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

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:

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

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

Therefore, multiplying the two equilibrium constants:

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

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).

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:

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 $c$):

1. Identify the hydrolysing anion ($CH_3COO^-$).
2. Look up $K_a$ of the parent acid, compute $K_b = K_w / K_a$.
3. Apply the weak-base ICE approximation: $[OH^-] \approx \sqrt{K_b \cdot c}$.
4. 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.

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.