The Ka-Kb Relation, Polyprotic Acids and Factors Affecting Acid Strength

Two Constants for One Pair

A conjugate acid-base pair differs by a single proton. Acetic acid and acetate ion are such a pair; ammonium ion and ammonia are another. Each member of the pair reacts with water, and each reaction has its own equilibrium constant.

The acid $\mathrm{HA}$ gives a proton to water:

$$\mathrm{HA(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{A^-(aq)} \qquad K_a = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$$

Its conjugate base $\mathrm{A^-}$ takes a proton from water:

$$\mathrm{A^-(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{HA(aq)} + \mathrm{OH^-(aq)} \qquad K_b = \frac{[\mathrm{HA}][\mathrm{OH^-}]}{[\mathrm{A^-}]}$$

Both leave $[\mathrm{H_2O}]$ out of the expression, since the solvent concentration is fixed at about $55.5\ \mathrm{mol\,L^{-1}}$.

These two constants are not independent. Add the two equations as chemical equations:

$$\mathrm{HA} + \mathrm{H_2O} + \mathrm{A^-} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{A^-} + \mathrm{HA} + \mathrm{OH^-}$$

$\mathrm{HA}$ appears on both sides. So does $\mathrm{A^-}$. Strike both out and what is left is the self-ionisation of water:

$$2\mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_3O^+(aq)} + \mathrm{OH^-(aq)} \qquad K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}]$$

Adding equations multiplies the equilibrium constants. That rule was established earlier for gas-phase equilibria and it holds here without change, because it is a property of the algebra of equilibrium expressions and not of the phase. So the constant for the net reaction is $K_a \times K_b$, and the net reaction is the water equilibrium.

$$K_a \times K_b = K_w$$

The multiplication is worth doing in full once:

$$K_a \times K_b = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]} \times \frac{[\mathrm{HA}][\mathrm{OH^-}]}{[\mathrm{A^-}]} = [\mathrm{H_3O^+}][\mathrm{OH^-}] = K_w$$

$[\mathrm{HA}]$ in the numerator of the second fraction meets $[\mathrm{HA}]$ in the denominator of the first. $[\mathrm{A^-}]$ does the same in reverse. Nothing survives except the ion product of water.

Key Point: For a conjugate acid-base pair in aqueous solution, $$K_a \times K_b = K_w = 1.0 \times 10^{-14} \quad \text{at } 298\ \mathrm{K}$$ Knowing either constant fixes the other. The pair must be conjugate: the base must be exactly the acid minus one proton.

The Same Relation in p-Form

Taking the negative logarithm of both sides of $K_a K_b = K_w$ turns a product into a sum. Since $\log(xy) = \log x + \log y$,

$$-\log K_a - \log K_b = -\log K_w$$

$$\mathrm{p}K_a + \mathrm{p}K_b = \mathrm{p}K_w = 14.00 \quad \text{at } 298\ \mathrm{K}$$

This is the form used in almost every calculation, because it replaces division by subtraction. A base with $\mathrm{p}K_b = 4.75$ has a conjugate acid with $\mathrm{p}K_a = 14.00 - 4.75 = 9.25$, and no exponent arithmetic is needed.

The temperature condition is real and it is tested. $K_w$ is an equilibrium constant, so it changes with temperature: the self-ionisation of water is endothermic, and heating drives it forward. At $298\ \mathrm{K}$, $K_w = 1.0 \times 10^{-14}$ and $\mathrm{p}K_w = 14.00$. At body temperature, $310\ \mathrm{K}$, $K_w$ is $2.7 \times 10^{-14}$, so $\mathrm{p}K_w$ is $13.57$. At that temperature $\mathrm{p}K_a + \mathrm{p}K_b = 13.57$, not $14$.

The general statement is the safe one to memorise: $\mathrm{p}K_a + \mathrm{p}K_b = \mathrm{p}K_w$, and $\mathrm{p}K_w = 14$ only at $298\ \mathrm{K}$. The same caution applies to $\mathrm{pH} + \mathrm{pOH} = \mathrm{p}K_w$ and to the neutral pH of $7$.

[JEE Main] A question that supplies $K_w$ at some temperature other than $298\ \mathrm{K}$ is not being decorative. It is checking whether $14$ was memorised as a number or understood as $\mathrm{p}K_w$.

The relation also makes the reciprocal nature of conjugate strength quantitative. Rearranging,

$$K_b = \frac{K_w}{K_a}$$

$K_w$ is a small fixed number. A large $K_a$ therefore forces a tiny $K_b$, and a tiny $K_a$ forces a large $K_b$. A strong acid has a very weak conjugate base; a very weak acid has a comparatively strong conjugate base. This is the statement met qualitatively in the Bronsted treatment, now with a number attached.

Perchloric, hydrochloric, sulphuric and nitric acids ionise completely in water, so $\mathrm{ClO_4^-}$, $\mathrm{Cl^-}$, $\mathrm{HSO_4^-}$ and $\mathrm{NO_3^-}$ have $K_b$ values too small to measure. A solution of sodium chloride is neutral for exactly this reason: $\mathrm{Cl^-}$ is such a feeble base that it does not take a proton from water at all. At the other end, $\mathrm{NH_2^-}$, $\mathrm{O^{2-}}$ and $\mathrm{H^-}$ are conjugate bases of extremely weak acids, and they are such powerful proton acceptors that they cannot exist in water — they strip a proton from the solvent itself.

Only for a Conjugate Pair

The single most common error with this relation is applying it to an acid and a base that are not conjugates. $K_a$ of acetic acid multiplied by $K_b$ of ammonia is

$$1.74 \times 10^{-5} \times 1.77 \times 10^{-5} = 3.08 \times 10^{-10}$$

which is not $K_w$ and is not anything. The product of the constants of two unrelated species means nothing, because the equations for those two species do not add up to the water equilibrium — the cancellation that produced $K_w$ never happens.

Test the pairing before using the relation. Write down the base. Add one proton. If the result is the acid you were given, the relation applies; otherwise it does not.

Acid	$K_a$ (298 K)	Its conjugate base	$K_b = K_w/K_a$
$\mathrm{HF}$	$3.5 \times 10^{-4}$	$\mathrm{F^-}$	$2.9 \times 10^{-11}$
$\mathrm{HCOOH}$	$1.8 \times 10^{-4}$	$\mathrm{HCOO^-}$	$5.6 \times 10^{-11}$
$\mathrm{CH_3COOH}$	$1.74 \times 10^{-5}$	$\mathrm{CH_3COO^-}$	$5.7 \times 10^{-10}$
$\mathrm{NH_4^+}$	$5.6 \times 10^{-10}$	$\mathrm{NH_3}$	$1.77 \times 10^{-5}$
$\mathrm{HCN}$	$4.9 \times 10^{-10}$	$\mathrm{CN^-}$	$2.0 \times 10^{-5}$

Read the last two rows carefully. The pair listed for ammonia is $\mathrm{NH_4^+}$ and $\mathrm{NH_3}$, not ammonia and hydroxide, and not ammonia and acetate. The $K_b$ of ammonia, $1.77 \times 10^{-5}$, is measured directly; the $K_a$ of ammonium ion, $5.6 \times 10^{-10}$, is what the relation delivers. That number is what makes an ammonium salt solution acidic, and it is the number needed for the hydrolysis calculations in the next section.

A second slip is a mismatch of forms. $K_a \times K_b = K_w$ multiplies constants; $\mathrm{p}K_a + \mathrm{p}K_b = 14$ adds logarithms. Writing $K_a + K_b = K_w$ or $\mathrm{p}K_a \times \mathrm{p}K_b = 14$ produces nonsense.

A third slip is direction. For the pair acetic acid and acetate, $K_a$ belongs to acetic acid and $K_b$ to acetate. Using $K_b$ of acetate where $K_a$ of acetic acid is wanted turns $1.74 \times 10^{-5}$ into $5.7 \times 10^{-10}$, a factor of thirty thousand, and turns the pH of a $0.1\ \mathrm{M}$ solution of the acid from $2.88$ into $5.12$.

Acids with More Than One Proton

Oxalic acid, sulphuric acid, carbonic acid and phosphoric acid have more than one ionisable proton per molecule. Such acids are polybasic or polyprotic. They do not lose their protons together. Each proton leaves in a separate step, with its own equilibrium and its own constant.

For a dibasic acid $\mathrm{H_2X}$:

$$\mathrm{H_2X(aq)} \rightleftharpoons \mathrm{H^+(aq)} + \mathrm{HX^-(aq)} \qquad K_{a_1} = \frac{[\mathrm{H^+}][\mathrm{HX^-}]}{[\mathrm{H_2X}]}$$

$$\mathrm{HX^-(aq)} \rightleftharpoons \mathrm{H^+(aq)} + \mathrm{X^{2-}(aq)} \qquad K_{a_2} = \frac{[\mathrm{H^+}][\mathrm{X^{2-}}]}{[\mathrm{HX^-}]}$$

$K_{a_1}$ and $K_{a_2}$ are the first and second ionisation constants. A tribasic acid such as $\mathrm{H_3PO_4}$ has three:

$$\mathrm{H_3PO_4} \rightleftharpoons \mathrm{H^+} + \mathrm{H_2PO_4^-} \qquad K_{a_1}$$

$$\mathrm{H_2PO_4^-} \rightleftharpoons \mathrm{H^+} + \mathrm{HPO_4^{2-}} \qquad K_{a_2}$$

$$\mathrm{HPO_4^{2-}} \rightleftharpoons \mathrm{H^+} + \mathrm{PO_4^{3-}} \qquad K_{a_3}$$

The intermediate species are worth naming. $\mathrm{H_2PO_4^-}$ and $\mathrm{HPO_4^{2-}}$ each appear as a product of one step and a reactant of the next, so each can donate a proton and accept one. Species of that kind are amphiprotic. $\mathrm{HCO_3^-}$, $\mathrm{HS^-}$ and $\mathrm{HSO_4^-}$ behave the same way.

Two consequences follow from the stepwise picture.

A solution of a diprotic acid contains three related solute species at once — $\mathrm{H_2X}$, $\mathrm{HX^-}$ and $\mathrm{X^{2-}}$ — in proportions set by the two constants and by the pH. It is never a solution of one substance.

The overall ionisation constant, for $\mathrm{H_2X} \rightleftharpoons 2\mathrm{H^+} + \mathrm{X^{2-}}$, is the product $K_{a_1} \times K_{a_2}$, since adding the two steps multiplies their constants. For carbonic acid that overall constant is $4.5 \times 10^{-7} \times 4.7 \times 10^{-11} = 2.1 \times 10^{-17}$.

The Values, and Why They Fall

Acid	$K_{a_1}$	$K_{a_2}$	$K_{a_3}$
Carbonic acid, $\mathrm{H_2CO_3}$	$4.5 \times 10^{-7}$	$4.7 \times 10^{-11}$
Sulphurous acid, $\mathrm{H_2SO_3}$	$1.7 \times 10^{-2}$	$6.4 \times 10^{-8}$
Sulphuric acid, $\mathrm{H_2SO_4}$	very large	$1.2 \times 10^{-2}$
Phosphoric acid, $\mathrm{H_3PO_4}$	$7.5 \times 10^{-3}$	$6.2 \times 10^{-8}$	$4.2 \times 10^{-13}$
Oxalic acid, $\mathrm{(COOH)_2}$	$5.9 \times 10^{-2}$	$6.4 \times 10^{-5}$
Hydrogen sulphide, $\mathrm{H_2S}$	$9.1 \times 10^{-8}$	$1.2 \times 10^{-13}$

Every row shows the same pattern. $K_{a_2}$ is smaller than $K_{a_1}$, and where there is a third step $K_{a_3}$ is smaller again. The gaps are large: four orders of magnitude for carbonic acid, five for sulphurous acid, five for phosphoric acid between the first and second steps and another five between the second and third, six for hydrogen sulphide. Oxalic acid has the narrowest gap at about three orders of magnitude.

The reason is electrostatic. The first proton leaves a neutral molecule. The second must leave an anion that already carries a negative charge, and a positive proton is held far more tightly by a negative ion than by a neutral molecule. Pulling $\mathrm{H^+}$ away from $\mathrm{HCO_3^-}$ means separating opposite charges, and that costs energy the first step never had to pay.

Compare the two carbonic acid steps directly. Removing a proton from uncharged $\mathrm{H_2CO_3}$ has $K_{a_1} = 4.5 \times 10^{-7}$. Removing one from $\mathrm{HCO_3^-}$ has $K_{a_2} = 4.7 \times 10^{-11}$, about ten thousand times harder. Phosphoric acid makes the point three times over: a proton comes off neutral $\mathrm{H_3PO_4}$ fairly readily, off singly charged $\mathrm{H_2PO_4^-}$ with much more difficulty, and off doubly charged $\mathrm{HPO_4^{2-}}$ barely at all, since $K_{a_3} = 4.2 \times 10^{-13}$ is close to the ionisation constant of water.

Key Point: For any polyprotic acid $K_{a_1} > K_{a_2} > K_{a_3}$, usually by several orders of magnitude, because removing a positively charged proton from an already negative ion is opposed by electrostatic attraction.

There is a second, smaller contribution. Each proton lost adds negative charge to the same central atom, and the remaining $\mathrm{O-H}$ bonds become less polar as the electron density on oxygen rises, so those protons are less willing to leave. The charge argument is the one to write in an answer.

Polyacidic bases behave the same way in reverse, with $K_{b_1} > K_{b_2}$.

What the Second Constant Does and Does Not Affect

A polyprotic acid solution looks complicated: several equilibria, several species. In practice one step dominates completely.

Take $0.1\ \mathrm{M}$ carbonic acid. The first step gives $[\mathrm{H^+}] = \sqrt{K_{a_1}c} = \sqrt{4.5 \times 10^{-7} \times 0.1} = 2.1 \times 10^{-4}\ \mathrm{M}$ and the same concentration of $\mathrm{HCO_3^-}$. The second step now has to work on that $\mathrm{HCO_3^-}$ in a solution which already contains $2.1 \times 10^{-4}\ \mathrm{M}$ of $\mathrm{H^+}$. That $\mathrm{H^+}$ is a common ion, and it pushes the second equilibrium back. Since $K_{a_2}$ is minute to begin with, the extra $\mathrm{H^+}$ produced by the second step is negligible.

Key Point: For most calculations the pH of a polyprotic acid solution is fixed by $K_{a_1}$ alone. The acid is treated as a monobasic acid of ionisation constant $K_{a_1}$, and $K_{a_2}$ is used only to find the concentration of the fully deprotonated ion.

The second constant does one useful job. For a diprotic acid where the first step supplies both the $\mathrm{H^+}$ and the $\mathrm{HX^-}$, those two concentrations are equal, so

$$K_{a_2} = \frac{[\mathrm{H^+}][\mathrm{X^{2-}}]}{[\mathrm{HX^-}]} \approx [\mathrm{X^{2-}}]$$

because $[\mathrm{H^+}]$ and $[\mathrm{HX^-}]$ cancel. The concentration of the doubly charged anion is numerically equal to $K_{a_2}$, whatever the concentration of the acid. In $0.1\ \mathrm{M}$ $\mathrm{H_2S}$, $[\mathrm{S^{2-}}] = 1.2 \times 10^{-13}\ \mathrm{M}$; in $0.05\ \mathrm{M}$ $\mathrm{H_2S}$ it is the same $1.2 \times 10^{-13}\ \mathrm{M}$. This result is used constantly in sulphide precipitation problems.

[JEE/NEET] Two standard answers worth keeping: for a diprotic acid, $[\mathrm{H^+}] \approx \sqrt{K_{a_1}c}$ and $[\mathrm{X^{2-}}] \approx K_{a_2}$.

Sulphuric acid is the exception that has to be handled separately. Its first ionisation is complete in dilute aqueous solution:

$$\mathrm{H_2SO_4(aq)} \rightarrow \mathrm{H^+(aq)} + \mathrm{HSO_4^-(aq)}$$

The arrow is single, not double. There is no equilibrium and no meaningful $K_{a_1}$; that is what "very large" in the table stands for. The second step is a real equilibrium of a genuinely weak acid:

$$\mathrm{HSO_4^-(aq)} \rightleftharpoons \mathrm{H^+(aq)} + \mathrm{SO_4^{2-}(aq)} \qquad K_{a_2} = 1.2 \times 10^{-2}$$

Sulphuric acid is a strong acid in its first proton and a weak acid in its second. So a $0.1\ \mathrm{M}$ solution does not give $0.2\ \mathrm{M}$ $\mathrm{H^+}$, and it does not give exactly $0.1\ \mathrm{M}$ either. The truth lies between, and $K_{a_2}$ decides where, as worked out below. Treating $\mathrm{H_2SO_4}$ as fully ionised in both steps and reporting $\mathrm{pH} = 0.70$ for a $0.1\ \mathrm{M}$ solution is a standard error; so is ignoring the second step entirely and reporting $1.00$.

What Makes One Acid Stronger Than Another

$K_a$ measures acid strength but does not explain it. For a binary acid $\mathrm{H-A}$ the explanation has two parts, and both concern the same event: a proton leaving.

$$\mathrm{H-A} \rightarrow \mathrm{H^+} + \mathrm{A^-}$$

The strength of the $\mathrm{H-A}$ bond. The bond must break. A weaker bond needs less energy to break, so the proton leaves more easily and the acid is stronger.

The stability of the conjugate base $\mathrm{A^-}$. The fragment left behind must accommodate the electron pair and the negative charge. A more stable $\mathrm{A^-}$ has less tendency to take the proton back, so the equilibrium sits further to the right and the acid is stronger. The two statements are linked through $K_a K_b = K_w$: a stable, unreactive $\mathrm{A^-}$ has a small $K_b$, which forces a large $K_a$.

Bond polarity feeds into the second factor. When $\mathrm{A}$ is more electronegative than hydrogen, the bonding pair sits closer to $\mathrm{A}$, charge is already separated along the bond, and the bond breaks in the required way — with both electrons going to $\mathrm{A}$ — more readily. A more polar $\mathrm{H-A}$ bond means a stronger acid, provided the bond strength is not what is changing.

Key Point: Acid strength of $\mathrm{H-A}$ rises when the $\mathrm{H-A}$ bond gets weaker and when the conjugate base $\mathrm{A^-}$ gets more stable. Increasing the electronegativity of $\mathrm{A}$ increases bond polarity and stabilises $\mathrm{A^-}$, and so increases acidity.

Both factors always operate, and in the periodic table they often point in opposite directions. Which one wins depends on the direction being compared, and that is the whole content of the next block.

One caution on vocabulary. "Bond strength" here is bond dissociation enthalpy, an energy. "Bond polarity" is charge separation, set by electronegativity difference. A bond can be very polar and very strong at the same time — $\mathrm{H-F}$ is exactly that — which is why the two factors have to be tracked separately.

Down a Group Bond Strength Wins, Across a Period Electronegativity Wins

Down a group, the atom $\mathrm{A}$ gets larger. Its valence orbital is bigger and more diffuse, its overlap with the small hydrogen $1s$ orbital gets poorer, and the $\mathrm{H-A}$ bond gets longer and weaker. Bond dissociation enthalpy falls steeply: $567\ \mathrm{kJ\,mol^{-1}}$ for $\mathrm{H-F}$, $431$ for $\mathrm{H-Cl}$, $366$ for $\mathrm{H-Br}$, $299$ for $\mathrm{H-I}$. Electronegativity also falls down the group, which would predict decreasing acidity, but the collapse in bond strength is much larger and it decides the outcome.

$$\mathrm{HF} \lll \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$$

$\mathrm{HCl}$, $\mathrm{HBr}$ and $\mathrm{HI}$ are all strong acids in water; $\mathrm{HF}$ is weak, with $K_a = 3.5 \times 10^{-4}$. Group 16 shows the same order, with all four being weak acids:

$$\mathrm{H_2O} < \mathrm{H_2S} < \mathrm{H_2Se} < \mathrm{H_2Te}$$

$\mathrm{H_2S}$ is a stronger acid than water even though oxygen is far more electronegative than sulphur, because the $\mathrm{S-H}$ bond is much weaker than the $\mathrm{O-H}$ bond. Size also stabilises the conjugate base: the negative charge on the large $\mathrm{I^-}$ or $\mathrm{HS^-}$ ion is spread over a bigger volume, so the charge density is lower and the ion is more stable than a small, charge-concentrated $\mathrm{F^-}$ or $\mathrm{OH^-}$.

Across a period, the atoms are all of similar size, so $\mathrm{H-A}$ bond strengths do not differ dramatically. What changes sharply is electronegativity, and it now decides.

$$\mathrm{CH_4} < \mathrm{NH_3} < \mathrm{H_2O} < \mathrm{HF}$$

Carbon, nitrogen, oxygen and fluorine rise in electronegativity from $2.5$ to $4.0$. The $\mathrm{H-A}$ bond becomes progressively more polar, and the conjugate base — $\mathrm{CH_3^-}$, $\mathrm{NH_2^-}$, $\mathrm{OH^-}$, $\mathrm{F^-}$ — becomes progressively better at holding the negative charge. Methane is not an acid in any practical sense; hydrogen fluoride is a recognisable, if weak, acid.

Key Point: Compare down a group using $\mathrm{H-A}$ bond strength — acidity increases downward. Compare across a period using the electronegativity of $\mathrm{A}$ — acidity increases rightward. Applying the wrong criterion reverses the answer.

The mistake this warning exists to prevent runs as follows: fluorine is the most electronegative element, so $\mathrm{HF}$ must be the strongest hydrohalic acid. The conclusion is wrong. In water, $\mathrm{HF}$ is the only weak acid of the four, weaker than $\mathrm{HCl}$ by many orders of magnitude — hydrochloric acid is ionised so completely that no $K_a$ can meaningfully be quoted for it at all. The $\mathrm{H-F}$ bond is exceptionally strong at $567\ \mathrm{kJ\,mol^{-1}}$, and in aqueous solution the small $\mathrm{F^-}$ ion hydrogen-bonds strongly to water and to undissociated $\mathrm{HF}$, both of which hold the proton back.

The two criteria coexist without contradiction because they answer different questions. Fluorine's electronegativity is why $\mathrm{HF}$ is a far stronger acid than $\mathrm{H_2O}$, $\mathrm{NH_3}$ or $\mathrm{CH_4}$ — that is the period comparison. Bond strength is why $\mathrm{HF}$ is a far weaker acid than $\mathrm{HI}$ — that is the group comparison. Both are correct, in their own direction.

[Board] Answers should name the factor, not just the order. "Acidity increases from $\mathrm{HF}$ to $\mathrm{HI}$ because the $\mathrm{H-X}$ bond dissociation enthalpy decreases down the group" earns the mark; the order by itself does not.

Question 1: Kb of fluoride ion

The ionisation constant of $\mathrm{HF}$ is $3.5 \times 10^{-4}$ at $298\ \mathrm{K}$. Calculate $K_b$ and $\mathrm{p}K_b$ for the fluoride ion.

Answer:

$\mathrm{F^-}$ is the conjugate base of $\mathrm{HF}$: remove one proton from $\mathrm{HF}$ and $\mathrm{F^-}$ is what remains. The pair is conjugate, so the relation applies.

$K_b = \dfrac{K_w}{K_a} = \dfrac{1.0 \times 10^{-14}}{3.5 \times 10^{-4}} = 2.86 \times 10^{-11}$

For the p-values, $\mathrm{p}K_a = -\log(3.5 \times 10^{-4}) = 4 - 0.544 = 3.46$.

$\mathrm{p}K_b = 14.00 - 3.46 = 10.54$

A check: $-\log(2.86 \times 10^{-11}) = 11 - 0.456 = 10.54$. The two routes agree.

Ans: $K_b = 2.9 \times 10^{-11}$; $\mathrm{p}K_b = 10.54$

Question 2: Ka of the conjugate acid of ammonia

The ionisation constant of ammonia is $1.77 \times 10^{-5}$. Find the ionisation constant and $\mathrm{p}K_a$ of its conjugate acid.

Answer:

The conjugate acid of $\mathrm{NH_3}$ is $\mathrm{NH_4^+}$, ammonia plus one proton.

$$\mathrm{NH_4^+(aq)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{NH_3(aq)} + \mathrm{H_3O^+(aq)}$$

$K_a = \dfrac{K_w}{K_b} = \dfrac{1.0 \times 10^{-14}}{1.77 \times 10^{-5}} = 5.6 \times 10^{-10}$

$\mathrm{p}K_b = -\log(1.77 \times 10^{-5}) = 5 - 0.248 = 4.75$, so $\mathrm{p}K_a = 14.00 - 4.75 = 9.25$.

Ans: $K_a = 5.6 \times 10^{-10}$; $\mathrm{p}K_a = 9.25$ Watch out: $K_a$ here belongs to $\mathrm{NH_4^+}$, not to $\mathrm{NH_3}$. Ammonia has no measurable $K_a$ in water. A question asking for "$K_a$ of ammonia solution" means the ammonium ion.

Question 3: Reading a pKa backwards

A weak base has $\mathrm{p}K_b = 8.30$ at $298\ \mathrm{K}$. Find $K_b$, and the $K_a$ and $\mathrm{p}K_a$ of its conjugate acid. Is that conjugate acid stronger or weaker than acetic acid, $\mathrm{p}K_a = 4.76$?

Answer:

$K_b = 10^{-8.30} = 10^{0.70} \times 10^{-9} = 5.0 \times 10^{-9}$

$\mathrm{p}K_a = 14.00 - 8.30 = 5.70$

$K_a = 10^{-5.70} = 10^{0.30} \times 10^{-6} = 2.0 \times 10^{-6}$

A larger $\mathrm{p}K_a$ means a weaker acid. Since $5.70 > 4.76$, the conjugate acid is weaker than acetic acid, by a factor of $10^{0.94} \approx 8.7$ in $K_a$.

Ans: $K_b = 5.0 \times 10^{-9}$; $K_a = 2.0 \times 10^{-6}$; $\mathrm{p}K_a = 5.70$; weaker than acetic acid Watch out: $\mathrm{p}K_a$ and $\mathrm{p}K_b$ run opposite to strength. Calling the acid with the larger $\mathrm{p}K_a$ the stronger one is the single most frequent slip in this topic.

Question 4: Ranking conjugate bases

Given $K_a(\mathrm{HCN}) = 4.9 \times 10^{-10}$, $K_a(\mathrm{CH_3COOH}) = 1.74 \times 10^{-5}$ and $K_{a_2}(\mathrm{H_2CO_3}) = 4.7 \times 10^{-11}$, arrange $\mathrm{CN^-}$, $\mathrm{CH_3COO^-}$ and $\mathrm{CO_3^{2-}}$ in increasing order of base strength.

Answer:

Each anion is the conjugate base of one of the three acids. $\mathrm{CO_3^{2-}}$ pairs with $\mathrm{HCO_3^-}$, whose constant is $K_{a_2}$ of carbonic acid.

$K_b(\mathrm{CH_3COO^-}) = \dfrac{10^{-14}}{1.74 \times 10^{-5}} = 5.7 \times 10^{-10}$

$K_b(\mathrm{CN^-}) = \dfrac{10^{-14}}{4.9 \times 10^{-10}} = 2.0 \times 10^{-5}$

$K_b(\mathrm{CO_3^{2-}}) = \dfrac{10^{-14}}{4.7 \times 10^{-11}} = 2.1 \times 10^{-4}$

Increasing $K_b$ is increasing base strength.

Ans: $\mathrm{CH_3COO^-} < \mathrm{CN^-} < \mathrm{CO_3^{2-}}$ Watch out: The order of the bases is the reverse of the order of their parent acids, so it can be written down without computing anything — but pairing $\mathrm{CO_3^{2-}}$ with $K_{a_1}$ instead of $K_{a_2}$ gives $K_b = 2.2 \times 10^{-8}$ and puts carbonate in the middle.

Question 5: pH of carbonic acid and the carbonate concentration

For $0.05\ \mathrm{M}$ carbonic acid, $K_{a_1} = 4.5 \times 10^{-7}$ and $K_{a_2} = 4.7 \times 10^{-11}$. Calculate the pH, $[\mathrm{HCO_3^-}]$ and $[\mathrm{CO_3^{2-}}]$.

Answer:

$K_{a_2}$ is four orders of magnitude below $K_{a_1}$, so the first step alone sets the pH. I treat the acid as monobasic with $K_a = 4.5 \times 10^{-7}$.

$c/K_a = 0.05/(4.5 \times 10^{-7}) = 1.1 \times 10^5$, far above $400$, so the square-root form is safe.

$[\mathrm{H^+}] = \sqrt{K_{a_1}c} = \sqrt{4.5 \times 10^{-7} \times 0.05} = \sqrt{2.25 \times 10^{-8}} = 1.5 \times 10^{-4}\ \mathrm{M}$

$\mathrm{pH} = -\log(1.5 \times 10^{-4}) = 4 - 0.176 = 3.82$

The first step produces $\mathrm{H^+}$ and $\mathrm{HCO_3^-}$ in equal amounts, so $[\mathrm{HCO_3^-}] = 1.5 \times 10^{-4}\ \mathrm{M}$.

For the carbonate ion I use the second equilibrium:

$K_{a_2} = \dfrac{[\mathrm{H^+}][\mathrm{CO_3^{2-}}]}{[\mathrm{HCO_3^-}]}$

Since $[\mathrm{H^+}] = [\mathrm{HCO_3^-}]$, they cancel and $[\mathrm{CO_3^{2-}}] = K_{a_2} = 4.7 \times 10^{-11}\ \mathrm{M}$.

Ans: $\mathrm{pH} = 3.82$; $[\mathrm{HCO_3^-}] = 1.5 \times 10^{-4}\ \mathrm{M}$; $[\mathrm{CO_3^{2-}}] = 4.7 \times 10^{-11}\ \mathrm{M}$ Watch out: Adding the two steps and using $2[\mathrm{H^+}]$, or using $K_{a_1}K_{a_2}$ for the pH, gives $[\mathrm{H^+}] = \sqrt{2.1 \times 10^{-17} \times 0.05} = \sqrt{1.05 \times 10^{-18}} = 1.0 \times 10^{-9}\ \mathrm{M}$ and a pH of $8.99$ — a basic pH for an acid solution, which should immediately look wrong.

Question 6: Sulphide ion concentration

Calculate $[\mathrm{H^+}]$, $[\mathrm{HS^-}]$ and $[\mathrm{S^{2-}}]$ in a $0.1\ \mathrm{M}$ solution of $\mathrm{H_2S}$, given $K_{a_1} = 9.1 \times 10^{-8}$ and $K_{a_2} = 1.2 \times 10^{-13}$.

Answer:

$[\mathrm{H^+}] = \sqrt{K_{a_1}c} = \sqrt{9.1 \times 10^{-8} \times 0.1} = \sqrt{9.1 \times 10^{-9}} = 9.54 \times 10^{-5}\ \mathrm{M}$

$[\mathrm{HS^-}] = [\mathrm{H^+}] = 9.54 \times 10^{-5}\ \mathrm{M}$

$[\mathrm{S^{2-}}] = K_{a_2} = 1.2 \times 10^{-13}\ \mathrm{M}$, since $[\mathrm{H^+}]$ and $[\mathrm{HS^-}]$ are equal and cancel in the second-step expression.

$\mathrm{pH} = -\log(9.54 \times 10^{-5}) = 4.02$

Ans: $[\mathrm{H^+}] = [\mathrm{HS^-}] = 9.5 \times 10^{-5}\ \mathrm{M}$; $[\mathrm{S^{2-}}] = 1.2 \times 10^{-13}\ \mathrm{M}$; $\mathrm{pH} = 4.02$ Watch out: $[\mathrm{S^{2-}}]$ does not depend on the concentration of the acid at all. It would still be $1.2 \times 10^{-13}\ \mathrm{M}$ in a $0.01\ \mathrm{M}$ solution. What does change $[\mathrm{S^{2-}}]$ is the pH, and that is how sulphide precipitations are controlled.

Question 7: pH of dilute sulphuric acid

Calculate the pH of a $0.020\ \mathrm{M}$ solution of $\mathrm{H_2SO_4}$, taking the first ionisation as complete and $K_{a_2} = 1.2 \times 10^{-2}$.

Answer:

The first step goes to completion, giving $0.020\ \mathrm{M}$ $\mathrm{H^+}$ and $0.020\ \mathrm{M}$ $\mathrm{HSO_4^-}$.

The second step is an equilibrium, and it starts in a solution that already contains $0.020\ \mathrm{M}$ $\mathrm{H^+}$. Let $x$ be the amount of $\mathrm{HSO_4^-}$ that ionises.

$\mathrm{HSO_4^-} \rightleftharpoons \mathrm{H^+} + \mathrm{SO_4^{2-}}$, with equilibrium concentrations $0.020-x$, $0.020+x$ and $x$.

$\dfrac{(0.020+x)x}{0.020-x} = 1.2 \times 10^{-2}$

$K_{a_2}$ is not small compared with the concentration, so no approximation is available. Expanding:

$x^2 + 0.020x = 2.4 \times 10^{-4} - 1.2 \times 10^{-2}x$

$x^2 + 0.032x - 2.4 \times 10^{-4} = 0$

$x = \dfrac{-0.032 + \sqrt{1.024 \times 10^{-3} + 9.6 \times 10^{-4}}}{2} = \dfrac{-0.032 + 0.04454}{2} = 6.27 \times 10^{-3}$

$[\mathrm{H^+}] = 0.020 + 6.27 \times 10^{-3} = 0.0263\ \mathrm{M}$

$\mathrm{pH} = -\log(0.0263) = 1.58$

Ans: $\mathrm{pH} = 1.58$; $[\mathrm{SO_4^{2-}}] = 6.3 \times 10^{-3}\ \mathrm{M}$; $[\mathrm{HSO_4^-}] = 1.4 \times 10^{-2}\ \mathrm{M}$ Watch out: Treating both protons as fully released gives $[\mathrm{H^+}] = 0.040\ \mathrm{M}$ and $\mathrm{pH} = 1.40$. Ignoring the second step gives $0.020\ \mathrm{M}$ and $\mathrm{pH} = 1.70$. The true value sits between them, and only the quadratic finds it.

Question 8: Phosphoric acid, all three steps

For $0.1\ \mathrm{M}$ $\mathrm{H_3PO_4}$ with $K_{a_1} = 7.5 \times 10^{-3}$, $K_{a_2} = 6.2 \times 10^{-8}$ and $K_{a_3} = 4.2 \times 10^{-13}$, calculate the pH and the concentrations of $\mathrm{H_2PO_4^-}$, $\mathrm{HPO_4^{2-}}$ and $\mathrm{PO_4^{3-}}$.

Answer:

Only the first step matters for the pH, but $K_{a_1}$ is fairly large, so I check the approximation: $c/K_{a_1} = 0.1/(7.5 \times 10^{-3}) = 13$, well below $400$. The quadratic is needed.

$\dfrac{x^2}{0.1-x} = 7.5 \times 10^{-3}$, so $x^2 + 7.5 \times 10^{-3}x - 7.5 \times 10^{-4} = 0$

$x = \dfrac{-7.5 \times 10^{-3} + \sqrt{5.625 \times 10^{-5} + 3.0 \times 10^{-3}}}{2} = \dfrac{-0.0075 + 0.05528}{2} = 2.39 \times 10^{-2}$

$[\mathrm{H^+}] = [\mathrm{H_2PO_4^-}] = 2.39 \times 10^{-2}\ \mathrm{M}$, and $\mathrm{pH} = -\log(2.39 \times 10^{-2}) = 1.62$

$[\mathrm{HPO_4^{2-}}] = K_{a_2} = 6.2 \times 10^{-8}\ \mathrm{M}$, since $[\mathrm{H^+}]$ and $[\mathrm{H_2PO_4^-}]$ cancel.

For the third step:

$[\mathrm{PO_4^{3-}}] = \dfrac{K_{a_3}[\mathrm{HPO_4^{2-}}]}{[\mathrm{H^+}]} = \dfrac{4.2 \times 10^{-13} \times 6.2 \times 10^{-8}}{2.39 \times 10^{-2}} = 1.1 \times 10^{-18}\ \mathrm{M}$

Ans: $\mathrm{pH} = 1.62$; $[\mathrm{H_2PO_4^-}] = 2.4 \times 10^{-2}\ \mathrm{M}$; $[\mathrm{HPO_4^{2-}}] = 6.2 \times 10^{-8}\ \mathrm{M}$; $[\mathrm{PO_4^{3-}}] = 1.1 \times 10^{-18}\ \mathrm{M}$ Watch out: The square-root shortcut gives $[\mathrm{H^+}] = \sqrt{7.5 \times 10^{-4}} = 2.74 \times 10^{-2}\ \mathrm{M}$ and $\mathrm{pH} = 1.56$, too low by $0.06$. Whenever $K_{a_1}$ is of the order of $10^{-2}$ or $10^{-3}$, check $c/K_a$ before using the shortcut.

Question 9: Ordering hydrides by acid strength

Arrange each set in increasing order of acid strength in water and name the factor that decides each set: (a) $\mathrm{H_2O}$, $\mathrm{H_2S}$, $\mathrm{H_2Se}$, $\mathrm{H_2Te}$; (b) $\mathrm{CH_4}$, $\mathrm{NH_3}$, $\mathrm{H_2O}$, $\mathrm{HF}$; (c) $\mathrm{HF}$, $\mathrm{HCl}$, $\mathrm{HBr}$, $\mathrm{HI}$.

Answer:

Set (a) runs down group 16. The size of the central atom increases, the $\mathrm{H-A}$ bond gets longer and weaker, and the acid strength rises.

$\mathrm{H_2O} < \mathrm{H_2S} < \mathrm{H_2Se} < \mathrm{H_2Te}$, decided by $\mathrm{H-A}$ bond strength.

Set (b) runs across period 2. Sizes are comparable, so bond strengths are comparable; electronegativity rises from carbon to fluorine, the bond becomes more polar and the conjugate base more stable.

$\mathrm{CH_4} < \mathrm{NH_3} < \mathrm{H_2O} < \mathrm{HF}$, decided by the electronegativity of $\mathrm{A}$.

Set (c) runs down group 17, so bond strength decides again.

$\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$, decided by $\mathrm{H-X}$ bond strength.

Ans: (a) $\mathrm{H_2O} < \mathrm{H_2S} < \mathrm{H_2Se} < \mathrm{H_2Te}$; (b) $\mathrm{CH_4} < \mathrm{NH_3} < \mathrm{H_2O} < \mathrm{HF}$; (c) $\mathrm{HF} < \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$ Watch out: Water appears in (a) as the weakest and in (b) as the second strongest. Both are right, because the comparison groups are different. Using electronegativity on set (c) reverses the order and puts $\mathrm{HF}$ first, which is the classic error.

Question 10: A conjugate pair from a measured pH

A $0.02\ \mathrm{M}$ solution of a weak monobasic acid $\mathrm{HA}$ has $\mathrm{pH} = 3.65$. Find $K_a$ of $\mathrm{HA}$ and $K_b$ of $\mathrm{A^-}$.

Answer:

$[\mathrm{H^+}] = 10^{-3.65} = 10^{0.35} \times 10^{-4} = 2.24 \times 10^{-4}\ \mathrm{M}$

The acid gives $\mathrm{H^+}$ and $\mathrm{A^-}$ together, so $[\mathrm{A^-}] = 2.24 \times 10^{-4}\ \mathrm{M}$. That is about one per cent of $0.02$, so $[\mathrm{HA}] \approx 0.02\ \mathrm{M}$.

$K_a = \dfrac{(2.24 \times 10^{-4})^2}{0.02} = \dfrac{5.02 \times 10^{-8}}{0.02} = 2.51 \times 10^{-6}$

$K_b(\mathrm{A^-}) = \dfrac{1.0 \times 10^{-14}}{2.51 \times 10^{-6}} = 3.98 \times 10^{-9}$

A check with p-values: $\mathrm{p}K_a = 5.60$ and $\mathrm{p}K_b = 14.00 - 5.60 = 8.40$, and $10^{-8.40} = 4.0 \times 10^{-9}$.

Ans: $K_a = 2.5 \times 10^{-6}$; $K_b(\mathrm{A^-}) = 4.0 \times 10^{-9}$ Watch out: Forgetting to square $[\mathrm{H^+}]$ gives $K_a = 2.24 \times 10^{-4}/0.02 = 1.1 \times 10^{-2}$, which would make $\mathrm{HA}$ a moderately strong acid inconsistent with a pH of $3.65$ at this concentration.

What to Carry Forward

Result	Statement	Condition
Conjugate relation	$K_a \times K_b = K_w$	conjugate pair only
p-form	$\mathrm{p}K_a + \mathrm{p}K_b = \mathrm{p}K_w$	any temperature
Numerical p-form	$\mathrm{p}K_a + \mathrm{p}K_b = 14.00$	$298\ \mathrm{K}$ only
Reciprocal strength	large $K_a$ forces small $K_b$	conjugate pair
Stepwise order	$K_{a_1} > K_{a_2} > K_{a_3}$	all polyprotic acids
pH of a polyprotic acid	$[\mathrm{H^+}] \approx \sqrt{K_{a_1}c}$	$K_{a_2} \lll K_{a_1}$, and $c/K_{a_1} > 400$
Doubly charged anion	$[\mathrm{X^{2-}}] \approx K_{a_2}$	diprotic acid alone in water
Down a group	bond strength decides; acidity rises downward	$\mathrm{HF} \lll \mathrm{HCl} < \mathrm{HBr} < \mathrm{HI}$
Across a period	electronegativity decides; acidity rises rightward	$\mathrm{CH_4} < \mathrm{NH_3} < \mathrm{H_2O} < \mathrm{HF}$

Six errors account for most of the lost marks here.

Multiplying $K_a$ and $K_b$ of species that are not a conjugate pair, and calling the product $K_w$.

Using $14$ where $\mathrm{p}K_w$ belongs, when the question has quietly changed the temperature.

Assigning $K_a$ to the base or $K_b$ to the acid within a correct pair, which shifts the answer by the full factor $K_a/K_b$.

Adding the ionisation steps of a polyprotic acid and using $K_{a_1}K_{a_2}$ to find the pH, which produces a basic pH for an acid.

Treating $\mathrm{H_2SO_4}$ as either fully strong in both steps or strong in the first step only, when it is strong in the first and weak in the second.

Using electronegativity to order the hydrogen halides, which puts $\mathrm{HF}$ at the strong end instead of the weak one.

The next section takes these constants into solutions that contain more than the acid alone: the common ion effect, the hydrolysis of salts, and buffers.