"Equilibrium" — Practice

Question 1

For $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ at 500 K, $K_c = 0.061$. Calculate $K_p$, taking $R = 0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$.

Accepted Answer

Gas moles fall from 4 to 2, so $\Delta n_g = -2$ and $K_p = K_c(RT)^{-2}$. Here $RT = 0.0821 	imes 500 = 41.05$, whose square is 1685, so $K_p = 0.061/1685 = 3.6 	imes 10^{-5}$. The drop of more than three orders of magnitude reflects how strongly the conversion depends on $\Delta n_g$, which is why the sign of that exponent must be checked before the arithmetic.

Question 2

For $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$, $K_c = 49$ at 717 K. Starting with 1 mole of each reactant in a 1 L vessel, find the equilibrium concentrations.

Accepted Answer

Let $x$ be the amount of each reactant consumed, so at equilibrium $[\mathrm{H_2}] = [\mathrm{I_2}] = 1-x$ and $[\mathrm{HI}] = 2x$, giving $49 = (2x)^2/(1-x)^2$. Both sides are perfect squares, so taking roots gives $7 = 2x/(1-x)$, hence $9x = 7$ and $x = 0.778$. Then $[\mathrm{H_2}] = [\mathrm{I_2}] = 0.222$ M and $[\mathrm{HI}] = 1.556$ M, which checks back as $1.556^2/0.222^2 = 49.1$.

Question 3

For an exothermic reaction that reduces the number of gas moles, state and justify the effect on yield of raising the temperature, raising the pressure, and adding a catalyst.

Accepted Answer

Raising the temperature lowers the yield, since the reverse reaction is endothermic and is favoured, and this is the only disturbance of the three that changes $K$ itself. Raising the pressure raises the yield, since the system relieves compression by shifting towards the side occupying less volume. Adding a catalyst leaves the yield exactly unchanged, because it multiplies both rates equally and only shortens the time taken to reach the same equilibrium.

Question 4

A reaction has $K_c = 4.0$ and a mixture is found to have $Q = 12$. State which way it shifts and why.

Accepted Answer

$Q$ exceeds $K$, meaning more products are present than equilibrium allows. The reaction therefore proceeds in the reverse direction, converting products back into reactants until $Q$ falls to 4.0. The comparison needs no chemical intuition, which is exactly why it is more reliable than reasoning about which species looks too abundant, and why a close call such as $Q = 5.0	imes10^{-3}$ against $K = 4.6	imes10^{-3}$ must be computed rather than estimated.

Question 5

Calculate the pH and degree of ionisation of 0.10 M acetic acid, given $K_a = 1.8 	imes 10^{-5}$.

Accepted Answer

The hydrogen ion concentration is $\sqrt{K_a c} = \sqrt{1.8 	imes 10^{-6}} = 1.34 	imes 10^{-3}$ M, so the pH is 2.87. The degree of ionisation is $\alpha = 1.34	imes10^{-3}/0.10 = 0.0134$, that is 1.34 per cent. At that level the approximation of ignoring $\alpha$ against 1 in the denominator of Ostwald's law is comfortably valid, since it is safe below about 5 per cent.

Question 6

Explain why the pH of pure water at 373 K is about 6.14 yet the water is still neutral.

Accepted Answer

Neutrality means $[\mathrm{H^+}] = [\mathrm{OH^-}]$, not that the pH is 7. Self-ionisation of water is endothermic, so raising the temperature shifts it forward and increases $K_w$ from $1.0	imes10^{-14}$ at 298 K to about $5.1	imes10^{-13}$ at 373 K. Both ion concentrations rise equally to about $7.2	imes10^{-7}$ M, so the pH falls to 6.14 while the water remains exactly neutral. The relation $\mathrm{pH} + \mathrm{pOH} = 14$ is equally specific to 298 K.

Question 7

Phosphoric acid has ionisation constants $7.5	imes10^{-3}$, $6.2	imes10^{-8}$ and $4.2	imes10^{-13}$. Explain the pattern and state which constant governs the pH.

Accepted Answer

Each constant is roughly $10^5$ times smaller than the one before, because removing a proton from a neutral molecule is far easier than pulling a positive charge away from a singly negative ion, and harder still from a doubly negative one. The electrostatic work rises at every stage. Since the later stages contribute negligibly to the hydrogen ion concentration, the pH of a phosphoric acid solution is computed from $K_{a1}$ alone; the later constants matter only for the concentrations of the intermediate anions.

Question 8

Calculate the pH of a buffer containing 0.20 M acetic acid and 0.30 M sodium acetate, with $K_a = 1.8 	imes 10^{-5}$.

Accepted Answer

The $\mathrm{p}K_a$ is $-\log(1.8	imes10^{-5}) = 4.74$. By Henderson-Hasselbalch, $\mathrm{pH} = 4.74 + \log(0.30/0.20) = 4.74 + 0.18 = 4.92$. The pH sits just above the $\mathrm{p}K_a$ because the conjugate base is in excess, and diluting the whole solution would leave it unchanged, since only the ratio of the two appears in the equation.

Question 9

Explain why a buffer works best when the pH is close to the $\mathrm{p}K_a$ of its acid, and give the useful working range.

Accepted Answer

Buffer capacity depends on holding substantial amounts of both the acid and its conjugate base, so that added base is neutralised by one and added acid by the other. At $\mathrm{pH} = \mathrm{p}K_a$ the two are present in equal amounts and capacity is maximal. Moving away makes one component scarce, and beyond about one pH unit either side the minor component is outnumbered ten to one, so the buffer fails quickly on that side. The useful range is therefore $\mathrm{p}K_a \pm 1$.

Question 10

Calculate the pH of 0.10 M sodium acetate, given $K_a = 1.8	imes10^{-5}$ for acetic acid.

Accepted Answer

Acetate is the conjugate base of a weak acid and hydrolyses to give a basic solution. The hydrolysis constant is $K_h = K_w/K_a = 1.0	imes10^{-14}/1.8	imes10^{-5} = 5.6	imes10^{-10}$. Then $[\mathrm{OH^-}] = \sqrt{K_h c} = \sqrt{5.6	imes10^{-11}} = 7.5	imes10^{-6}$ M, so the pOH is 5.13 and the pH is 8.87. The compact formula $\mathrm{pH} = 7 + 	frac{1}{2}(\mathrm{p}K_a + \log c)$ gives the same result in one line, and the sodium ion is a spectator throughout.

Question 11

Silver chloride has $K_{sp} = 1.8	imes10^{-10}$. Find its solubility in pure water and in 0.010 M sodium chloride, and comment.

Accepted Answer

In pure water $K_{sp} = s^2$, so $s = \sqrt{1.8	imes10^{-10}} = 1.34	imes10^{-5}$ M. In 0.010 M sodium chloride the chloride concentration is already 0.010 M, far exceeding anything the salt itself contributes, so $s = K_{sp}/0.010 = 1.8	imes10^{-8}$ M. Solubility falls by a factor of about 750, which is why a precipitate is washed with a dilute solution of a common ion rather than with pure water, which would dissolve part of it.

Question 12

Two salts AB and AB$_2$ both have $K_{sp} = 4.0	imes10^{-12}$. Which is more soluble?

Accepted Answer

For AB, $K_{sp} = s^2$, so $s = 2.0	imes10^{-6}$ M. For AB$_2$, $K_{sp} = 4s^3$, so $s^3 = 1.0	imes10^{-12}$ and $s = 1.0	imes10^{-4}$ M. The 1:2 salt is about fifty times more soluble despite an identical solubility product, because its expression carries a higher power of $s$. Comparing $K_{sp}$ values directly is legitimate only within the same formula type.

Practice — "Equilibrium"

Sign in to save attempts to your dashboard and track your accuracy over time

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities