, so the solution is buffered on both sides of the equivalence point.

The result is a curve with no steep section: the pH creeps through the equivalence region rather than jumping.

Since every indicator changes over roughly two pH units, none can change sharply enough to mark a point on a gently sloping curve.

Such titrations are followed conductometrically or potentiometrically instead, where the equivalence point appears as a change of slope rather than a colour change.

Question 10

The solubility of calcium fluoride is $2.0	imes10^{-4}$ M. Find its solubility product, and its solubility in $0.01$ M sodium fluoride.

Accepted Answer

$	ext{CaF}_2ightleftharpoons	ext{Ca}^{2+}+2	ext{F}^-$, so $K_{sp}=s\left(2sight)^{2}=4s^{3}$.

$K_{sp}=4\left(2.0	imes10^{-4}ight)^{3}=3.2	imes10^{-11}$

In $0.01$ M fluoride the added ion dominates: $s\left(0.01ight)^{2}=3.2	imes10^{-11}$

$s=3.2	imes10^{-7}$ M, suppressed by a factor of over $600$.

Question 11

A solution is $0.01$ M in both $	ext{Ag}^+$ and $	ext{Pb}^{2+}$. Iodide is added slowly. Which precipitates first, and what fraction of it remains when the second begins? Take $K_{sp}$ as $8.5	imes10^{-17}$ for $	ext{AgI}$ and $7.1	imes10^{-9}$ for $	ext{PbI}_2$.

Accepted Answer

$	ext{AgI}$ needs $\left[	ext{I}^-ight]=\dfrac{8.5	imes10^{-17}}{0.01}=8.5	imes10^{-15}$ M.

$	ext{PbI}_2$ needs $\left[	ext{I}^-ight]=\sqrt{\dfrac{7.1	imes10^{-9}}{0.01}}=8.4	imes10^{-4}$ M.

Silver iodide precipitates first, by eleven orders of magnitude.

When lead begins, $\left[	ext{Ag}^+ight]=\dfrac{8.5	imes10^{-17}}{8.4	imes10^{-4}}=1.0	imes10^{-13}$ M — a fraction of $10^{-11}$ remains, so the separation is essentially perfect.

Question 12

Explain why silver chloride is insoluble in water, dissolves in aqueous ammonia, and is reprecipitated by nitric acid.

Accepted Answer

In water the ionic product exceeds the small solubility product almost immediately, so the salt precipitates.

Ammonia forms the stable complex $\left[	ext{Ag}\left(	ext{NH}_3ight)_2ight]^+$, removing free silver ions from solution. The ionic product falls below $K_{sp}$ and the solid dissolves.

Adding nitric acid protonates the ammonia to ammonium, which is not a ligand. The complex collapses, free silver ions reappear, and the chloride reprecipitates.

Each step works by shifting one equilibrium and letting the coupled one follow.

Practice — "Equilibrium"

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