NSWChemistrySyllabus dot point

Inquiry Question 3: It is all about hydrogen ions

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

A focused answer to the HSC Chemistry Module 6 dot point on conjugate acid-base pair strength. The inverse relationship between conjugate strengths, the Ka times Kb equals Kw identity, salt hydrolysis predictions, and worked HSC past exam questions.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-18

Have a quick question? Jump to the Q&A page

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	IMATH_15	Conjugate base	IMATH_16
IMATH_17	very large	IMATH_18	negligible
IMATH_19	IMATH_20	IMATH_21	IMATH_22
IMATH_23	IMATH_24	IMATH_25	IMATH_26
IMATH_27	IMATH_28	IMATH_29	IMATH_30
IMATH_31	IMATH_32	IMATH_33	IMATH_34

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 IMATH_53

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.

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 $K \cdot c$ .

In one sentence

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.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC4 marksEthanoic 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 HSC3 marksState, 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.