Acid-base reactions — neutralization, salt formation: Worked examples

Before any symbol appears, here is the whole vocabulary, in plain words:

One more tool must be defined before it appears in any pH calculation:

The scenario matrix

Every neutralization problem is one of these cells. The right column names the worked example that lands on it.

#	Case class	What is special about it	Covered by
A	Strong monoprotic + strong monobasic, 1:1	$n_{a} = n_{b} = 1$ , pH at equivalence $= 7$	Example 1
B	Diprotic acid ( $n_{a} = 2$ )	the $n$ -factor doubles the base needed	Example 2
C	Dibasic base ( $n_{b} = 2$ )	mirror of B; base supplies 2 $O H^{-}$	Example 3
D	Both poly: $n_{a} \neq = 1, n_{b} \neq = 1$	ratios stack; salt formula from charge balance	Example 4
E	Weak acid + strong base salt → basic	anion hydrolyzes; equivalence pH $> 7$	Example 5
F	Degenerate / zero input ( $V_{a} = 0$ , or pure water)	limiting behaviour, no reaction	Example 6
G	Half-neutralization (buffer point)	only half the acid consumed → $pH = p K_{a}$	Example 7
H	Real-world word problem (vinegar strength)	translate grocery words into $M, V, n$	Example 8
I	Exam twist / excess reagent	one reagent left over → final pH from leftovers	Example 9
J	Weak base + strong acid salt → acidic	cation hydrolyzes; equivalence pH $< 7$	Example 10

The figure below is the picture behind the master rule — we dissect it piece by piece right after.

Figure — Acid-base reactions — neutralization, salt formation

Let us read the figure symbol by symbol, so it is not just decoration but the equation made visible:

Left pink stack — each block is one acid formula unit, and the "H+" written on it is a proton that unit can donate. If the acid were diprotic every block would carry two protons — that "protons per block" number is exactly $n_{a}$ . The height of the stack (how many blocks) is set by $M_{a} V_{a}$ (crowdedness × how much liquid). So the whole pink pile counts $n_{a} M_{a} V_{a}$ protons.
Right blue stack — mirror image: each block is one base formula unit, "OH-" is a hydroxide it can donate, "hydroxides per block" is $n_{b}$ , and stack height is $M_{b} V_{b}$ . Total blue pile $= n_{b} M_{b} V_{b}$ hydroxides.
Yellow double arrow + big "=" — this is the balance. Neutralization is finished precisely when the pink pile of protons and the blue pile of hydroxides are the same size, because each proton pairs off with exactly one hydroxide to make one water. Set the two counts equal and you have literally written $n_{a} M_{a} V_{a} = n_{b} M_{b} V_{b}$ — the master rule is nothing more than "make the two piles match."

Keep this picture in mind: every example below is just counting one pile and matching the other.

Example 1 — Cell A: the clean 1:1 titration

Worked example Strong acid, strong base, one proton each

Q: $25.0 mL$ of $H C l$ is exactly neutralized by $30.0 mL$ of $0.100 M$ $N a O H$ . Find the molarity of the $H C l$ , and state the pH at equivalence.

Forecast: guess — will $[H C l]$ be above or below $0.100$ M? (It took more base volume than the acid volume, so the acid must be a bit more crowded than the base. Expect above $0.100$ .)

Identify the $n$ -factors. $n_{a} = 1$ ( $H C l$ gives one $H^{+}$ ), $n_{b} = 1$ ( $N a O H$ gives one $O H^{-}$ ). Why this step? The master rule only simplifies to $M_{a} V_{a} = M_{b} V_{b}$ when both $n$ 's are $1$ ; we must check that first, not assume it.
Write the rule and cancel the equal $n$ 's. $1 \cdot M_{a} \cdot 25.0 = 1 \cdot 0.100 \cdot 30.0$ Why this step? Volumes are both in mL, so the mL cancels — no need to convert to litres.
Solve. $M_{a} = \frac{0.100 \times 30.0}{25.0} = \frac{3.00}{25.0} = 0.120 M$ . Why this step? We isolate the single unknown $M_{a}$ by dividing both sides by its coefficient $25.0$ — the only algebra the balance rule ever needs.

Verify: $0.120 > 0.100$ ✓ matches the forecast. Moles check: acid $H^{+} = 0.120 \times 0.0250 = 3.00 \times 1 0^{- 3}$ ; base $O H^{-} = 0.100 \times 0.0300 = 3.00 \times 1 0^{- 3}$ . Equal ✓. pH at equivalence $= 7$ because the salt $N a C l$ comes from a strong acid and strong base — neither ion hydrolyzes (see Salt hydrolysis).

Example 2 — Cell B: the diprotic acid

Worked example

$H_{2} S O_{4}$ — the $n$ -factor bites Q: How many mL of $0.10 M$ $N a O H$ neutralize $20.0 mL$ of $0.10 M$ $H_{2} S O_{4}$ ?

Forecast: sulfuric acid hands out two protons per molecule. So it should demand more base than a 1:1 case. Guess roughly double.

$n$ -factors: $n_{a} = 2$ ( $H_{2} S O_{4} \to 2 H^{+}$ ), $n_{b} = 1$ . Why this step? The whole point of this cell is that $n_{a} \neq = n_{b}$ ; skipping it is the classic error.
Master rule with the 2 kept in: $2 \cdot 0.10 \cdot 20.0 = 1 \cdot 0.10 \cdot V_{b}$ Why this step? Both volumes are in mL, so the unit cancels across the equals sign and $V_{b}$ comes out in mL directly — no litre conversion needed.
Solve: left side $= 4.0$ ; so $V_{b} = \frac{4.0}{0.10} = 40.0 mL$ . Why this step? Everything on the left is known, so it collapses to one number; dividing by the coefficient $0.10$ of the lone unknown $V_{b}$ isolates it.

Verify: $40.0$ mL is exactly double the $20.0$ mL of acid at equal molarity ✓ matches forecast. Proton count: $2 \times 0.10 \times 0.020 = 4.0 \times 1 0^{- 3}$ mol $H^{+}$ ; hydroxide count: $0.10 \times 0.040 = 4.0 \times 1 0^{- 3}$ mol $O H^{-}$ . Balanced ✓.

Example 3 — Cell C: the dibasic base (mirror of Example 2)

Worked example

$C a (O H)_{2}$ — two hydroxides per unit Q: What volume of $0.20 M$ $H C l$ neutralizes $50.0 mL$ of $0.10 M$ $C a (O H)_{2}$ ? Name the salt.

Forecast: now the base is the greedy one — each unit offers two $O H^{-}$ . So we should need extra acid.

$n$ -factors: $n_{a} = 1$ ( $H C l$ ), $n_{b} = 2$ ( $C a (O H)_{2} \to 2 O H^{-}$ ). Why this step? Mirror of Example 2 — the doubling now sits on the base side.
Rule: $1 \cdot 0.20 \cdot V_{a} = 2 \cdot 0.10 \cdot 50.0$ . Why this step? Both sides use mL for volume, so the mL unit cancels and $V_{a}$ will come out in mL directly.
Solve: right side $= 10.0$ ; $V_{a} = \frac{10.0}{0.20} = 50.0 mL$ . Why this step? The right side is fully known, so it is one number; dividing by the coefficient $0.20$ of the unknown $V_{a}$ isolates it.
Salt: cation $C a^{2 +}$ from base, anion $C l^{-}$ from acid. Charge balance needs two $C l^{-}$ per $C a^{2 +}$ → $C a C l_{2}$ , calcium chloride. Full: $C a (O H)_{2} + 2 H C l \to C a C l_{2} + 2 H_{2} O$ .

Verify: $O H^{-}$ supplied $= 2 \times 0.10 \times 0.050 = 0.010$ mol; $H^{+}$ supplied $= 0.20 \times 0.050 = 0.010$ mol. Equal ✓.

Example 4 — Cell D: both reagents poly

Worked example

$H_{2} S O_{4}$ meets $K O H$ — ratios stack, salt from charges Q: Write the balanced equation for $H_{2} S O_{4} + K O H$ , name the salt, and find the mL of $0.050 M$ $K O H$ needed for $10.0 mL$ of $0.10 M$ $H_{2} S O_{4}$ .

Forecast: the acid needs 2 base units per acid unit; the base itself is only monobasic, so expect a lot of base volume.

$n$ -factors: $n_{a} = 2$ , $n_{b} = 1$ .
Salt from charge balance: $K^{+}$ (charge $+ 1$ ) with $S O_{4}^{2 -}$ (charge $- 2$ ) → need two $K^{+}$ → $K_{2} S O_{4}$ , potassium sulfate. Equation: $H_{2} S O_{4} + 2 K O H \to K_{2} S O_{4} + 2 H_{2} O$ . Why this step? The coefficient 2 in front of $K O H$ is the $n_{a} = 2$ made visible — two protons need two hydroxides.
Volume: $2 \cdot 0.10 \cdot 10.0 = 1 \cdot 0.050 \cdot V_{b} \Rightarrow 2.0 = 0.050 V_{b} \Rightarrow V_{b} = 40.0 mL$ . Why this step? Both volumes are in mL (unit cancels); the known left side reduces to $2.0$ , then dividing by the unknown's coefficient $0.050$ gives $V_{b}$ in mL.

Verify: $H^{+} = 2 \times 0.10 \times 0.010 = 2.0 \times 1 0^{- 3}$ ; $O H^{-} = 0.050 \times 0.040 = 2.0 \times 1 0^{- 3}$ . Equal ✓. Salt $K_{2} S O_{4}$ is strong-acid + strong-base → neutral solution.

Example 5 — Cell E: weak-acid salt, pH above 7

Worked example Sodium acetate at equivalence

Q: $50.0 mL$ of $0.100 M$ acetic acid ( $C H_{3} C O O H$ , $K_{a} = 1.8 \times 1 0^{- 5}$ ) is titrated to the equivalence point with $0.100 M$ $N a O H$ . Find the volume of base used, and the pH at equivalence.

Forecast: acetic is monoprotic, $N a O H$ monobasic → volumes should match (1:1). But the leftover salt is $C H_{3} C O O N a$ , whose anion is the conjugate base of a weak acid — so equivalence pH should be above 7 (basic), not exactly 7.

Volume (mole balance): $n_{a} = n_{b} = 1$ , so $M_{a} V_{a} = M_{b} V_{b} \Rightarrow 0.100 \times 50.0 = 0.100 \times V_{b} \Rightarrow V_{b} = 50.0 mL$ . Why this step? Strength does not change the counting; it only changes the resulting pH. Both volumes in mL, so the unit cancels and $V_{b}$ is in mL.
Concentration of the salt after mixing. Total volume $= 50.0 + 50.0 = 100.0$ mL. Moles of acetate $= 0.100 \times 0.0500 = 5.00 \times 1 0^{- 3}$ . So the salt concentration is $C = [C H_{3} C O O^{-}] = \frac{5.00 \times 1 0 ^{- 3}}{0.100} = 0.0500 M$ . Why this step? Hydrolysis depends on the diluted salt concentration $C$ , not the original acid's — the two liquids doubled the volume. We name this number $C$ so we can plug it into step 4.
Hydrolysis constant. $K_{b} = \frac{K _{w}}{K _{a}} = \frac{1.0 \times 1 0 ^{- 14}}{1.8 \times 1 0 ^{- 5}} = 5.56 \times 1 0^{- 10}$ (see Salt hydrolysis).
Set up the hydrolysis equilibrium and derive $[O H^{-}]$ . The acetate reacts: $C H_{3} C O O^{-} + H_{2} O ⇌ C H_{3} C O O H + O H^{-}$ . Let $x = [O H^{-}]$ produced. Then $[C H_{3} C O O H] = x$ too, and the leftover acetate is $C - x$ . By definition of $K_{b}$ : $K_{b} = \frac{[ C H _{3} C O O H ] [ O H ^{-} ]}{[ C H _{3} C O O ^{-} ]} = \frac{x \cdot x}{C - x} = \frac{x ^{2}}{C - x}$ Why this step? This is just the equilibrium constant written for the hydrolysis reaction — no new idea, only bookkeeping of what each species becomes. The approximation: because $K_{b}$ is tiny ( $5.56 \times 1 0^{- 10}$ ), only a whisper of acetate reacts, so $x ≪ C$ and $C - x \approx C$ . Then $K_{b} \approx \frac{x ^{2}}{C}$ , giving $x = [O H^{-}] = K_{b} C$ . Why is this allowed? We may drop $x$ next to $C$ only if $x$ is under ~5% of $C$ ; here $x = 5.3 \times 1 0^{- 6}$ vs $C = 0.05$ — about $0.01%$ , so the approximation is excellent. $[O H^{-}] = 5.56 \times 1 0^{- 10} \times 0.0500 = 2.78 \times 1 0^{- 11} = 5.27 \times 1 0^{- 6} M$
pH via pOH. Using the definition above, $pOH = - lo g (5.27 \times 1 0^{- 6}) = 5.28$ , so $pH = 14 - pOH = 14 - 5.28 = 8.72$ .

Verify: check the approximation: $x / C = 5.27 \times 1 0^{- 6} /0.05 = 1.1 \times 1 0^{- 4} = 0.011% ≪ 5%$ ✓. pH $= 8.72 > 7$ ✓ basic, as forecast — a strong-base + weak-acid salt. This is exactly the mistake warned about in the parent: "neutralization ≠ pH 7."

Example 6 — Cell F: degenerate / zero input

Worked example The limiting cases the formula must survive

Q (a): You add $0$ mL of acid to a beaker of $N a O H$ . What does the master rule predict? Q (b): You mix pure water with pure water. What is the "salt"?

Forecast: with no acid, nothing should react; the rule should collapse to a trivial truth.

Case (a): put $V_{a} = 0$ into $n_{a} M_{a} V_{a} = n_{b} M_{b} V_{b}$ : the left side is $0$ , so it says $n_{b} M_{b} V_{b} = 0$ — i.e. no base has been consumed. Correct: an unreacted $N a O H$ solution stays basic; the equivalence rule is simply not yet satisfied. Why this step? A formula must give the right nonsense at the boundary — here, "nothing reacted."
Case (b): water is both an extremely weak acid and an extremely weak base (Brønsted-Lowry conjugate acid-base pairs). Its self-reaction $2 H_{2} O ⇌ H_{3} O^{+} + O H^{-}$ has $K_{w} = 1.0 \times 1 0^{- 14}$ , giving $[H^{+}] = [O H^{-}] = 1.0 \times 1 0^{- 7}$ M. There is no external salt — the "products" are just trace ions. pH $= 7.00$ .

Verify: (a) $0 = 0$ is consistent, no over-titration predicted ✓. (b) $- lo g (1.0 \times 1 0^{- 7}) = 7.00$ ✓ — the neutral anchor of the whole pH scale.

Example 7 — Cell G: half-neutralization (the buffer point)

Before this example, two small tools must be built, because the parent note never defined them:

Worked example Stop the titration halfway

Q: In the acetic-acid titration of Example 5, you add only $25.0 mL$ of the $0.100 M$ $N a O H$ — exactly half the equivalence volume. What is the pH? ( $K_{a} = 1.8 \times 1 0^{- 5}$ .)

Forecast: half the acid is converted to acetate, half remains as acetic acid → equal amounts of acid and conjugate base. At that special point the pH should equal $p K_{a}$ .

Moles before. Convert each mL to litres ( $25.0 mL = 0.0250 L$ , $50.0 mL = 0.0500 L$ ) because molarity is per litre, so mol $= M \times V$ only works with $V$ in litres. Acid $= 0.100 \times 0.0500 = 5.00 \times 1 0^{- 3}$ mol; base added $= 0.100 \times 0.0250 = 2.50 \times 1 0^{- 3}$ mol. Why this step? Here we need actual amounts (moles), not a volume ratio — so unlike the balance-rule examples, the mL do not cancel and must be turned into litres.
After reaction: base eats $2.50 \times 1 0^{- 3}$ mol of acid → leaves $2.50 \times 1 0^{- 3}$ mol $C H_{3} C O O H$ and makes $2.50 \times 1 0^{- 3}$ mol $C H_{3} C O O^{-}$ . Equal amounts. Why this step? Equal acid and conjugate base is the definition of the half-way (buffer) point.
Henderson–Hasselbalch: $pH = p K_{a} + lo g \frac{[ base ]}{[ acid ]} = p K_{a} + lo g 1 = p K_{a}$ . Why this step? Equal amounts make the ratio $1$ , and $lo g 1 = 0$ , so the whole correction term vanishes — that is why the half-point is famous. (The ratio is of moles; the shared total volume cancels top and bottom, so we needn't even compute it.)
Compute: $p K_{a} = - lo g (1.8 \times 1 0^{- 5}) = 4.74$ . So $pH = 4.74$ .

Verify: $- lo g (1.8 \times 1 0^{- 5}) = 4.745$ ✓. Note pH $= 4.74 < 7$ here but rose to $8.72$ at full equivalence (Example 5) — the curve climbs, exactly as a titration should.

Example 8 — Cell H: real-world word problem

Worked example How strong is your vinegar?

Q: A bottle claims "5.0% acetic acid by mass." You titrate $10.0 mL$ of it (density $\approx 1.01 g/mL$ ) and it takes $84.0 mL$ of $0.100 M$ $N a O H$ to reach the endpoint. Does the label check out? ( $M_{C H_{3} C O O H} = 60.0 g/mol$ .)

Forecast: translate "grocery words" to moles. Guess the label is roughly right if the two acid-mass estimates agree.

Moles of acid by titration ( $n_{a} = n_{b} = 1$ ): $M_{a} V_{a} = M_{b} V_{b} \Rightarrow M_{a} \times 10.0 = 0.100 \times 84.0 \Rightarrow M_{a} = 0.840 M$ . Why this step? The titration measures the acid concentration directly; the label is what we test against. Both volumes in mL, so the unit cancels and $M_{a}$ is in mol/L.
Grams of acetic acid in the $10.0$ mL sample: moles $= 0.840 \times 0.0100 = 8.40 \times 1 0^{- 3}$ mol; mass $= 8.40 \times 1 0^{- 3} \times 60.0 = 0.504 g$ . Why this step? Percent-by-mass needs grams, so we convert mol → g using the molar mass; volume here must be in litres because $M$ is per litre.
Mass of the $10.0$ mL vinegar: $10.0 mL \times 1.01 g/mL = 10.1 g$ .
Percent by mass: $\frac{0.504}{10.1} \times 100 = 4.99% \approx 5.0%$ . Why this step? Percent by mass is (mass of solute ÷ mass of whole solution) × 100 — that is the definition the label uses.

Verify: $4.99%$ ✓ matches the $5.0%$ label. Units track: (mol/L)(L)=mol, (mol)(g/mol)=g, (g/g) is dimensionless → % ✓.

Example 9 — Cell I: exam twist, excess reagent

Worked example One reagent left over — find the final pH

Q: Mix $50.0 mL$ of $0.100 M$ $H C l$ with $30.0 mL$ of $0.100 M$ $N a O H$ . This is not the equivalence point — find the pH of the mixture.

Forecast: more acid moles than base moles → acid wins → leftover $H^{+}$ → pH $< 7$ .

Moles each: $H^{+} = 0.100 \times 0.0500 = 5.00 \times 1 0^{- 3}$ ; $O H^{-} = 0.100 \times 0.0300 = 3.00 \times 1 0^{- 3}$ . Why this step? When volumes/molarities are given directly (not "find the volume"), you compare moles, not solve the balance formula.
Who wins: $H^{+}$ exceeds $O H^{-}$ by $5.00 \times 1 0^{- 3} - 3.00 \times 1 0^{- 3} = 2.00 \times 1 0^{- 3}$ mol of leftover $H^{+}$ . Why this step? Every $O H^{-}$ pairs off one $H^{+}$ into water; the difference is what physically remains in the beaker.
Dilute over total volume: $50.0 + 30.0 = 80.0 mL = 0.0800 L$ , so $[H^{+}] = \frac{2.00 \times 1 0 ^{- 3}}{0.0800} = 0.0250 M$ . Why this step? Concentration is moles per litre of the combined solution, so we divide the leftover moles by the total (now in litres, since pH needs mol/L).
pH: $- lo g (0.0250) = 1.60$ .

Verify: pH $= 1.60 < 7$ ✓ acidic, as forecast. Sanity: if we had used equal moles we'd be back at Example 1's neutral case — the excess is what pushes it acidic.

Example 10 — Cell J: weak base titrated by strong acid

Worked example Ammonia meets HCl — acidic equivalence point

Q: $50.0 mL$ of $0.100 M$ ammonia ( $N H_{3}$ , $K_{b} = 1.8 \times 1 0^{- 5}$ ) is titrated to the equivalence point with $0.100 M$ $H C l$ . Find the volume of acid used and the pH at equivalence.

Forecast: $N H_{3}$ is monobasic and $H C l$ monoprotic → 1:1, volumes should match. But the salt formed is $N H_{4} C l$ , whose $N H_{4}^{+}$ is the conjugate acid of a weak base — so the equivalence pH should be below 7 (acidic). This is the mirror of Example 5.

Volume (mole balance): $n_{a} = n_{b} = 1$ , so $M_{a} V_{a} = M_{b} V_{b} \Rightarrow 0.100 \times V_{a} = 0.100 \times 50.0 \Rightarrow V_{a} = 50.0 mL$ . Why this step? Counting is unaffected by strength; both volumes in mL, so the unit cancels.
Salt concentration after mixing. Total volume $= 50.0 + 50.0 = 100.0$ mL. Moles of $N H_{4}^{+} = 0.100 \times 0.0500 = 5.00 \times 1 0^{- 3}$ , so $C = [N H_{4}^{+}] = \frac{5.00 \times 1 0 ^{- 3}}{0.100} = 0.0500 M$ . Why this step? The hydrolysing ion is diluted to twice the volume; $C$ is the concentration we feed into step 4.
Hydrolysis constant of the cation. $K_{a} = \frac{K _{w}}{K _{b}} = \frac{1.0 \times 1 0 ^{- 14}}{1.8 \times 1 0 ^{- 5}} = 5.56 \times 1 0^{- 10}$ . Why this step? $N H_{4}^{+}$ acts as a weak acid; its strength is $K_{w} / K_{b}$ (see Salt hydrolysis).
Find $[H^{+}]$ from the same equilibrium logic as Example 5. Here $N H_{4}^{+} ⇌ N H_{3} + H^{+}$ ; letting $x = [H^{+}]$ gives $K_{a} = \frac{x ^{2}}{C - x} \approx \frac{x ^{2}}{C}$ (again $x ≪ C$ since $K_{a}$ is tiny), so $[H^{+}] = K_{a} C = 5.56 \times 1 0^{- 10} \times 0.0500 = 5.27 \times 1 0^{- 6} M$ . Why this step? This is the standard weak-acid approximation applied to the cation now that we have $K_{a}$ and $C$ .
pH: $pH = - lo g (5.27 \times 1 0^{- 6}) = 5.28$ .

Verify: pH $= 5.28 < 7$ ✓ acidic, exactly the mirror of Example 5's basic $8.72$ — strong acid + weak base → acidic salt.

Active recall

Recall Which cell needs the

$n$ -factor kept? Any polyprotic acid ( $H_{2} S O_{4}$ ) or polybasic base ( $C a (O H)_{2}$ ) — Cells B, C, D. ::: Because one formula unit trades more than one proton/hydroxide.

Recall At the equivalence point of weak acid + strong base, is pH 7?

No — it is basic (Example 5 gave 8.72), because the salt's anion hydrolyzes. ::: Salt hydrolysis

Recall At the equivalence point of weak base + strong acid, is pH 7?

No — it is acidic (Example 10 gave 5.28), because the salt's cation hydrolyzes. ::: Salt hydrolysis

Recall What is special about the half-neutralization point?

pH = pKₐ, because [acid] = [conjugate base]. ::: Example 7.

What does

V_{a} = 0

predict in the master rule?

That no base has yet been consumed — the trivial, correct boundary behaviour.

Excess-acid mixture pH depends on what?

The leftover moles of

H^{+}

divided by the total combined volume.

Connections

Parent: Acid–base reactions
Molarity and solution stoichiometry
pH and pOH scale
Titration and indicators
Salt hydrolysis
Brønsted-Lowry conjugate acid-base pairs
Enthalpy of reaction