Conjugate acid-base pairs: Worked examples — Chemistry deep dive

This page is a drill hall. The parent note built the idea: two species differing by exactly one proton ( $H^{+}$ ) form a conjugate pair, and their strengths are tied by $K_{a} \times K_{b} = K_{w}$ . Here we hunt down every case class the topic can throw at you and work one clean example for each — including the awkward ones (zero H⁺ to move, doubly-charged species, water fighting itself, exam traps).

Before we start, one reminder of the vocabulary we will reuse constantly:

We take base-10 logarithms as $p K = - lo g_{10} K$ — a compression trick so we can say " $p K_{a} = 5$ " instead of " $K_{a} = 1 0^{- 5}$ ". Whenever I write $lo g$ I mean $lo g_{10}$ . Applying this to $K_{w}$ itself gives $p K_{w} = - lo g_{10} K_{w} = - lo g_{10} (1 0^{- 14}) = 14$ — a name we will use in Example 6.

One more symbol appears in the word problem (Example 9), so we define it now rather than surprise you later:

The scenario matrix

Every problem this topic asks lives in one of these cells. The examples below are labelled with the cell they hit.

#	Case class	What makes it tricky	Example
A	Neutral acid → anion base	the baseline case	Ex 1
B	Cation acid → neutral base	positive parent	Ex 2
C	Doubly/multiply charged pair	both members carry charge	Ex 3
D	Amphoteric species (both roles)	one molecule, two conjugates	Ex 4
E	Number crunch: $K_{a} \to K_{b}$	using $K_{a} K_{b} = K_{w}$	Ex 5
F	$p K_{a} + p K_{b} = 14$ shortcut	the log form of the law	Ex 6
G	Degenerate: strong acid, " $K_{b} \to 0$ "	limiting/near-zero conjugate	Ex 7
H	Spectator-ion trap	a species that is NOT a conjugate	Ex 8
I	Real-world word problem	pick the right pair in a mess	Ex 9
J	Exam twist: rank by strength	reasoning, not just plugging	Ex 10

Figure 1 (below) is our master map of the "add-a-proton / remove-a-proton" axis. The horizontal axis is charge, running from negative (left) to positive (right); each yellow arrow is one $H^{+}$ hop, which always shifts a species one step to the right (charge $+ 1$ ). Water sits dead centre at charge $0$ , with its conjugate base $OH^{-}$ one step left ( $- 1$ ) and its conjugate acid $H_{3} O^{+}$ one step right ( $+ 1$ ). Every example below is just a walk one step left or right on this single line.

Worked examples

Cell A — neutral acid becomes an anion

Worked example Example 1 — the baseline pair

Given: $HF + H_{2} O ⇌ H_{3} O^{+} + F^{-}$ . Identify the conjugate acid–base pair centred on HF, and state each species' charge.

Forecast: Guess now — does the charge on fluorine go up or down when HF loses its proton?

Spot who lost the proton. HF (has one H) becomes $F^{-}$ (no H). So HF is the acid, $F^{-}$ is its conjugate base. Why this step? A conjugate pair is defined by "differs by one $H^{+}$ ". Find the two species that differ by exactly one H — that IS the pair.
Track the charge. Remove $H^{+}$ (charge $+ 1$ ) from neutral HF. To keep total charge balanced, what's left must carry $- 1$ . So $F^{-}$ . Why this step? Removing a $+ 1$ particle from a $0$ molecule forces the remainder to $- 1$ . Charge is conserved, always.

Verify: Left side of the pair charge: $0$ . Right side: $(+ 1)_{H^{+}} + (- 1)_{F^{-}} = 0$ . Balanced. ✓ The pair is $(HF / F^{-})$ .

Notice on Figure 1: HF sits one step right of $F^{-}$ , because "adding a proton" moves you rightward on the axis and increases charge by exactly $+ 1$ .

Cell B — a positive acid becomes a neutral base

Contrast A vs B: In A the acid was neutral and the base negative; in B the acid was positive and the base neutral. The difference is always one $H^{+}$ and one $+ 1$ charge — never a fixed final sign. This kills the "conjugates have opposite signs" myth.

Cell C — both members are charged (multi-proton acid)

On Figure 1 this pair sits far on the left (already anionic), but the step between them is the same single-proton hop as every other pair.

Cell D — amphoteric: one molecule, two conjugates

Worked example Example 4 — water plays both sides

Given: water. Write BOTH its conjugate acid and conjugate base, and place them on the axis.

Forecast: Water is in the middle. Which way is "acid", which way is "base"?

Water as a base (accepts $H^{+}$ , step right): $H_{2} O + H^{+} \to H_{3} O^{+}$ . Charge $0 \to + 1$ . Conjugate acid = $H_{3} O^{+}$ . Why this step? Gaining a proton = moving right = charge up by one. The gained species is the conjugate acid.
Water as an acid (donates $H^{+}$ , step left): $H_{2} O \to H^{+} + OH^{-}$ . Charge $0 \to - 1$ . Conjugate base = $OH^{-}$ . Why this step? Losing a proton = moving left = charge down by one. The remainder is the conjugate base.

Verify: The chain $H_{3} O^{+} - H^{+} H_{2} O - H^{+} OH^{-}$ has charges $+ 1, 0, - 1$ : each step drops charge by exactly $1$ . ✓ Water is amphoteric — it appears in TWO pairs.

Amphoteric = a species that can donate or accept a proton, so it has both a conjugate acid (to its right) and a conjugate base (to its left) on our axis. Look at Figure 1: water sits dead centre.

Cell E — turn a $K_{a}$ into a $K_{b}$

Worked example Example 5 — the number-crunch

Given: hydrofluoric acid, $K_{a} (HF) = 6.6 \times 1 0^{- 4}$ at $2 5^{\circ} C$ . Find $K_{b}$ of its conjugate base $F^{-}$ .

Forecast: HF is a fairly weak acid. Will $F^{-}$ be a strong or weak base?

Invoke the law. $K_{a} \times K_{b} = K_{w} = 1.0 \times 1 0^{- 14}$ , so $K_{b} = \frac{K _{w}}{K _{a}}$ . Why this step? This is the derived relationship from the parent note — the only bridge from an acid's number to its conjugate's number.
Plug in. $K_{b} = \frac{1.0 \times 1 0 ^{- 14}}{6.6 \times 1 0 ^{- 4}} = 1.52 \times 1 0^{- 11}$ . Why this step? Direct substitution. Watch the exponent arithmetic: $- 14 - (- 4) = - 10$ , then $1/6.6 \approx 0.15$ shifts it to $\approx 1.5 \times 1 0^{- 11}$ .
Interpret. $K_{b} \approx 1.5 \times 1 0^{- 11} ≪ 1$ : $F^{-}$ is a weak base. Why this step? Small $K_{b}$ = poor proton grabber. Weak acid ↔ weak-ish conjugate base, as expected.

Verify: $K_{a} \cdot K_{b} = 6.6 \times 1 0^{- 4} \times 1.52 \times 1 0^{- 11} = 1.0 \times 1 0^{- 14} = K_{w}$ . ✓

Cell F — the log shortcut

Worked example Example 6 —

$p K_{a} + p K_{b} = 14$ Given: an acid with $p K_{a} = 4.76$ (this is acetic acid). Find $p K_{b}$ of its conjugate base $CH_{3} COO^{-}$ .

Forecast: Guess the answer before doing any algebra — the two $p K$ values must sum to a familiar number.

Take $lo g$ of the master law. From $K_{a} K_{b} = K_{w}$ , apply $- lo g$ to both sides: $- lo g K_{a} - lo g K_{b} = - lo g K_{w}$ , i.e. $p K_{a} + p K_{b} = p K_{w} = 14$ . (Recall $p K_{w} = - lo g_{10} K_{w} = 14$ , defined above.) Why this step? $lo g$ turns a product into a sum. Multiplying $K$ 's becomes adding $p K$ 's — much friendlier arithmetic. That's the whole reason $p K$ notation exists.
Solve. $p K_{b} = 14 - 4.76 = 9.24$ . Why this step? Just rearrange the sum.

Verify: Convert back: $K_{a} = 1 0^{- 4.76} = 1.74 \times 1 0^{- 5}$ , $K_{b} = 1 0^{- 9.24} = 5.75 \times 1 0^{- 10}$ . Product $= 1.0 \times 1 0^{- 14}$ ✓ (matches the parent note's Example 2 value for acetate, $5.6 \times 1 0^{- 10}$ , to rounding).

Cell G — the degenerate limit (strong acid)

Worked example Example 7 — when

$K_{a}$ blows up and $K_{b} \to 0$ Given: hydrochloric acid HCl is a strong acid — in water it donates its proton essentially completely, so its $K_{a}$ is not a finite tabulated number but is taken as "very large," formally $K_{a} \to \infty$ . Describe the limiting behaviour of $K_{b}$ of $Cl^{-}$ .

Forecast: As an acid's $K_{a}$ grows without bound, what happens to its conjugate's $K_{b}$ ?

Write $K_{b}$ as a function of $K_{a}$ . $K_{b} = \frac{K _{w}}{K _{a}} = \frac{1.0 \times 1 0 ^{- 14}}{K _{a}}$ . Why this step? Same bridge law as Example 5, but now we keep $K_{a}$ symbolic so we can watch the trend as it grows, rather than pinning an arbitrary value.
Take the limit. As $K_{a} \to \infty$ , $K_{b} = K_{w} / K_{a} \to 0$ . A perfect proton-thrower has a conjugate that essentially never grabs the proton back. Why this step? This is the mathematical form of "strong acid ⇒ negligibly weak conjugate base" — the inverse relationship, now provably a $1/ x$ curve heading to zero.
Sanity number. Even a modest "strong" estimate $K_{a} \sim 1 0^{6}$ already gives $K_{b} \sim 1 0^{- 14} /1 0^{6} = 1 0^{- 20}$ — vanishingly small. Why this step? Shows the limit is not just abstract: any large $K_{a}$ you pick pushes $K_{b}$ far below anything measurable.

Verify: The function $K_{w} / K_{a}$ is strictly decreasing and its limit as $K_{a} \to \infty$ is $0$ ✓. Practically, $Cl^{-}$ is so weak a base ( $K_{b} \to 0$ ) we treat it as a non-base spectator in water.

Figure 2 (below) shows this inverse curve. Its horizontal axis is $K_{a}$ of the acid on a log scale, its vertical axis is $K_{b}$ of the conjugate base, also on a log scale; on such log–log axes the relation $K_{b} = K_{w} / K_{a}$ appears as a straight descending line. Walk-through: find any acid on the horizontal axis, go straight up to the blue line, then left to read its conjugate's $K_{b}$ . Move rightward (stronger acid, bigger $K_{a}$ ) and the line drops (weaker conjugate base). The four labelled dots are our examples — HCl (far right, strongest acid) sits lowest, HCN (far left, weakest acid) sits highest.

Cell H — the spectator trap

Worked example Example 8 — spot the non-conjugate

Given: $NaOH + HCl \to NaCl + H_{2} O$ . A student claims $(Na^{+} / NaOH)$ is a conjugate pair. True or false, and why?

Forecast: Do $Na^{+}$ and NaOH differ by exactly one $H^{+}$ ?

Test the definition. NaOH minus $Na^{+}$ leaves $OH^{-}$ , not a bare proton removal. $Na^{+}$ and NaOH differ by an entire $OH^{-}$ group, not by one $H^{+}$ . Why this step? The ONLY test for a conjugate pair is "differ by one $H^{+}$ ." Anything else — a whole $OH$ , a metal ion — disqualifies it.
Find the real pairs. The genuine proton transfer is $H^{+}$ (from HCl) meeting $OH^{-}$ (from NaOH) to make $H_{2} O$ . Real pairs: $(HCl / Cl^{-})$ and $(H_{2} O / OH^{-})$ . Why this step? $Na^{+}$ never gains or loses a proton — it is a spectator ion, watching the transfer without joining.

Verify: Does $H_{2} O$ differ from $OH^{-}$ by one $H^{+}$ ? $H_{2} O - H^{+} = OH^{-}$ ✓. Does $Na^{+}$ ? No. Claim is false.

Cell I — real-world word problem

Worked example Example 9 — blood buffer (word problem)

Given: human blood is buffered near $p H 7.4$ (recall: $p H = - lo g_{10} [H^{+}]$ , so $7.4$ means blood is very mildly basic) by the pair $H_{2} CO_{3} / HCO_{3}^{-}$ (carbonic acid / bicarbonate). Carbonic acid has $K_{a} = 4.5 \times 1 0^{- 7}$ . (a) Is $HCO_{3}^{-}$ its conjugate acid or base? (b) Find $K_{b}$ of $HCO_{3}^{-}$ .

Forecast: Carbonic acid has two H's; bicarbonate has one. Which lost the proton?

Assign roles. $H_{2} CO_{3}$ (acid) donates one $H^{+}$ to become $HCO_{3}^{-}$ . So $HCO_{3}^{-}$ is the conjugate base. Why this step? We apply the one universal test: the two species differ by exactly one $H^{+}$ , and the member with fewer H's and more negative charge is always the one that lost the proton — hence it is the conjugate base. This tells us which formula plays which role before we touch any numbers. (See 2.6.12-Buffer-solutions for why both members must be present to buffer.)
Compute $K_{b}$ . $K_{b} = \frac{K _{w}}{K _{a}} = \frac{1.0 \times 1 0 ^{- 14}}{4.5 \times 1 0 ^{- 7}} = 2.22 \times 1 0^{- 8}$ . Why this step? Same conjugate-bridge law; a buffer works precisely because this pair coexists in equilibrium (see 1.5.08-Chemical-equilibrium).

Verify: $4.5 \times 1 0^{- 7} \times 2.22 \times 1 0^{- 8} = 1.0 \times 1 0^{- 14} = K_{w}$ ✓. Both $K_{a}$ and $K_{b}$ are tiny ⇒ a weak acid paired with a weak base ⇒ good buffer.

Cell J — exam twist: rank by conjugate strength

Worked example Example 10 — rank the conjugate bases

Given: three acids with $K_{a}$ values — HF ( $6.6 \times 1 0^{- 4}$ ), $CH_{3} COOH$ ( $1.8 \times 1 0^{- 5}$ ), HCN ( $6.2 \times 1 0^{- 10}$ ). Rank their conjugate bases ( $F^{-}, CH_{3} COO^{-}, CN^{-}$ ) from strongest base to weakest.

Forecast: Reason it out first — which acid is weakest, and what does that imply about its conjugate?

Recall the inverse law. Strong acid ↔ weak conjugate base. Formally, $K_{b} = K_{w} / K_{a}$ is a decreasing function of $K_{a}$ : a smaller $K_{a}$ makes a bigger $K_{b}$ . Why this step? It tells us we can rank the bases purely from the acids' $K_{a}$ order — no full computation needed. The weakest acid must yield the strongest conjugate base.
Order the acids by $K_{a}$ , smallest first. $6.2 \times 1 0^{- 10}$ (HCN) $<$ $1.8 \times 1 0^{- 5}$ (CH₃COOH) $<$ $6.6 \times 1 0^{- 4}$ (HF). Why this step? Sorting the given numbers puts the weakest acid (HCN) at the front — that is the one whose conjugate base is strongest.
Flip the order for the conjugate bases. Because $K_{b}$ runs opposite to $K_{a}$ , the base ranking (strongest first) is the reverse of the acid ranking (weakest first) — but since we sorted acids weakest-first, the bases come out in the same listed order: $CN^{-} > CH_{3} COO^{-} > F^{-}$ . Why this step? Weakest acid → strongest base, so HCN's conjugate $CN^{-}$ tops the list and HF's conjugate $F^{-}$ is weakest.
Confirm with two actual numbers. $K_{b} (CN^{-}) = \frac{1 0 ^{- 14}}{6.2 \times 1 0 ^{- 10}} = 1.61 \times 1 0^{- 5}$ and $K_{b} (F^{-}) = \frac{1 0 ^{- 14}}{6.6 \times 1 0 ^{- 4}} = 1.52 \times 1 0^{- 11}$ . Why this step? Two endpoints are enough to prove the spread: the strongest base is about a million times stronger than the weakest.

Conclusion: From strongest base to weakest: $CN^{-} > CH_{3} COO^{-} > F^{-}$ .

Verify: $K_{b} (CN^{-}) = 1.61 \times 1 0^{- 5} ≫ K_{b} (F^{-}) = 1.52 \times 1 0^{- 11}$ , so $CN^{-}$ really is the strongest base and $F^{-}$ the weakest — matching the ordering derived from $K_{a}$ alone. ✓

Wrap-up recall

Recall One rule generated every answer above

What single test decides whether two species form a conjugate pair? ::: They differ by exactly one $H^{+}$ (which shifts charge by exactly $+ 1$ ). Nothing about the final sign.

Recall Conjugate acid vs conjugate base

If you remove one proton from a species, do you get its conjugate acid or its conjugate base? ::: Its conjugate base (charge drops by one). Adding a proton gives its conjugate acid (charge rises by one).

Recall The bridge formula

Given $K_{a}$ of an acid, how do you get $K_{b}$ of its conjugate base? ::: $K_{b} = K_{w} / K_{a} = 1.0 \times 1 0^{- 14} / K_{a}$ at $2 5^{\circ} C$ ; equivalently $p K_{a} + p K_{b} = p K_{w} = 14$ .

See also: 2.6.02-Brønsted-Lowry-theory (where the donor/acceptor idea comes from), 2.6.09-pH-calculations (using these $K_{a}$ / $K_{b}$ numbers to get actual $p H$ ), 2.6.15-Lewis-acids-bases (the broader electron-pair view), 2.6.01-Arrhenius-acids-bases (the older, narrower definition).