Conjugate acid-base pairs (Ka, Kb, Kw relationship) explained: HSC Chemistry Module 6

Syllabus dot point

“Investigate quantitatively the relationship between the strength of conjugate acid-base pairs, including the relationship Ka times Kb equals Kw”

9 min readUpdated 22 Jun 2026

Quick answer

For any conjugate acid-base pair $K_a \cdot K_b = K_w = 10^{-14}$ at 25 degrees C, so the weaker the acid the stronger its conjugate base, and to predict the pH of a salt solution identify the conjugate origin of each ion (spectator if from a strong parent, hydrolysing if from a weak parent) and apply the appropriate $K_a$ or $K_b$ ICE shortcut.

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In plain English

Think of an acid and its conjugate base as two ends of a seesaw balanced on a fixed pivot, where the pivot height is set by water itself (that is what $K_w$ represents). If one end sits very high (a strong acid, a huge $K_a$ , meaning it gives away its proton very easily), the other end MUST sit very low (a tiny $K_b$ , meaning the leftover species is a hopeless proton-grabber). If the acid is only middling (a weak acid, a small $K_a$ ), its conjugate base sits middling-high too, meaning it can grab a proton back from water often enough to nudge the pH. The seesaw rule is fixed: $K_a$ times $K_b$ always multiplies out to the same number, $K_w$ , at a given temperature. So once you know how "strong" one side of a conjugate pair is, the other side is already decided, you never need to measure it separately, just divide $K_w$ by the one you know. This is exactly why dissolving a salt in water can shift the pH away from 7: whichever ion in the salt is the "weak" end of its own conjugate pair will hydrolyse water and nudge the seesaw.

What this dot point is asking

NESA wants you to relate the strength of an acid to the strength of its conjugate base (and vice versa), use the identity $K_a \cdot K_b = K_w$ for any conjugate pair, predict whether a given salt solution will be acidic, basic, or neutral by identifying the conjugate origins of its ions, and calculate the pH of salt solutions when asked. This builds on Bronsted-Lowry theory and strong vs weak acid concepts.

The answer

The inverse relationship

When an acid ionises, its conjugate base forms:

HA_{(aq)} \rightleftharpoons H^+_{(aq)} + A^-_{(aq)}

The conjugate base $A^-$ can re-accept a proton from water:

A^-_{(aq)} + H_2O_{(l)} \rightleftharpoons HA_{(aq)} + OH^-_{(aq)}

These two equilibria are linked. Adding them gives the auto-ionisation of water:

2H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)}

Therefore, multiplying the two equilibrium constants:

K_a \cdot K_b = K_w = 1.0 \times 10^{-14} \text{ at 25 degrees C}

This is the key identity for the dot point. Rearranged:

pK_a + pK_b = 14

What the identity tells you

A strong acid has a very large $K_a$ , so its conjugate base has a vanishingly small $K_b$ . The conjugate base of a strong acid is essentially a non-base (does not hydrolyse).
A weak acid has a small $K_a$ , so its conjugate base has a meaningful $K_b$ . The conjugate base of a weak acid is a measurable weak base.
The weaker the acid, the stronger its conjugate base (and the more it hydrolyses water).

Acid	$K_a$	Conjugate base	$K_b$
$HCl$	very large	$Cl^-$	negligible
$HF$	$6.8 \times 10^{-4}$	$F^-$	$1.5 \times 10^{-11}$
$CH_3COOH$	$1.8 \times 10^{-5}$	$CH_3COO^-$	$5.6 \times 10^{-10}$
$NH_4^+$	$5.6 \times 10^{-10}$	$NH_3$	$1.8 \times 10^{-5}$
$HCO_3^-$	$4.7 \times 10^{-11}$	$CO_3^{2-}$	$2.1 \times 10^{-4}$

The conjugate-pair identity

$K_a \cdot K_b = K_w$ for any conjugate acid-base pair, at the SAME temperature.
At 25 degrees C, $K_w = 1.0 \times 10^{-14}$ , so $pK_a + pK_b = 14$ .
The weaker the acid, the stronger its conjugate base, and vice versa; strong-acid conjugate bases are effectively non-basic (negligible hydrolysis).
Only ions from a WEAK parent acid or base hydrolyse water and shift a salt solution's pH away from 7; ions from a strong parent are spectators.

Salt hydrolysis: predicting pH

A salt is named by its cation and anion. Each ion comes from an acid or a base.

Cation from a strong base (Na+, K+, $Ca^{2+}$ , $Ba^{2+}$ ): spectator, no hydrolysis.
Cation from a weak base ( $NH_4^+$ , $Al^{3+}$ , transition metal cations like $Fe^{3+}$ ): acidic, hydrolyses to release $H^+$ .
Anion from a strong acid ( $Cl^-$ , $NO_3^-$ , $ClO_4^-$ , $Br^-$ , $I^-$ ): spectator, no hydrolysis.
Anion from a weak acid ( $CH_3COO^-$ , $F^-$ , $CO_3^{2-}$ , $HCO_3^-$ , $CN^-$ ): basic, hydrolyses to release $OH^-$ .

Combine the two ions to predict the pH:

Cation	Anion	Salt pH
Strong base cation	Strong acid anion	Neutral (pH = 7)
Strong base cation	Weak acid anion	Basic (pH > 7)
Weak base cation	Strong acid anion	Acidic (pH < 7)
Weak base cation	Weak acid anion	Depends on $K_a$ vs $K_b$

For the last case, compare $K_a$ of the cation to $K_b$ of the anion. If $K_a > K_b$ the solution is acidic; if $K_b > K_a$ it is basic; if equal, near neutral. For ammonium ethanoate, $K_a(NH_4^+) = 5.6 \times 10^{-10}$ and $K_b(CH_3COO^-) = 5.6 \times 10^{-10}$ , so the solution is approximately neutral.

Calculating the pH of a salt solution

For a salt of a strong base and a weak acid (say, sodium ethanoate at concentration $c$ ):

Identify the hydrolysing anion ( $CH_3COO^-$ ).
Look up $K_a$ of the parent acid, compute $K_b = K_w / K_a$ .
Apply the weak-base ICE approximation: $[OH^-] \approx \sqrt{K_b \cdot c}$ .
Convert to pH: $pOH = -\log_{10}[OH^-]$ , then $pH = 14 - pOH$ .

For a salt of a weak base and a strong acid, the symmetric calculation gives $[H^+] \approx \sqrt{K_a \cdot c}$ where $K_a$ refers to the conjugate acid cation.

A titration curve makes the conjugate-pair consequence visible: at the equivalence point, the acid has been converted entirely into its conjugate base (as a salt), so the pH at equivalence is decided by whether that conjugate base hydrolyses water, not simply set to 7.

Reading a titration curve for conjugate-pair reasoning

The equivalence point is the STEEPEST part of the curve (the biggest pH jump per mL added), not necessarily pH 7. Read the pH AT that point to decide which conjugate species controls the solution: if it sits above 7, the analyte was a weak acid whose conjugate base now hydrolyses; if it sits below 7, the analyte was a weak base whose conjugate acid now hydrolyses; only a strong-acid/strong-base titration gives an equivalence point at exactly pH 7. Always state the equivalence VOLUME and the equivalence PH separately when a question asks you to "describe" a curve.

Worked example

Calculate the pH of a 0.20 mol/L solution of ammonium chloride at 25 degrees C, given that $K_b$ for ammonia is $1.8 \times 10^{-5}$ .

Step 1: Identify the hydrolysing ion. $NH_4^+$ is the conjugate acid of $NH_3$ (weak base). $Cl^-$ is a spectator (conjugate base of strong acid).

Step 2: $K_a$ of $NH_4^+$ .

K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}

Step 3: Hydrolysis.

NH_4^+_{(aq)} + H_2O_{(l)} \rightleftharpoons NH_{3(aq)} + H_3O^+_{(aq)}

Step 4: $[H^+]$ .

[H^+] \approx \sqrt{K_a \cdot c} = \sqrt{(5.56 \times 10^{-10})(0.20)} = 1.05 \times 10^{-5} \text{ mol/L}

Step 5: pH.

pH = -\log_{10}(1.05 \times 10^{-5}) = 4.98

The solution is slightly acidic, as expected for a salt of a weak base and a strong acid.

Common traps

Treating $Cl^-$ as a weak base: $Cl^-$ is the conjugate of a strong acid. Its $K_b$ is too small to influence pH. Do not write a hydrolysis equation for it.
Forgetting that $Na^+$ and $K^+$ are spectators: They never affect pH at HSC.
Mixing up $K_a$ and $K_b$ in the identity: $K_a$ refers to the acid; $K_b$ refers to its conjugate base. Use $K_a \cdot K_b = K_w$ on the same pair only.

Using $K_b$ of $NH_3$ when you mean $K_a$ of $NH_4^+$ . They are different sides of the same conjugate pair. When you have a salt like $NH_4Cl$ , the ion in solution is $NH_4^+$ , so you need its $K_a$ . Convert with $K_w / K_b$ .

Ignoring the second hydrolysis of $CO_3^{2-}$ . Carbonate is the conjugate base of $HCO_3^-$ (not $H_2CO_3$ directly). Use $K_a$ of $HCO_3^-$ ( $K_{a2}$ of carbonic acid, about $4.7 \times 10^{-11}$ ). A common slip is to use $K_{a1}$ instead.

Forgetting to take square root. The weak-acid or weak-base ICE shortcut gives $x = \sqrt{K \cdot c}$ , not $K \cdot c$ .

Examples in context

Example 1. Ammonium chloride as a fertiliser additive on NSW canola farms. Incitec Pivot supplies ammonium chloride as a nitrogen source to canola growers across the central west of NSW. Once dissolved in soil moisture the ammonium ion hydrolyses: $NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$ . Because ammonia is a weak base with $K_b = 1.8 \times 10^{-5}$ , its conjugate acid $NH_4^+$ has $K_a = K_w / K_b = 5.6 \times 10^{-10}$ , giving soil solutions around pH 5.5 from a 0.1 mol L $^{-1}$ application. Repeated heavy use acidifies paddocks over decades, an agronomic problem that NSW DPI advises growers to monitor and correct with lime. The HSC $K_a \cdot K_b = K_w$ identity quantifies the soil-pH consequence directly.

Example 2. Sodium fluoride dosing at Sydney Water Prospect. Sydney Water adds sodium fluoride to drinking water at 1.0 ppm to reduce dental caries. The fluoride ion is the conjugate base of HF ( $K_a = 6.8 \times 10^{-4}$ ), so $K_b(F^-) = K_w / K_a = 1.5 \times 10^{-11}$ , a very weak base. A 0.5 mmol L $^{-1}$ dose produces a pH shift of less than 0.01 units, well within the operating envelope of the distribution network. If fluoride were the conjugate base of a much weaker acid, hydrolysis would push the pH up significantly and operators would need to compensate with acid dosing. The HSC identity tells the dosing engineer the size of the effect before the chemistry is run in the plant.

Try this

Q1. State the relationship between $K_a$ , $K_b$ and $K_w$ for a conjugate acid-base pair, and explain in words what the relationship implies. [3 marks]

Cue. $K_a \times K_b = K_w$ ; the weaker the acid, the stronger its conjugate base, and vice versa.

Q2. The $K_a$ of formic acid (HCOOH) is $1.8 \times 10^{-4}$ . Calculate $K_b$ for the formate ion ( $HCOO^-$ ) and state the pH of a 0.10 mol L $^{-1}$ sodium formate solution. [3 marks]

Cue. $K_b = 10^{-14} / 1.8 \times 10^{-4} = 5.6 \times 10^{-11}$ ; ICE gives $[OH^-] = \sqrt{K_b \times c} = 2.4 \times 10^{-6}$ , $pOH = 5.6$ , $pH = 8.4$ .

Q3. Predict whether each of the following salt solutions is acidic, basic or neutral and justify with the conjugate-pair analysis: (a) $NaCl$ , (b) $NH_4NO_3$ , (c) $Na_2CO_3$ . [2+2+2 marks]

Cue. (a) Neutral: both ions from strong parents. (b) Acidic: $NH_4^+$ hydrolyses, $NO_3^-$ spectator. (c) Basic: $CO_3^{2-}$ hydrolyses, $Na^+$ spectator.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2022 HSC-style4 marks

Ethanoic acid (CH₃COOH) has a Ka of 1.8 × 10⁻⁵ at 25°C. Calculate the Kb of the ethanoate ion (CH₃COO⁻) at 25°C and use the value to predict whether a 0.10 mol/L solution of sodium ethanoate will be acidic, neutral or basic. Justify your answer.

Show worked answer →

A 4 mark answer needs the Kb calculation, the hydrolysis equation, the prediction, and a quantitative justification.

Step 1: $K_b$ of ethanoate.

K_a \cdot K_b = K_w

K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}

Step 2: Hydrolysis equation.

CH_3COO^-_{(aq)} + H_2O_{(l)} \rightleftharpoons CH_3COOH_{(aq)} + OH^-_{(aq)}

Step 3: $[OH^-]$ from $K_b$ .

[OH^-] \approx \sqrt{K_b \cdot c} = \sqrt{(5.56 \times 10^{-10})(0.10)} = 7.45 \times 10^{-6} \text{ mol/L}

pOH = -\log_{10}(7.45 \times 10^{-6}) = 5.13

pH = 14 - 5.13 = 8.87

Prediction. pH > 7, so the solution is slightly basic. The ethanoate ion is the conjugate base of a weak acid and hydrolyses water to release $OH^-$ . The $Na^+$ spectator does not affect pH.

Markers reward (1) correct use of $K_a \cdot K_b = K_w$ , (2) the hydrolysis equation, (3) numerical calculation of pH, (4) the explicit prediction with justification.

2018 HSC-style3 marks

State, with reasons, whether each of the following salts will give an acidic, basic or neutral aqueous solution: (a) NH₄Cl, (b) KNO₃, (c) Na₂CO₃.

Show worked answer →

A salt is the product of a neutralisation. The aqueous pH depends on whether the cation and anion are conjugates of strong or weak acids/bases.

(a) $NH_4Cl$ . Acidic: $NH_4^+$ is the conjugate acid of the weak base $NH_3$ , so it hydrolyses: $NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$ , releasing $H^+$ . $Cl^-$ is the conjugate base of a strong acid ( $HCl$ ) and is too weak a base to affect pH. Overall acidic.
(b) $KNO_3$ . Neutral: $K^+$ is the conjugate acid of a strong base ( $KOH$ ); $NO_3^-$ is the conjugate base of a strong acid ( $HNO_3$ ). Neither hydrolyses. pH = 7.
(c) $Na_2CO_3$ . Basic: $Na^+$ is a spectator. $CO_3^{2-}$ is the conjugate base of the weak acid $HCO_3^-$ , so it hydrolyses: $CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-$ , releasing $OH^-$ . Overall basic.

Markers reward (1) identifying conjugate origins of each ion, (2) writing a hydrolysis equation where relevant, (3) the correct acidic/basic/neutral conclusion.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

Identify the conjugate base of

HSO_4^-

and the conjugate acid of

HSO_4^-

Show worked solution →

A 2-mark identify needs both species named correctly.

Conjugate base (loses a proton): $HSO_4^- \rightarrow SO_4^{2-} + H^+$ , so the conjugate base is $SO_4^{2-}$ .

Conjugate acid (gains a proton): $HSO_4^- + H^+ \rightarrow H_2SO_4$ , so the conjugate acid is $H_2SO_4$ .

Marking criteria: 1 mark for the correct conjugate base, 1 mark for the correct conjugate acid. $HSO_4^-$ is amphiprotic, which is why it has both.

foundation3 marks

The

K_a

of hydrofluoric acid (HF) is

6.8 \times 10^{-4}

at 25 degrees C. Calculate

K_b

for the fluoride ion,

F^-

, to 2 significant figures.

Show worked solution →

Step 1: identity.

K_a \cdot K_b = K_w

Step 2: rearrange and substitute.

K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.8 \times 10^{-4}} = 1.4706 \times 10^{-11}

Step 3: round to 2 significant figures (matching the 2 s.f. of the given $K_a$ ).

K_b(F^-) = 1.5 \times 10^{-11}

Marking criteria: 1 mark for stating/using $K_a \cdot K_b = K_w$ , 1 mark for the correct substitution, 1 mark for the correctly rounded answer with no units ( $K_b$ is dimensionless on the standard state convention used at HSC level).

core5 marks

A 0.15 mol/L solution of sodium benzoate (

C_6H_5COONa

) is prepared, where benzoic acid (

C_6H_5COOH

) has

K_a = 6.3 \times 10^{-5}

at 25 degrees C. Calculate the pH of the solution to 2 decimal places, showing all working.

Show worked solution →

Step 1: identify the hydrolysing ion. $Na^+$ is a spectator (conjugate acid of the strong base NaOH). The benzoate ion, $C_6H_5COO^-$ , is the conjugate base of the weak acid benzoic acid, so it hydrolyses.

Step 2: calculate $K_b$ of benzoate.

K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{6.3 \times 10^{-5}} = 1.587 \times 10^{-10}

Step 3: hydrolysis equation.

C_6H_5COO^-_{(aq)} + H_2O_{(l)} \rightleftharpoons C_6H_5COOH_{(aq)} + OH^-_{(aq)}

Step 4: ICE approximation for $[OH^-]$ .

[OH^-] \approx \sqrt{K_b \cdot c} = \sqrt{(1.587 \times 10^{-10})(0.15)} = \sqrt{2.381 \times 10^{-11}} = 4.879 \times 10^{-6}\ \text{mol/L}

Step 5: pOH, then pH.

pOH = -\log_{10}(4.879 \times 10^{-6}) = 5.312

pH = 14.00 - 5.312 = 8.69

Step 6: round to 2 decimal places (as requested).

pH = 8.69

Marking criteria: 1 mark for identifying benzoate as the hydrolysing species and $Na^+$ as a spectator, 1 mark for correct $K_b$ , 1 mark for the hydrolysis equation, 1 mark for the $[OH^-]$ calculation, 1 mark for the correct final pH to 2 decimal places. The solution is basic, consistent with a salt of a strong base and a weak acid.

core4 marks

The titration curve below shows pH against volume of 0.10 mol/L NaOH added to 25.0 mL of a weak monoprotic acid HA. Using the curve, (a) estimate the equivalence point volume, (b) state whether the solution AT the equivalence point is acidic, basic or neutral and justify using conjugate-pair reasoning, and (c) name a suitable indicator.

Show worked solution →

(a) Equivalence point volume. The steepest part of the curve (the inflection, where pH jumps most sharply per mL added) is centred at 25.0 mL, read from the midpoint of the vertical jump on the graph.

(b) pH at equivalence. At equivalence, all of the HA has been converted to its conjugate base $A^-$ (as the sodium salt $NaA$ ) dissolved in water; there is no excess HA or NaOH left. $Na^+$ is a spectator, but $A^-$ is the conjugate base of a WEAK acid, so it hydrolyses:

A^-_{(aq)} + H_2O_{(l)} \rightleftharpoons HA_{(aq)} + OH^-_{(aq)}

This hydrolysis releases $OH^-$ , so the solution at equivalence is basic (pH greater than 7), consistent with the curve showing the equivalence point above pH 7, not at exactly pH 7.

(c) Indicator. Phenolphthalein (colour change range approximately pH 8.2 to 10.0) is suitable because its colour-change range brackets the basic equivalence-point pH; methyl orange (range approximately pH 3.1 to 4.4) would change colour too early and give an inaccurate endpoint.

Marking criteria: 1 mark for reading the equivalence volume correctly from the steepest part of the curve, 1 mark for correctly identifying "basic" at equivalence, 1 mark for the conjugate-base hydrolysis reasoning (not just an assertion), 1 mark for a correctly justified indicator choice matching the basic equivalence point.

exam6 marks

Assess the claim that 'a salt formed from a strong acid and a strong base is always neutral, so testing a salt's pH is unnecessary once you know its formula.' Use conjugate-pair reasoning and at least one calculation in your answer.

Show worked solution →

This is a 6-mark ASSESS: markers reward a judgement supported by conjugate-pair reasoning and a worked calculation, not just a list of examples.

Band 6 PLAN.

Thesis: the claim is true only for the strong-acid/strong-base case itself, but is dangerously overgeneralised as written; salts of a weak acid and a weak base, or of a strong parent with a weak partner, are NOT reliably neutral, so pH cannot be assumed from formula alone without conjugate-pair analysis.
Confirm the true part: for a salt like $NaCl$ (strong base $NaOH$ + strong acid $HCl$ ), both $Na^+$ and $Cl^-$ are spectators (conjugate acid of a strong base; conjugate base of a strong acid), so hydrolysis is negligible and pH = 7. This part of the claim is correct.
Show the claim breaks down: for a salt of a weak acid and a weak base, e.g. ammonium ethanoate $NH_4CH_3COO$ , BOTH ions hydrolyse ( $NH_4^+$ releases $H^+$ ; $CH_3COO^-$ releases $OH^-$ ), and the resulting pH depends on comparing $K_a(NH_4^+)$ to $K_b(CH_3COO^-)$ , not on a simple rule.
Worked calculation: $K_a(NH_4^+) = K_w / K_b(NH_3) = 1.0\times10^{-14} / 1.8\times10^{-5} = 5.6\times10^{-10}$ . Compare to $K_b(CH_3COO^-) = K_w/K_a(CH_3COOH) = 1.0\times10^{-14}/1.8\times10^{-5} = 5.6\times10^{-10}$ . Since $K_a \approx K_b$ , the solution is approximately neutral in this particular case, but this is a coincidence of the two parent acids having equal strength, not a general rule.
Judgement: the claim is an unsafe generalisation; only the strong/strong combination can be assumed neutral without further work. Any salt involving at least one weak parent requires explicit conjugate-pair and $K_a$ / $K_b$ comparison, so "testing is unnecessary" is false in general.

Model paragraph (excerpt). The claim holds only for the specific case it describes: a salt of a strong acid and a strong base, such as sodium chloride, is reliably neutral because neither ion is capable of hydrolysing water, both being conjugates of strong parents. However, extending this to "testing a salt's pH is unnecessary" overreaches, because salts with at least one weak parent, such as ammonium ethanoate, require a direct comparison of $K_a$ and $K_b$ values to predict pH; in this case $K_a(NH_4^+) = 5.6\times10^{-10}$ and $K_b(CH_3COO^-) = 5.6\times10^{-10}$ happen to be equal, giving an approximately neutral solution, but a different pairing of weak acid and weak base would not give the same result. The general rule, therefore, is that pH prediction always requires identifying the conjugate origin of each ion, and the claim as stated is only safely true for the strong/strong case.

Marker's note: top-band answers (1) correctly confirm the true strong/strong case, (2) correctly identify at least one counter-case with both ions hydrolysing, (3) include an explicit $K_a$ / $K_b$ numeric comparison rather than a qualitative claim only, and (4) end with an explicit judgement on the scope of the original claim, not a restatement of the facts.

Practise this

Sources & how we know this

Chemistry Stage 6 Syllabus (2017) — NESA
Chemistry HSC exam papers — NESA