Oxoacids of halogens — HClO, HClO₂, HClO₃, HClO₄ — acidity trend: Worked examples

This page is a workout. The parent note gave you the ideas: more terminal oxygens → more delocalised charge → more stable anion → stronger acid, plus the shortcut $p K_{a} \approx 8 - 5 q$ . Here we grind through every kind of question that trend can generate — so no exam scenario surprises you.

Before we start, let us re-anchor the one symbol everything rests on.

Figure — Oxoacids of halogens — HClO, HClO₂, HClO₃, HClO₄ — acidity trend

Look at the figure: in $C l O_{4}^{-}$ (right) the same single lump of negative charge is shared by four oxygens, so each carries only a quarter. In $C l O^{-}$ (left) one oxygen carries the whole load. Sharing = stability. That is the entire physics — every example below is just this idea in a new costume.

The scenario matrix

Here is the full menu of case-classes this topic can throw at you. Every cell is covered by a worked example (E-number) below.

#	Case class	What makes it tricky	Example
C1	Same central atom, differing O (HClO series)	pure oxygen-count trend	E1
C2	Zero terminal O ( $q = 0$ )	degenerate/limiting: weakest possible	E2
C3	Same O, different halogen (HClO₃ vs HBrO₃)	electronegativity tie-breaker	E3
C4	Cross-element (sulfur/nitrogen oxoacids)	same rule, new atom	E4
C5	Pauling number vs experiment	when the estimate is "off" and why	E5
C6	Acidity vs oxidising power (opposite trends)	sign-flip trap	E6
C7	Polyprotic ( $p > 1$ : $K_{a 1}$ vs $K_{a 2}$ )	which H, and the "+5 per step" rule	E7
C8	Real-world word problem (pool sanitiser)	translate chemistry ↔ context	E8
C9	Exam twist: rank a mixed random set	apply all rules together	E9

Read the matrix, then attempt each example's Forecast before opening the steps.

E1 — Case C1: the pure oxygen-count ladder

Worked example Rank

$H C l O, H C l O_{2}, H C l O_{3}, H C l O_{4}$ by acid strength and give each $p K_{a}$ estimate.

Forecast: Which direction does strength go, and by how much per oxygen? Guess before reading.

Step 1 — Find $q$ for each. Write $(H O)_{1} C l O_{q}$ . Total O minus the one in $O - H$ gives $q$ .

$H C l O$ : 1 O total − 1 (the $O - H$ ) = $q = 0$
$H C l O_{2}$ : 2 − 1 = $q = 1$
$H C l O_{3}$ : 3 − 1 = $q = 2$
$H C l O_{4}$ : 4 − 1 = $q = 3$

Why this step? $q$ is the only input Pauling's rule needs; it counts the charge-sharing seats.

Step 2 — Apply $p K_{a} \approx 8 - 5 q$ .

$q = 0 \Rightarrow 8$
$q = 1 \Rightarrow 3$
$q = 2 \Rightarrow - 2$
$q = 3 \Rightarrow - 7$

Why this step? Each terminal O empirically drops $p K_{a}$ by ~5 (raises $K_{a}$ ~ $1 0^{5}$ ).

Step 3 — Translate to strength. Lower $p K_{a}$ ⇒ larger $K_{a}$ ⇒ stronger acid. $H C l O < H C l O_{2} < H C l O_{3} < H C l O_{4}$

Verify: the estimates $8, 3, - 2, - 7$ march downward by exactly 5 each step, matching the experimental ladder $7.5, 2.0, \sim - 1, \sim - 10$ . Order confirmed. ✔

E2 — Case C2: the degenerate $q = 0$ floor

Worked example What is the

weakest an oxoacid with formula $(H O)_{p} X O_{q}$ can be, and which real acid sits there? Estimate $p K_{a}$ of water using the same rule as a sanity check.

Forecast: If more oxygen = stronger, what happens at the bottom of the ladder, $q = 0$ ?

Step 1 — Set $q = 0$ in Pauling's rule. $p K_{a} \approx 8 - 5 (0) = 8$ Why this step? $q = 0$ means no terminal oxygens to share the charge — the anion's negative charge is stuck on a single oxygen, the least stable case, hence the weakest acid.

Step 2 — Name a $q = 0$ acid. $H C l O$ (i.e. $H - O - C l$ ) has $q = 0$ ; its measured $p K_{a}$ is $7.5$ , right at the floor.

Step 3 — Limiting check with water. Water is $H - O - H$ : no terminal O, $q = 0$ , so the rule predicts $p K_{a} \approx 8$ . Water's actual acid $p K_{a}$ is $\approx 15.7$ — the rule under-shoots because H is far less electron-withdrawing than Cl, but it correctly places water in the "very weak" bin.

Verify: $8 - 5 \cdot 0 = 8$ ; both $H C l O$ (7.5) and water (weak) land near the predicted floor. The rule's floor is $p K_{a} \approx 8$ , and nothing with $q = 0$ is a strong acid. ✔

E3 — Case C3: same oxygen, the electronegativity tie-breaker

Worked example Which is the stronger acid,

$H C l O_{3}$ or $H B r O_{3}$ ? Explain the decider.

Forecast: They have the same number of oxygens. Can Pauling's rule alone separate them? What is the second-order factor?

Step 1 — Compare $q$ . Both are $(H O) X O_{2}$ , so $q = 2$ for each. Pauling: $p K_{a} \approx 8 - 10 = - 2$ for both. The rule declares a tie.

Why this step? Establishes that the dominant factor (oxygen count) is neutral here, so we must go to the next factor.

Step 2 — Bring in electronegativity (Inductive effect). The central atom pulls electron density from the $O - H$ oxygen and helps stabilise the anion. Cl (electronegativity $\approx 3.16$ ) pulls harder than Br ( $\approx 2.96$ ).

Why this step? When oxygen count ties, the more electronegative central atom withdraws charge more strongly → more stable anion → stronger acid.

Step 3 — Conclude. $H C l O_{3}$ is slightly stronger than $H B r O_{3}$ . $H C l O_{3} > H B r O_{3} (small margin)$

Verify: Experimental values agree — $H C l O_{3}$ is a somewhat stronger acid than $H B r O_{3}$ ; the gap is small precisely because the big lever (oxygen count) is equal. Electronegativity is only the tie-breaker. ✔

E4 — Case C4: cross-element, same rule (sulfur & nitrogen)

Worked example Use the

same Pauling logic to rank $H_{2} S O_{4}$ vs $H_{2} S O_{3}$ , and $H N O_{3}$ vs $H N O_{2}$ . (See Oxoacids of sulfur and nitrogen.)

Forecast: The rule was built on Cl. Will it survive a jump to S and N? Which member of each pair should win?

Step 1 — Count $q$ for the sulfur pair.

$H_{2} S O_{4} = (H O)_{2} S O_{2} \Rightarrow q = 2$
$H_{2} S O_{3} = (H O)_{2} S O_{1} \Rightarrow q = 1$

Why this step? $p = 2$ here (two $O - H$ ), but only $q$ (terminal O) sets strength; the two $O - H$ groups are just the H-count.

Step 2 — Apply the rule (first ionisation, $K_{a 1}$ ).

$H_{2} S O_{4}$ : $8 - 5 (2) = - 2$
$H_{2} S O_{3}$ : $8 - 5 (1) = + 3$ So $H_{2} S O_{4} > H_{2} S O_{3}$ . ✔ (Experiment: $H_{2} S O_{4}$ strong, $H_{2} S O_{3}$ weak, $p K_{a 1} \approx 1.8$ .)

Step 3 — Nitrogen pair.

$H N O_{3} = (H O) N O_{2} \Rightarrow q = 2 \Rightarrow p K_{a} \approx - 2$ (strong)
$H N O_{2} = (H O) N O_{1} \Rightarrow q = 1 \Rightarrow p K_{a} \approx + 3$ (weak; actual $\approx 3.3$ ) So $H N O_{3} > H N O_{2}$ .

Verify: More terminal O wins every time — $H_{2} S O_{4} > H_{2} S O_{3}$ and $H N O_{3} > H N O_{2}$ , exactly as observed. The rule is element-agnostic; only $q$ matters. ✔

E5 — Case C5: estimate vs experiment — where the rule bends

Worked example Pauling predicts

$p K_{a} (H C l O) \approx 8$ but the measured value is $7.5$ . Compute the error, and explain why $H C l O_{4}$ 's estimate ( $- 7$ ) is even further from experiment ( $\sim - 10$ ).

Forecast: Is the rule an equality or an approximation? In which direction do you expect it to drift for the strongest acids?

Step 1 — Error for $H C l O$ . $Δ = ∣ - 8 - (- 7.5) ∣ = ∣8 - 7.5∣ = 0.5$ Small: within half a $p K_{a}$ unit. Good.

Step 2 — Error for $H C l O_{4}$ . $Δ = ∣ - 7 - (- 10) ∣ = 3$ Why the drift? For very strong acids, extra resonance/relativistic and solvent effects add stability the linear rule doesn't capture; the true acid is even stronger than " $8 - 5 q$ " predicts. The rule under-estimates strength at the extreme.

Step 3 — Takeaway. Use $p K_{a} \approx 8 - 5 q$ for ordering and rough magnitude, not exact values. It never mis-orders the series; it only mis-sizes the extremes.

Verify: errors $0.5$ (mild acid, rule is tight) and $3$ (strong acid, rule under-shoots) — both keep the correct rank. ✔

E6 — Case C6: the sign-flip trap — acidity vs oxidising power

Worked example Among

$H C l O$ and $H C l O_{4}$ : which is the stronger acid, and which is the stronger oxidiser? Watch the direction.

Forecast: Your gut may say "the strong acid is the strong everything." Is that true here? (See Oxidising power of oxoacids.)

Step 1 — Acidity direction. From E1: $H C l O_{4} > H C l O$ as an acid (more O → stronger acid).

Step 2 — Oxidising direction. Oxidising power = eagerness to gain electrons = tied to how badly Cl wants to drop to a lower oxidation state. In $H C l O$ , Cl is $+ 1$ (far from its stable state and easy to reduce); in $H C l O_{4}$ , Cl is $+ 7$ locked tightly by four oxygens.

Why this step? Lower oxidation state + fewer stabilising oxygens ⇒ more eager to grab electrons ⇒ stronger oxidiser.

Step 3 — Conclude the opposite pair. $Acid: H C l O_{4} > H C l O Oxidiser: H C l O > H C l O_{4}$

Verify: The two trends run in opposite directions — this is the classic trap. Standard reduction potentials confirm $H C l O$ is the stronger oxidiser. Direction flip confirmed. ✔

E7 — Case C7: polyprotic — $K_{a 1}$ vs $K_{a 2}$

Worked example For

$H_{2} S O_{4}$ ( $p = 2$ ), estimate both $p K_{a 1}$ and $p K_{a 2}$ . Which proton leaves easier and why?

Forecast: After the first H⁺ leaves, is the second one easier or harder to remove?

Step 1 — First ionisation. $H_{2} S O_{4}$ , $q = 2$ : $p K_{a 1} \approx 8 - 5 (2) = - 2$ . Very easy — the neutral molecule is happy to shed the first H⁺.

Step 2 — Second ionisation, the "+5 rule". Removing a proton from an already negative ion ( $H S O_{4}^{-}$ ) is harder because you are pulling H⁺ off a species that is already $- 1$ ; empirically each successive ionisation adds about +5 to $p K_{a}$ . $p K_{a 2} \approx p K_{a 1} + 5 = - 2 + 5 = 3$

Why this step? Electrostatics: a second $H^{+}$ must escape an anion that already holds negative charge, which resists losing more.

Step 3 — Conclude. $p K_{a 1} \approx - 2$ (strong), $p K_{a 2} \approx 3$ (weak). The first proton leaves far more easily.

Verify: Experiment: $H_{2} S O_{4}$ has $p K_{a 1} \approx - 3$ and $p K_{a 2} \approx 1.99$ — same story, first ≪ second in $p K_{a}$ . Order and "+5 step" confirmed. ✔

E8 — Case C8: real-world word problem (pool sanitiser)

Worked example A pool is treated with "chlorine". At the working pH, the disinfecting species is

$H C l O$ , whose $p K_{a} \approx 7.5$ . A test shows pool pH $= 7.5$ . What fraction of the "free chlorine" is the active acid $H C l O$ vs its conjugate base $C l O^{-}$ ? And is $H C l O$ or $C l O^{-}$ the stronger germ-killer?

Forecast: At $pH = p K_{a}$ , guess the split before computing.

Step 1 — Henderson–Hasselbalch. For an acid $H A ⇌ H^{+} + A^{-}$ : $pH = p K_{a} + lo g_{10} \frac{[ A ^{-} ]}{[ H A ]}$ Why this tool? It converts a pH and a $p K_{a}$ directly into the base-to-acid ratio — exactly what "what fraction" asks.

Step 2 — Plug in $pH = p K_{a} = 7.5$ . $7.5 = 7.5 + lo g_{10} \frac{[ C l O ^{-} ]}{[ H C l O ]} \Rightarrow lo g_{10} \frac{[ C l O ^{-} ]}{[ H C l O ]} = 0 \Rightarrow \frac{[ C l O ^{-} ]}{[ H C l O ]} = 1$ So it is a 50/50 split: $50% H C l O$ , $50% C l O^{-}$ .

Step 3 — Which kills germs better? Neutral $H C l O$ penetrates microbial cell walls far more effectively than the charged $C l O^{-}$ , so pool operators keep pH below $p K_{a}$ to push the equilibrium toward $H C l O$ .

Verify: At $pH = p K_{a}$ the ratio must be $1 0^{0} = 1$ , giving exactly $50%$ each. Lower pH ⇒ more $H C l O$ ⇒ better sanitising. Chemistry ↔ context matches. ✔

E9 — Case C9: the mixed exam twist

Worked example Rank these five acids strongest → weakest:

$H C l O_{2}, H N O_{3}, H_{2} S O_{3}, H C l O_{4}, H B r O$ .

Forecast: Different central atoms, different O counts — which factor do you apply first?

Step 1 — Count $q$ for each (dominant factor first).

$H C l O_{4}$ : $q = 3$
$H N O_{3}$ : $q = 2$
$H C l O_{2}$ : $q = 1$
$H_{2} S O_{3}$ : $q = 1$
$H B r O$ : $q = 0$

Why this step? Oxygen count is the biggest lever; sort by it before anything else.

Step 2 — Estimate $p K_{a} = 8 - 5 q$ .

$H C l O_{4}$ : $- 7$
$H N O_{3}$ : $- 2$
$H C l O_{2}$ : $+ 3$
$H_{2} S O_{3}$ : $+ 3$ (tie with $H C l O_{2}$ )
$H B r O$ : $+ 8$

Step 3 — Break the $q = 1$ tie ( $H C l O_{2}$ vs $H_{2} S O_{3}$ ). Central-atom electronegativity: Cl ( $3.16$ ) > S ( $2.58$ ), so $H C l O_{2}$ withdraws charge harder ⇒ slightly stronger.

Final order (strongest → weakest): $H C l O_{4} > H N O_{3} > H C l O_{2} > H_{2} S O_{3} > H B r O$

Verify: $p K_{a}$ estimates $- 7 < - 2 < 3 = 3 < 8$ give the ranking; the tie is split by electronegativity toward $H C l O_{2}$ . Consistent with all earlier examples. ✔

Active recall

Recall In one line, what single number decides the acid-strength

order? The terminal-oxygen count $q$ — via $p K_{a} \approx 8 - 5 q$ ; everything else is a tie-breaker.

Recall Why do we count terminal oxygens and not hydrogens?

Acidity = stability of the conjugate base; only terminal O's spread the anion's negative charge. H-count just tells you how many protons exist (see Resonance and charge delocalisation).

Recall For

$H_{2} S O_{4}$ , why is the second proton harder to remove? It must leave an already-negative $H S O_{4}^{-}$ ; each successive ionisation adds ~+5 to $p K_{a}$ ( $p K_{a 2} \approx p K_{a 1} + 5$ ).

Connections

Hinglish parent note
p-Block — halogen chemistry home
Pauling rules for oxoacids — the $p K_{a} \approx 8 - 5 q$ engine used throughout
Inductive effect — the electronegativity tie-breaker (E3, E9)
Resonance and charge delocalisation — why sharing charge stabilises (E1, s01)
Conjugate acid–base pairs — strength = base stability
Oxidising power of oxoacids — the opposite trend (E6)
Oxoacids of sulfur and nitrogen — cross-element practice (E4, E9)