Ionisation Constants of Weak Acids and Weak Bases

A Weak Acid Stops Part-Way

Dissolve $0.1\ \mathrm{mol}$ of $\mathrm{HCl}$ in a litre of water and no $\mathrm{HCl}$ molecules survive. Dissolve $0.1\ \mathrm{mol}$ of acetic acid and almost all of it survives: about thirteen molecules in a thousand have given up a proton, since $\alpha = 0.0132$ at this concentration, and the hydronium concentration is about seventy-five times smaller.

The difference is one of position, not of kind. A weak acid in water reaches a genuine equilibrium — proton transfer forward, proton transfer back, both continuing, concentrations steady. Writing the acid as $\mathrm{HA}$:

$$\mathrm{HA(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{A^-(aq)}$$

Water is the base; it accepts the proton and becomes hydronium. The species $\mathrm{A^-}$ is the conjugate base of $\mathrm{HA}$, and it is the reason the reverse arrow matters: $\mathrm{A^-}$ is strong enough as a base to take the proton back from $\mathrm{H_3O^+}$, so the reaction settles part-way instead of running to completion.

The equilibrium constant for that proton transfer, written the way any equilibrium constant is written, would be

$$K = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}][\mathrm{H_2O}]}$$

Water is the solvent, present at about $55.5\ \mathrm{mol\,L^{-1}}$. Ionising a few millimoles of acid changes that number in the fourth decimal place at best, and a quantity that does not change belongs on the same side as the constant, so $[\mathrm{H_2O}]$ is absorbed into $K$.

Key Point (Definition): The ionisation constant (or acid dissociation constant) of a weak acid $\mathrm{HA}$ is $$K_a = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$$ where every concentration is the equilibrium concentration in $\mathrm{mol\,L^{-1}}$. Water does not appear.

Three properties of $K_a$ decide how it is used.

It is fixed at a fixed temperature. Diluting the solution does not change $K_a$; adding a common ion does not change $K_a$; only heating or cooling does.

It is a pure number. Each concentration is understood to be divided by the standard-state concentration of $1\ \mathrm{M}$, so the units cancel. Quoting $K_a$ for acetic acid as $1.74 \times 10^{-5}\ \mathrm{M}$ is a habit, not a requirement.

Its size measures acid strength directly. A large $K_a$ puts the equilibrium to the right and means a large hydronium concentration for a given amount of acid dissolved.

$$\text{larger } K_a \;\Longrightarrow\; \text{stronger acid}$$

The shorthand $\mathrm{HA} \rightleftharpoons \mathrm{H^+} + \mathrm{A^-}$ with $K_a = [\mathrm{H^+}][\mathrm{A^-}]/[\mathrm{HA}]$ gives identical numbers, because a bare proton in water is a hydronium ion. Write the full version whenever a question asks for the equilibrium; the short version is a calculation device.

[Board] A question asking for the expression for the ionisation constant wants the water-free ratio. Including $[\mathrm{H_2O}]$ in the denominator loses the mark.

pKa Turns a Range of Eight Powers into a Range of Eight Units

Ionisation constants run from about $10^{-2}$ down to about $10^{-10}$. Comparing numbers across eight decades is awkward, so the same logarithmic operator used for $[\mathrm{H_3O^+}]$ is applied to $K_a$.

Key Point (Definition): $\mathrm{p}K_a = -\log K_a$ and $\mathrm{p}K_b = -\log K_b$.

The minus sign flips the direction of the comparison, and that is where marks are lost.

$$\text{smaller } \mathrm{p}K_a \;\Longrightarrow\; \text{larger } K_a \;\Longrightarrow\; \text{stronger acid}$$

A drop of one unit in $\mathrm{p}K_a$ is a factor of ten in $K_a$. Nitrous acid at $3.35$ is about twenty-six times stronger than acetic acid at $4.76$, and about $10^{6}$ times stronger than hydrocyanic acid at $9.31$.

The constants worth carrying in memory, at 298 K:

Acid	Formula	$K_a$	$\mathrm{p}K_a$
Nitrous acid	$\mathrm{HNO_2}$	$4.5 \times 10^{-4}$	3.35
Hydrofluoric acid	$\mathrm{HF}$	$3.5 \times 10^{-4}$	3.46
Formic acid	$\mathrm{HCOOH}$	$1.8 \times 10^{-4}$	3.74
Benzoic acid	$\mathrm{C_6H_5COOH}$	$6.5 \times 10^{-5}$	4.19
Acetic acid	$\mathrm{CH_3COOH}$	$1.74 \times 10^{-5}$	4.76
Niacin	$\mathrm{C_5H_4NCOOH}$	$1.5 \times 10^{-5}$	4.82
Hypochlorous acid	$\mathrm{HOCl}$	$3.0 \times 10^{-8}$	7.52
Hydrocyanic acid	$\mathrm{HCN}$	$4.9 \times 10^{-10}$	9.31
Phenol	$\mathrm{C_6H_5OH}$	$1.3 \times 10^{-10}$	9.89

Three features of this table do real work in problems.

The strongest entries are all around $10^{-4}$, and $\mathrm{p}K_a$ separates them where the powers of ten do not. Formic acid beats acetic acid by a full unit, and benzoic acid also sits above acetic acid because the ring stabilises the conjugate base.

Hypochlorous acid, hydrocyanic acid and phenol are all extremely weak. A $0.1\ \mathrm{M}$ solution of $\mathrm{HCN}$ has a pH near 5, not far from neutral, and yet the acid is present at a tenth molar.

The ordering by $K_a$ and the ordering by $\mathrm{p}K_a$ are exact reverses. Any question that hands you $\mathrm{p}K_a$ values and asks for the strongest acid wants the smallest number.

Converting between the two forms needs nothing beyond a log table. Acetic acid: $K_a = 1.74 \times 10^{-5}$, so $\mathrm{p}K_a = 5 - \log(1.74) = 4.76$. Backwards: $K_a = 10^{-4.76} = 10^{0.24} \times 10^{-5} = 1.74 \times 10^{-5}$. Split the number into mantissa and power of ten before touching the logarithm and the arithmetic never goes wrong.

Weak Bases, Kb and pKb

A weak base is the same argument with the role of water reversed. The base $\mathrm{B}$ takes a proton from water and leaves hydroxide behind:

$$\mathrm{B(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{BH^+(aq)} + \mathrm{OH^-(aq)}$$

$$K_b = \frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]}$$

Water is the acid here, and it is again omitted as the pure solvent. The constant $K_b$ is the base ionisation constant, and a larger $K_b$ means a stronger base, exactly as a larger $K_a$ means a stronger acid.

Some bases are written as hydroxides that ionise rather than as proton acceptors:

$$\mathrm{MOH(aq)} \rightleftharpoons \mathrm{M^+(aq)} + \mathrm{OH^-(aq)}, \qquad K_b = \frac{[\mathrm{M^+}][\mathrm{OH^-}]}{[\mathrm{MOH}]}$$

Ammonia is quoted both ways, and the constant is the same $1.77 \times 10^{-5}$ either way.

Base	Formula	$K_b$	$\mathrm{p}K_b$
Dimethylamine	$\mathrm{(CH_3)_2NH}$	$5.4 \times 10^{-4}$	3.27
Methylamine	$\mathrm{CH_3NH_2}$	$4.4 \times 10^{-4}$	3.36
Triethylamine	$\mathrm{(C_2H_5)_3N}$	$6.45 \times 10^{-5}$	4.19
Ammonia	$\mathrm{NH_3}$	$1.77 \times 10^{-5}$	4.75
Quinine	plant alkaloid	$1.10 \times 10^{-6}$	5.96
Pyridine	$\mathrm{C_5H_5N}$	$1.77 \times 10^{-9}$	8.75
Aniline	$\mathrm{C_6H_5NH_2}$	$4.27 \times 10^{-10}$	9.37
Urea	$\mathrm{CO(NH_2)_2}$	$1.3 \times 10^{-14}$	13.89

The amines dominate the list. They are ammonia with one or more hydrogens replaced by carbon groups, and the nitrogen lone pair is what accepts the proton. Alkyl groups push electron density towards that lone pair and make the base stronger, which is why methylamine and dimethylamine outrank ammonia.

Aniline and pyridine sit four to five orders of magnitude below ammonia, but for two different reasons. In aniline the nitrogen lone pair is conjugated with the ring: it is delocalised into the $\pi$ system and is much less available to a proton, which is why aniline is roughly forty thousand times weaker a base than ammonia. Pyridine's lone pair is not delocalised at all — it lies in an $sp^2$ orbital in the plane of the ring, at right angles to the $\pi$ system — and its weakness comes instead from the high $s$ character of that orbital, which holds the pair closer to the nitrogen nucleus and makes it less willing to be shared with a proton. Urea at $1.3 \times 10^{-14}$ is barely a base at all.

A weak base calculation runs on hydroxide, so it has one extra step at the end. Find $[\mathrm{OH^-}]$ from $K_b$, take $\mathrm{pOH} = -\log[\mathrm{OH^-}]$, then $\mathrm{pH} = 14.00 - \mathrm{pOH}$ at 298 K. Skipping that conversion turns a pH of 11 into a pH of 3, the single most common error in this part of the chapter.

[NEET] Given $\mathrm{p}K_b$, the fastest route to pH is $\mathrm{pOH} = \tfrac{1}{2}(\mathrm{p}K_b - \log c)$ followed by $\mathrm{pH} = 14 - \mathrm{pOH}$.

The Degree of Ionisation

$K_a$ tells you about the acid. It says nothing on its own about how much of a particular solution has ionised, because that depends on the concentration as well. The quantity that answers the second question is the degree of ionisation.

Key Point (Definition): The degree of ionisation $\alpha$ is the fraction of the dissolved acid or base that has ionised at equilibrium: $$\alpha = \frac{\text{moles ionised}}{\text{moles taken initially}}$$ It is a pure fraction between 0 and 1. Multiplied by 100 it is the percent ionisation or percent dissociation.

For an acid $\mathrm{HA}$ taken at initial concentration $c$, the amounts follow directly.

	$\mathrm{HA}$	$+\ \mathrm{H_2O}$	$\rightleftharpoons$	$\mathrm{H_3O^+}$	$+\ \mathrm{A^-}$
Initial (M)	$c$	—		0	0
Change (M)	$-c\alpha$	—		$+c\alpha$	$+c\alpha$
Equilibrium (M)	$c - c\alpha = c(1-\alpha)$	—		$c\alpha$	$c\alpha$

Two readings come out of the bottom row and both get used constantly.

$$[\mathrm{H_3O^+}] = c\alpha \qquad\text{and}\qquad \alpha = \frac{[\mathrm{H_3O^+}]}{c}$$

The second form is how $\alpha$ is measured: take the pH of a solution of known concentration, convert to $[\mathrm{H_3O^+}]$, divide by $c$. No assumption has been made anywhere.

Percent dissociation is the same number in different clothes:

$$\text{percent dissociation} = \frac{[\mathrm{HA}]_{\text{dissociated}}}{[\mathrm{HA}]_{\text{initial}}} \times 100 = \alpha \times 100$$

A $0.08\ \mathrm{M}$ solution of hypochlorous acid produces $[\mathrm{H_3O^+}] = 1.41 \times 10^{-3}\ \mathrm{M}$ — the figure that follows from the $K_a = 2.5 \times 10^{-5}$ used in this standard problem, rather than from the tabulated $3.0 \times 10^{-8}$ — so $\alpha = 1.41 \times 10^{-3}/0.08 = 0.0176$, which is $1.76$ per cent dissociated.

For a base the table is identical with $\mathrm{OH^-}$ in place of $\mathrm{H_3O^+}$: $[\mathrm{OH^-}] = c\alpha$ and $[\mathrm{B}] = c(1-\alpha)$.

Two warnings before the next step.

$\alpha$ is not a property of the acid. Acetic acid has one $K_a$ at 298 K but a different $\alpha$ at every concentration. A question that asks for the degree of ionisation of acetic acid without giving a concentration is incomplete.

$\alpha$ has no units and cannot exceed 1. An answer of $\alpha = 3.2$ or $\alpha = 140$ per cent means an algebra error, usually a forgotten square root.

Ostwald's Dilution Law

Substituting the equilibrium row into the expression for $K_a$ connects the constant of the acid to the fraction ionised in this particular solution.

$$K_a = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]} = \frac{(c\alpha)(c\alpha)}{c(1-\alpha)} = \frac{c^2\alpha^2}{c(1-\alpha)}$$

One factor of $c$ cancels between numerator and denominator.

Key Point: Ostwald's dilution law. For a weak monobasic acid of ionisation constant $K_a$ at initial concentration $c$, $$K_a = \frac{c\alpha^2}{1-\alpha}$$ and for a weak monoacidic base, $$K_b = \frac{c\alpha^2}{1-\alpha}$$

The law is exact. No approximation has entered — only the balanced stoichiometry and the definition of $\alpha$. It is a quadratic in $\alpha$:

$$c\alpha^2 + K_a\alpha - K_a = 0, \qquad \alpha = \frac{-K_a + \sqrt{K_a^2 + 4K_ac}}{2c}$$

The other root is negative and is discarded: a fraction of the acid that has ionised cannot be less than zero.

Three consequences follow from the shape of the expression, before any numbers are put in.

$\alpha$ rises when $c$ falls. The product $c\alpha^2$ must stay fixed at $K_a(1-\alpha)$, so shrinking $c$ forces $\alpha$ up. This is Le Chatelier's principle applied to a reaction that produces more solute particles than it consumes.

$\alpha$ rises when $K_a$ rises. A stronger acid ionises further at the same concentration.

$\alpha \rightarrow 1$ as $c \rightarrow 0$. At infinite dilution every weak electrolyte is completely ionised, which is why $\alpha$ is sometimes obtained as the ratio $\Lambda_m / \Lambda_m^{\infty}$ of molar conductivities.

The same algebra for a weak base with $[\mathrm{OH^-}] = c\alpha$ gives $K_b = c\alpha^2/(1-\alpha)$ unchanged, so everything proved for acids transfers directly; only the final conversion from $\mathrm{pOH}$ to $\mathrm{pH}$ differs.

The Square-Root Shortcut

Solving a quadratic for every weak acid problem is slow, and for most of them unnecessary, because $\alpha$ is tiny and $1 - \alpha$ is barely distinguishable from 1.

For acetic acid at $0.1\ \mathrm{M}$, $\alpha$ is about $0.013$. Then $1 - \alpha = 0.987$, and treating it as 1 changes the answer by about one per cent — far less than the uncertainty in the tabulated $K_a$.

Setting $1 - \alpha \approx 1$ collapses the law:

$$K_a \approx c\alpha^2 \qquad\Longrightarrow\qquad \alpha = \sqrt{\frac{K_a}{c}}$$

Multiplying by $c$ gives the hydronium concentration:

$$[\mathrm{H_3O^+}] = c\alpha = c\sqrt{\frac{K_a}{c}} = \sqrt{K_ac}$$

Key Point: For a weak monobasic acid with small $\alpha$, $$\alpha = \sqrt{\frac{K_a}{c}}, \qquad [\mathrm{H_3O^+}] = \sqrt{K_ac}, \qquad \mathrm{pH} = \frac{1}{2}\left(\mathrm{p}K_a - \log c\right)$$ and for a weak monoacidic base, $$\alpha = \sqrt{\frac{K_b}{c}}, \qquad [\mathrm{OH^-}] = \sqrt{K_bc}, \qquad \mathrm{pOH} = \frac{1}{2}\left(\mathrm{p}K_b - \log c\right)$$

The pH form is worth deriving once so it is never misremembered. Take logarithms of $[\mathrm{H_3O^+}] = \sqrt{K_ac} = (K_ac)^{1/2}$:

$$\log[\mathrm{H_3O^+}] = \tfrac{1}{2}\log K_a + \tfrac{1}{2}\log c$$

Multiply through by $-1$:

$$\mathrm{pH} = \tfrac{1}{2}\mathrm{p}K_a - \tfrac{1}{2}\log c = \tfrac{1}{2}\left(\mathrm{p}K_a - \log c\right)$$

For the usual solutions $c<1$, so $\log c$ is negative and the formula adds to $\tfrac{1}{2}\mathrm{p}K_a$. For $0.01\ \mathrm{M}$ acetic acid, $\mathrm{pH} = \tfrac{1}{2}(4.76 + 2) = 3.38$.

Two facts about the shortcut are tested without any computing at all.

$[\mathrm{H_3O^+}]$ goes as the square root of the concentration, not as the concentration. Diluting a weak acid ten-fold divides $[\mathrm{H_3O^+}]$ by $\sqrt{10} = 3.16$, so the pH rises by $0.5$, not by 1. For a strong acid the same dilution raises the pH by a full unit.

$\alpha$ goes as the inverse square root of concentration, so a hundred-fold dilution multiplies $\alpha$ by 10.

When the Shortcut Breaks

The approximation replaced $1 - \alpha$ by 1. That is safe while $\alpha$ is small, and unsafe when it is not.

Key Point: The approximation $\alpha = \sqrt{K_a/c}$ is acceptable when $\alpha$ comes out below about $0.05$ (5 per cent), equivalently when $$\frac{c}{K_a} > 400$$ Outside that range the quadratic $c\alpha^2 + K_a\alpha - K_a = 0$ must be solved.

The two conditions are one condition. If $\alpha = \sqrt{K_a/c} = 0.05$ then $K_a/c = 0.0025$ and $c/K_a = 400$. Checking $c/K_a$ takes one division and is done before any square root, so it tells you in advance whether the shortcut is allowed.

The error grows quickly once the ratio drops.

$c/K_a$	Exact $\alpha$	Approximate $\alpha$	Error in $\alpha$	Error in pH
400	4.88 %	5.00 %	2.5 %	0.01
100	9.51 %	10.00 %	5.1 %	0.02
57.1	12.38 %	13.23 %	6.8 %	0.03
10	27.02 %	31.62 %	17.1 %	0.07
4	39.04 %	50.00 %	28.1 %	0.11

The approximate value is always too large. Dropping $\alpha$ from the denominator makes the denominator bigger than it should be, which makes the calculated $\alpha$ bigger than it should be.

pH is far more forgiving than $\alpha$, because the logarithm compresses everything. Even a 28 per cent error in $\alpha$ moves the pH by only $0.11$ units. A question asking for pH to two decimals will usually forgive the shortcut; one asking for the degree of ionisation will not.

A case where it fails. Hydrofluoric acid, $K_a = 3.5 \times 10^{-4}$, at $c = 0.02\ \mathrm{M}$. The check first: $c/K_a = 57.1$, well under 400, so the shortcut is not allowed.

Shortcut anyway, to see the damage: $\alpha = \sqrt{0.0175} = 0.1323$, so $13.23$ per cent, and $[\mathrm{H_3O^+}] = 2.65 \times 10^{-3}\ \mathrm{M}$, $\mathrm{pH} = 2.58$.

Exact treatment, from Ostwald's law without simplification:

$$\frac{0.02\alpha^2}{1-\alpha} = 3.5 \times 10^{-4} \qquad\Longrightarrow\qquad \alpha^2 + 1.75 \times 10^{-2}\alpha - 1.75 \times 10^{-2} = 0$$

$$\alpha = \frac{-1.75 \times 10^{-2} + \sqrt{3.06 \times 10^{-4} + 7.0 \times 10^{-2}}}{2} = \frac{-0.0175 + 0.2652}{2} = 0.1238$$

So $\alpha = 0.124$, that is $12.4$ per cent, and $[\mathrm{H_3O^+}] = 0.02 \times 0.1238 = 2.48 \times 10^{-3}\ \mathrm{M}$, giving $\mathrm{pH} = 2.61$.

The two degrees of ionisation differ by $6.8$ per cent — enough to change a reported value from $12$ to $13$ per cent. The two pH values differ by $0.03$, which no marking scheme would notice. That asymmetry is the practical rule: check $c/K_a$, and if it is small, solve the quadratic whenever the answer wanted is $\alpha$ itself.

[JEE Main] Very dilute solutions of very weak acids break a second assumption. When $\sqrt{K_ac}$ comes out near $10^{-7}\ \mathrm{M}$, the hydronium supplied by water is no longer negligible and even the quadratic above is insufficient.

Dilution Pulls the Two Quantities in Opposite Directions

The pair of results $\alpha = \sqrt{K_a/c}$ and $[\mathrm{H_3O^+}] = \sqrt{K_ac}$ contain a contradiction that is only apparent. One rises on dilution, the other falls.

$$\alpha \propto \frac{1}{\sqrt{c}} \qquad\qquad [\mathrm{H_3O^+}] \propto \sqrt{c}$$

Acetic acid, $K_a = 1.74 \times 10^{-5}$, makes it concrete.

$c$ (M)	$\alpha$	Percent ionised	$[\mathrm{H_3O^+}]$ (M)	pH
1.0	0.00417	0.42 %	$4.17 \times 10^{-3}$	2.38
0.1	0.0132	1.32 %	$1.32 \times 10^{-3}$	2.88
0.01	0.0417	4.17 %	$4.17 \times 10^{-4}$	3.38
0.001	0.123	12.3 %	$1.24 \times 10^{-4}$	3.91

A thousand-fold dilution raises the fraction ionised from under half a per cent to over twelve per cent, and cuts $[\mathrm{H_3O^+}]$ by a factor of about thirty-four. Both are true because $\alpha$ is a fraction and $[\mathrm{H_3O^+}]$ is an amount. A larger fraction of a much smaller total is still a smaller total.

The mechanism is Le Chatelier. Ionisation converts one dissolved particle into two, so adding water shifts the system towards the side with more particles and more $\mathrm{HA}$ ionises. It shifts forward, but not far enough to keep $[\mathrm{H_3O^+}]$ where it was, because the added water has diluted the ions too.

The last row carries its own warning. At $0.001\ \mathrm{M}$, $\alpha$ has passed 5 per cent, so the shortcut is no longer clean: $\sqrt{K_a/c}$ gives $0.132$ where the quadratic gives $0.123$, and the pH shifts from $3.88$ to $3.91$. Only that last row has been solved from the quadratic; the three rows above it come from the shortcut, which is accurate to better than one per cent there.

Key Point: On dilution of a weak acid: $\alpha$ increases, $[\mathrm{H_3O^+}]$ decreases, pH increases, and $K_a$ does not change.

Adding "$K_a$ does not change" is not padding. A common wrong answer holds that the ionisation constant increases on dilution because more ionisation has occurred. What changed was the position of the equilibrium, not the constant.

[NEET] For a strong acid, dilution by a factor of ten raises the pH by exactly 1 until the solution approaches $10^{-6}\ \mathrm{M}$. For a weak acid the same dilution raises it by only $0.5$.

Question 1: Degree of ionisation of hydrofluoric acid

The ionisation constant of $\mathrm{HF}$ is $3.5 \times 10^{-4}$. Calculate the degree of dissociation of $\mathrm{HF}$ in its $0.02\ \mathrm{M}$ solution, the concentrations of $\mathrm{H_3O^+}$, $\mathrm{F^-}$ and $\mathrm{HF}$, and the pH.

Answer:

Two proton transfers are possible: $\mathrm{HF}$ with water, $K_a = 3.5 \times 10^{-4}$, and water with itself, $K_w = 1.0 \times 10^{-14}$. Since $K_a \gg K_w$ the first is the principal reaction.

$$\mathrm{HF} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{F^-}$$

Equilibrium concentrations are $0.02(1-\alpha)$, $0.02\alpha$ and $0.02\alpha$.

$K_a = \dfrac{(0.02\alpha)^2}{0.02(1-\alpha)} = \dfrac{0.02\alpha^2}{1-\alpha} = 3.5 \times 10^{-4}$

I check whether the shortcut is allowed: $c/K_a = 0.02/(3.5 \times 10^{-4}) = 57.1$, which is below 400, so I solve the quadratic.

$\alpha^2 + 1.75 \times 10^{-2}\alpha - 1.75 \times 10^{-2} = 0$

$\alpha = \dfrac{-0.0175 + \sqrt{0.000306 + 0.07}}{2} = \dfrac{-0.0175 + 0.2652}{2} = 0.124$

The negative root $-0.141$ is rejected.

$[\mathrm{H_3O^+}] = [\mathrm{F^-}] = c\alpha = 0.02 \times 0.1238 = 2.48 \times 10^{-3}\ \mathrm{M}$

$[\mathrm{HF}] = c(1-\alpha) = 0.02 \times 0.876 = 1.75 \times 10^{-2}\ \mathrm{M}$

$\mathrm{pH} = -\log(2.48 \times 10^{-3}) = 2.61$

Ans: $\alpha = 0.124$; $[\mathrm{H_3O^+}] = [\mathrm{F^-}] = 2.48 \times 10^{-3}\ \mathrm{M}$; $[\mathrm{HF}] = 1.75 \times 10^{-2}\ \mathrm{M}$; $\mathrm{pH} = 2.61$ Watch out: The shortcut gives $\alpha = 0.132$, about $7$ per cent too high. Its pH, $2.58$, is close enough to pass; its degree of ionisation is not.

Question 2: pH and percent dissociation of hypochlorous acid

Calculate the pH of a $0.08\ \mathrm{M}$ solution of $\mathrm{HOCl}$, ionisation constant $2.5 \times 10^{-5}$, and its percent dissociation.

Answer:

$$\mathrm{HOCl(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{ClO^-(aq)}$$

The constant quoted here is the $2.5 \times 10^{-5}$ used in this standard problem, not the $3.0 \times 10^{-8}$ listed for $\mathrm{HOCl}$ in the table earlier; I work with the value the question supplies.

The ratio first: $c/K_a = 0.08/(2.5 \times 10^{-5}) = 3200$, comfortably above 400, so the shortcut is safe.

Writing $x$ for $[\mathrm{H_3O^+}]$ at equilibrium, with $0.08 - x \approx 0.08$:

$\dfrac{x^2}{0.08} = 2.5 \times 10^{-5}$

$x^2 = 2.0 \times 10^{-6}$, so $x = 1.41 \times 10^{-3}\ \mathrm{M}$

$\mathrm{pH} = -\log(1.41 \times 10^{-3}) = 2.85$

Percent dissociation $= \dfrac{1.41 \times 10^{-3}}{0.08} \times 100 = 1.76$ per cent

Ans: $\mathrm{pH} = 2.85$; percent dissociation $= 1.76$ per cent

Question 3: Ka from a measured pH

The pH of a $0.1\ \mathrm{M}$ solution of a monobasic acid is $4.50$. Find $[\mathrm{H^+}]$, $[\mathrm{A^-}]$ and $[\mathrm{HA}]$ at equilibrium, and the values of $K_a$ and $\mathrm{p}K_a$.

Answer:

From the pH:

$[\mathrm{H^+}] = 10^{-4.50} = 10^{0.50} \times 10^{-5} = 3.16 \times 10^{-5}\ \mathrm{M}$

The acid gives one $\mathrm{H^+}$ and one $\mathrm{A^-}$ together, so $[\mathrm{A^-}] = 3.16 \times 10^{-5}\ \mathrm{M}$. That is negligible against $0.1\ \mathrm{M}$, so $[\mathrm{HA}] \approx 0.1\ \mathrm{M}$.

$K_a = \dfrac{(3.16 \times 10^{-5})^2}{0.1} = \dfrac{1.0 \times 10^{-9}}{0.1} = 1.0 \times 10^{-8}$

$\mathrm{p}K_a = -\log(10^{-8}) = 8$

Ans: $[\mathrm{H^+}] = [\mathrm{A^-}] = 3.16 \times 10^{-5}\ \mathrm{M}$; $[\mathrm{HA}] = 0.1\ \mathrm{M}$; $K_a = 1.0 \times 10^{-8}$; $\mathrm{p}K_a = 8$ Watch out: $10^{-4.50}$ is not $4.5 \times 10^{-5}$. Split the exponent as $10^{0.50} \times 10^{-5}$ and take the antilog of the positive part.

Question 4: Degree of ionisation and pH of ammonia

Determine the degree of ionisation and the pH of a $0.05\ \mathrm{M}$ solution of ammonia, $K_b = 1.77 \times 10^{-5}$. Also find $K_a$ for the ammonium ion.

Answer:

$$\mathrm{NH_3} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_4^+} + \mathrm{OH^-}$$

Here $c/K_b = 0.05/(1.77 \times 10^{-5}) = 2825$, far above 400, so I drop $\alpha$ against 1 in Ostwald's law.

$\alpha = \sqrt{\dfrac{1.77 \times 10^{-5}}{0.05}} = \sqrt{3.54 \times 10^{-4}} = 0.0188$

$[\mathrm{OH^-}] = c\alpha = 0.05 \times 0.0188 = 9.4 \times 10^{-4}\ \mathrm{M}$

$[\mathrm{H^+}] = \dfrac{1.0 \times 10^{-14}}{9.4 \times 10^{-4}} = 1.06 \times 10^{-11}\ \mathrm{M}$

$\mathrm{pH} = -\log(1.06 \times 10^{-11}) = 10.97$

For the conjugate acid, $K_a = K_w/K_b = (1.0 \times 10^{-14})/(1.77 \times 10^{-5}) = 5.6 \times 10^{-10}$.

Ans: $\alpha = 0.0188$ (1.88 per cent); $\mathrm{pH} = 10.97$; $K_a(\mathrm{NH_4^+}) = 5.6 \times 10^{-10}$ Watch out: $-\log(9.4 \times 10^{-4}) = 3.03$ is the pOH. Reporting it turns a solution of ammonia into an acid.

Question 5: Kb and pKb from a measured pH

The pH of a $0.004\ \mathrm{M}$ hydrazine solution is $9.7$. Calculate its ionisation constant $K_b$ and $\mathrm{p}K_b$.

Answer:

$$\mathrm{NH_2NH_2} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_2NH_3^+} + \mathrm{OH^-}$$

I go from pH to hydronium, then to hydroxide.

$[\mathrm{H^+}] = 10^{-9.7} = 2.0 \times 10^{-10}\ \mathrm{M}$

$[\mathrm{OH^-}] = \dfrac{1.0 \times 10^{-14}}{2.0 \times 10^{-10}} = 5.0 \times 10^{-5}\ \mathrm{M}$

The hydrazinium ion is produced in equal amount, and both are far below $0.004\ \mathrm{M}$, so the undissociated base is still $0.004\ \mathrm{M}$.

$K_b = \dfrac{(5.0 \times 10^{-5})^2}{0.004} = \dfrac{2.5 \times 10^{-9}}{0.004} = 6.3 \times 10^{-7}$

$\mathrm{p}K_b = -\log(6.3 \times 10^{-7}) = 6.20$

Ans: $K_b = 6.3 \times 10^{-7}$; $\mathrm{p}K_b = 6.20$ Watch out: Using $[\mathrm{H^+}]$ in place of $[\mathrm{OH^-}]$ gives about $10^{-17}$, a value no base could have. $K_b$ is built from the hydroxide side.

Question 6: Degree of ionisation from pKa

Calculate the degree of ionisation of $0.05\ \mathrm{M}$ acetic acid if its $\mathrm{p}K_a$ is $4.76$.

Answer:

First I convert $\mathrm{p}K_a$ to $K_a$.

$K_a = 10^{-4.76} = 10^{0.24} \times 10^{-5} = 1.74 \times 10^{-5}$

Check: $c/K_a = 0.05/(1.74 \times 10^{-5}) = 2874$, above 400.

$\alpha = \sqrt{\dfrac{1.74 \times 10^{-5}}{0.05}} = \sqrt{3.48 \times 10^{-4}} = 1.87 \times 10^{-2}$

Ans: $\alpha = 1.87 \times 10^{-2}$, that is $1.87$ per cent Watch out: Using $\mathrm{p}K_a$ in place of $K_a$ gives $\sqrt{4.76/0.05} = 9.8$. Any $\alpha$ above 1 signals that the conversion was skipped.

Question 7: pKa from a large degree of ionisation

The degree of ionisation of a $0.1\ \mathrm{M}$ bromoacetic acid solution is $0.132$. Calculate the pH of the solution and the $\mathrm{p}K_a$ of the acid.

Answer:

$[\mathrm{H_3O^+}] = c\alpha = 0.1 \times 0.132 = 1.32 \times 10^{-2}\ \mathrm{M}$

$\mathrm{pH} = -\log(1.32 \times 10^{-2}) = 1.88$

For $K_a$, $\alpha$ is $13.2$ per cent, well over the 5 per cent limit, so I keep the $(1-\alpha)$ term.

$K_a = \dfrac{c\alpha^2}{1-\alpha} = \dfrac{0.1 \times (0.132)^2}{1 - 0.132} = \dfrac{1.742 \times 10^{-3}}{0.868} = 2.01 \times 10^{-3}$

$\mathrm{p}K_a = -\log(2.01 \times 10^{-3}) = 2.70$

Ans: $\mathrm{pH} = 1.88$; $K_a = 2.01 \times 10^{-3}$; $\mathrm{p}K_a = 2.70$ Watch out: Dropping $(1-\alpha)$ gives $K_a = 1.74 \times 10^{-3}$ and $\mathrm{p}K_a = 2.76$. The pH is unaffected because it comes from $c\alpha$ directly.

Question 8: pH of a very weak base

Calculate the degree of ionisation and the pH of $0.01\ \mathrm{M}$ aniline, $K_b = 4.27 \times 10^{-10}$.

Answer:

$$\mathrm{C_6H_5NH_2} + \mathrm{H_2O} \rightleftharpoons \mathrm{C_6H_5NH_3^+} + \mathrm{OH^-}$$

$\alpha = \sqrt{\dfrac{4.27 \times 10^{-10}}{0.01}} = \sqrt{4.27 \times 10^{-8}} = 2.07 \times 10^{-4}$

$[\mathrm{OH^-}] = c\alpha = 0.01 \times 2.07 \times 10^{-4} = 2.07 \times 10^{-6}\ \mathrm{M}$

$\mathrm{pOH} = -\log(2.07 \times 10^{-6}) = 5.68$

$\mathrm{pH} = 14.00 - 5.68 = 8.32$

Ans: $\alpha = 2.07 \times 10^{-4}$; $\mathrm{pH} = 8.32$ Watch out: A pH of $8.32$ is only just alkaline, and that is correct — aniline is roughly forty thousand times weaker a base than ammonia.

Question 9: Comparing the shortcut with the quadratic

Chloroacetic acid has $K_a = 1.35 \times 10^{-3}$. For a $0.01\ \mathrm{M}$ solution, find $\alpha$ and the pH both by the approximation and exactly, and state the error.

Answer:

The check: $c/K_a = 0.01/(1.35 \times 10^{-3}) = 7.4$, far below 400.

By the shortcut: $\alpha = \sqrt{1.35 \times 10^{-3}/0.01} = \sqrt{0.135} = 0.367$, and $[\mathrm{H_3O^+}] = 3.67 \times 10^{-3}\ \mathrm{M}$, $\mathrm{pH} = 2.43$.

Exactly, from $\dfrac{0.01\alpha^2}{1-\alpha} = 1.35 \times 10^{-3}$:

$\alpha^2 + 0.135\alpha - 0.135 = 0$

$\alpha = \dfrac{-0.135 + \sqrt{0.0182 + 0.54}}{2} = \dfrac{-0.135 + 0.7471}{2} = 0.306$

$[\mathrm{H_3O^+}] = 0.01 \times 0.306 = 3.06 \times 10^{-3}\ \mathrm{M}$, $\mathrm{pH} = 2.51$.

The approximate $\alpha$ is too large by $(0.367-0.306)/0.306 = 20$ per cent. The pH is wrong by $0.08$ units.

Ans: Exact $\alpha = 0.306$, $\mathrm{pH} = 2.51$; the approximation gives $0.367$ and $2.43$, an error of $20$ per cent in $\alpha$ and $0.08$ in pH Watch out: The approximate answer is always the larger one. A calculated $\alpha$ above $0.05$ is an overestimate, and the quadratic is the fix.

Question 10: pH straight from the formula

Calculate the pH of $0.01\ \mathrm{M}$ acetic acid, $\mathrm{p}K_a = 4.76$.

Answer:

$\mathrm{pH} = \tfrac{1}{2}(\mathrm{p}K_a - \log c)$, and $\log(0.01) = -2$.

$\mathrm{pH} = \tfrac{1}{2}(4.76 + 2) = 3.38$

Ans: $\mathrm{pH} = 3.38$ Watch out: The formula subtracts $\log c$, it does not add it. With $c<1$, $\log c$ is negative, so subtracting raises the pH. Adding gives $1.38$, a value stronger than the same concentration of $\mathrm{HCl}$.

Question 11: Ka from percent ionisation

A $0.1\ \mathrm{M}$ solution of a weak monobasic acid is $1.32$ per cent ionised at 298 K. Calculate $K_a$, $\mathrm{p}K_a$ and the pH.

Answer:

$\alpha = 0.0132$

$[\mathrm{H_3O^+}] = c\alpha = 0.1 \times 0.0132 = 1.32 \times 10^{-3}\ \mathrm{M}$

$\mathrm{pH} = -\log(1.32 \times 10^{-3}) = 2.88$

$\alpha$ is under 5 per cent, so $K_a \approx c\alpha^2 = 0.1 \times (0.0132)^2 = 1.74 \times 10^{-5}$. Keeping the $(1-\alpha)$ term gives $1.77 \times 10^{-5}$, a difference of about one per cent.

$\mathrm{p}K_a = -\log(1.74 \times 10^{-5}) = 4.76$

Ans: $K_a = 1.74 \times 10^{-5}$; $\mathrm{p}K_a = 4.76$; $\mathrm{pH} = 2.88$ Watch out: Percent ionisation must be divided by 100 first. Using $1.32$ for $0.0132$ gives $K_a = 0.17$, a strong acid.

Question 12: Effect of dilution

A $0.1\ \mathrm{M}$ solution of acetic acid is diluted a hundred-fold. State what happens to $K_a$, $\alpha$, $[\mathrm{H_3O^+}]$ and the pH.

Answer:

$K_a$ depends only on temperature, so it stays at $1.74 \times 10^{-5}$.

$\alpha = \sqrt{K_a/c}$, and $c$ has fallen by 100, so $\alpha$ rises by $\sqrt{100} = 10$: from $0.0132$ to about $0.13$.

$[\mathrm{H_3O^+}] = \sqrt{K_ac}$, so the hydronium concentration falls by 10: from $1.32 \times 10^{-3}\ \mathrm{M}$ to about $1.3 \times 10^{-4}\ \mathrm{M}$, and the pH rises by 1 unit.

At the final concentration $\alpha$ has crossed 5 per cent, so the exact values are $\alpha = 0.123$, $[\mathrm{H_3O^+}] = 1.24 \times 10^{-4}\ \mathrm{M}$ and $\mathrm{pH} = 3.91$.

Ans: $K_a$ unchanged; $\alpha$ up ten-fold to about $0.12$; $[\mathrm{H_3O^+}]$ down ten-fold to $1.24 \times 10^{-4}\ \mathrm{M}$; pH up by about 1 unit to $3.91$ Watch out: "More ionisation" and "more acidic" are different claims. The fraction ionised went up; the hydronium concentration went down.

What to Carry Forward

Every result here comes from one equilibrium and one definition.

Quantity	Expression	Condition
$K_a$	$\dfrac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$	any weak acid; water omitted
$K_b$	$\dfrac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]}$	any weak base; water omitted
$\mathrm{p}K_a$, $\mathrm{p}K_b$	$-\log K_a$, $-\log K_b$	always; smaller means stronger
$\alpha$	$[\mathrm{H_3O^+}]/c$	always; exact
Ostwald's law	$K_a = \dfrac{c\alpha^2}{1-\alpha}$	always; exact
$\alpha = \sqrt{K_a/c}$		only when $\alpha < 0.05$, i.e. $c/K_a > 400$
$[\mathrm{H_3O^+}] = \sqrt{K_ac}$		same condition
$\mathrm{pH} = \tfrac{1}{2}(\mathrm{p}K_a - \log c)$		same condition
$\mathrm{pOH} = \tfrac{1}{2}(\mathrm{p}K_b - \log c)$		same condition, weak base

Five errors account for almost all the lost marks.

Reading $\mathrm{p}K_a$ the wrong way and calling the acid with the larger $\mathrm{p}K_a$ the stronger one.

Using $\mathrm{p}K_a$ where $K_a$ belongs, which produces a degree of ionisation greater than 1.

Applying $\alpha = \sqrt{K_a/c}$ without checking $c/K_a$, and reporting $37$ per cent where the true value is $31$.

Forgetting the $\mathrm{pOH}$ to $\mathrm{pH}$ conversion for a base, so $10.97$ is reported as $3.03$.

Confusing $\alpha$ with $[\mathrm{H_3O^+}]$ on dilution, or claiming $K_a$ itself rises. The fraction rises, the concentration falls, and the constant does neither.

The next section takes $K_a$ and $K_b$ for a conjugate pair and shows that their product is $K_w$, and extends the treatment to acids that can lose more than one proton.