Solution stoichiometry — titrations, dilutions: Worked examples

Before anything, one plain-English reminder of the symbols, so line one makes sense to a newcomer:

Recall The symbols, in words

$n$ = number of moles = a count of particles measured in "packs" of $6.022 \times 1 0^{23}$ . See Mole concept and Avogadro's number.
$c$ = concentration (molarity) = how many moles sit in one litre of solution. Big $c$ = strong, small $c$ = weak.
$V$ = volume = how much liquid, in litres unless we deliberately keep both volumes in mL.
$M_{r}$ = molar mass = the mass in grams of one mole of a substance (unit $g mol^{- 1}$ ). It's the "exchange rate" between grams and moles.
$a, b$ = stoichiometric coefficients = the plain numbers written in front of two substances in a balanced equation, e.g. the " $2$ " in " $+ 2 NaOH$ ". They say how many of one react per how many of the other. See Balancing chemical equations. The master equation $n = c V$ reads: "count = strength × amount of liquid."

The scenario matrix

Every problem in this topic is one of these cells. Our job below is to hit all of them.

#	Cell (case class)	What's tricky about it	Example that hits it
A	Dilution, find final $c$	volume grows, moles fixed	Ex 1
B	Dilution, find water to add	solve for the added volume, not total	Ex 2
C	1:1 titration ( $a = b = 1$ )	shortcut $c_{A} V_{A} = c_{B} V_{B}$ works	Ex 3
D	Non-1:1 titration ( $a \neq = b$ )	must divide by coefficients	Ex 4
E	Degenerate / zero input	add zero water, or concentration $= 0$	Ex 5
F	Limiting / extreme value	huge dilution → $c \to 0$ ; what it means	Ex 5
G	Mass ↔ moles bridge	start from grams, not molarity	Ex 6
H	Mixing two solutions of the SAME solute	moles add, volumes add	Ex 7
I	Back-titration (excess method)	reacted = added − leftover	Ex 8
J	Real-world word problem	strip the words to find $c, V, a, b$	Ex 9
K	Exam twist (percent purity)	extra layer on top of stoichiometry	Ex 10

Figure — Solution stoichiometry — titrations, dilutions

What this figure shows (read before continuing): it is the map of the whole page. The two left boxes are starting corners — either a known $c, V$ pair (violet) or a mass in grams (orange). The two right boxes are target corners — an unknown $c$ or $V$ , or a mass/purity. Every road you can travel runs through the central magenta "moles $n$ " hub: you convert into moles with $n = c V$ (or $n = m / M_{r}$ ), cross the magenta ratio bridge given by the balanced equation's coefficients $a : b$ , then convert back out. The single lesson of the picture: you can never jump corner-to-corner directly — you must pass through moles. Keep glancing back at it as each example travels one of these roads.

Example 1 — Cell A: dilution, find final concentration

Example 2 — Cell B: dilution, find the water to ADD

Worked example How much water?

You need to turn $25.0 mL$ of $1.20 M$ KMnO₄ into a $0.300 M$ solution. How much water must you add?

Forecast: Concentration drops ×4, so final volume must be ×4 = $100$ mL. But the question asks water added, not total. Guess the water first.

Find the final total volume $V_{2}$ : $V_{2} = \frac{c _{1} V _{1}}{c _{2}} = \frac{1.20 \times 25.0}{0.300} = 100 mL$ Why this step? $c_{1} V_{1} = c_{2} V_{2}$ gives the total volume the solution must reach.
Subtract the starting volume to get water added: $V_{water} = V_{2} - V_{1} = 100 - 25.0 = 75.0 mL$ Why subtract? You already have $25$ mL; only the extra $75$ mL is new water.

Verify: Total $= 25 + 75 = 100$ mL, and $1.20 \times 25 = 0.300 \times 100 = 30$ (mmol both sides). Moles conserved. ✓

Example 3 — Cell C: the clean 1:1 titration

Example 4 — Cell D: non-1:1, coefficients matter

Worked example Sulfuric acid by NaOH

$15.0 mL$ of $H_{2} SO_{4}$ (unknown $c$ ) needs $24.0 mL$ of $0.250 M$ NaOH. Find $c_{H_{2} SO_{4}}$ .

Forecast: Each H₂SO₄ eats two NaOH. So the acid is "twice as powerful per molecule" — expect its concentration to be well below the NaOH's. Guess before reading.

Reaction: $H_{2} SO_{4} + 2 NaOH \to Na_{2} SO_{4} + 2 H_{2} O$ . Here the coefficients are $a = 1$ (acid), $b = 2$ (base).

Moles of NaOH: $n_{NaOH} = 0.250 \times 0.0240 = 6.00 \times 1 0^{- 3} mol$ .
Ratio bridge with coefficients: $\frac{n _{acid}}{1} = \frac{n _{NaOH}}{2} \Rightarrow n_{acid} = \frac{1}{2} (6.00 \times 1 0^{- 3}) = 3.00 \times 1 0^{- 3} mol$ Why halve? Two NaOH are used up per single acid molecule, so acid moles are half the base moles.
Concentration: $c = \frac{3.00 \times 1 0 ^{- 3}}{0.0150} = 0.200 M$ .

Verify: $0.200$ M acid vs $0.250$ M base — acid lower, as forecast. If you'd forgotten the 2 you'd get $0.400$ M — double the truth, the classic trap. See Balancing chemical equations. ✓

Example 5 — Cells E & F: zero and limiting inputs

Worked example The degenerate corners

(a) You add $0 mL$ of water to $50.0 mL$ of $0.800 M$ NaCl. Final $c$ ? (b) You dilute $10.0 mL$ of $2.00 M$ NaCl to $10.0 L$ . Final $c$ ? What happens as the final volume $\to \infty$ ?

Forecast: (a) Adding nothing should change nothing. (b) A ×1000 dilution should make it tiny — and infinite water should make it zero.

(a) Plug in $V_{2} = V_{1}$ : $c_{2} = \frac{c _{1} V _{1}}{V _{2}} = \frac{0.800 \times 50.0}{50.0} = 0.800 M$ Why does this matter? It's a sanity check on the formula itself: zero added water ⇒ $V_{2} = V_{1}$ ⇒ $c$ unchanged. The equation degrades gracefully.
(b) Convert to matching units ( $10.0 mL = 0.0100 L$ ): $c_{2} = \frac{2.00 \times 0.0100}{10.0} = 2.00 \times 1 0^{- 3} M$ Why convert here? The two volumes are in different units (mL and L); we must make them agree before dividing.
The limit $V_{2} \to \infty$ : since $c_{2} = \frac{c _{1} V _{1}}{V _{2}}$ and the top ( $n_{1} = c_{1} V_{1}$ , a fixed mole count) is constant, growing the bottom without bound drives $c_{2} \to 0$ . Why does this make physical sense? The same finite pile of solute smeared through an infinite ocean is infinitely dilute — but the moles never vanish, only the concentration does.

Verify: (a) $0.800$ M unchanged. (b) $2.00 \times 1 0^{- 3}$ M is exactly $2.00/1000$ of the start (×1000 dilution). Limiting behaviour: numerator fixed, denominator → ∞ ⇒ ratio → 0. ✓

Example 6 — Cell G: start from grams (mass ↔ moles)

Worked example Making a standard solution from solid

You dissolve $5.30 g$ of anhydrous sodium carbonate ( $Na_{2} CO_{3}$ , molar mass $M_{r} = 106 g mol^{- 1}$ ) and make it up to exactly $250 mL$ . What is its molarity? This is how Standard solutions and primary standards are prepared.

Recall $M_{r}$ (defined in the symbols recall above) is the molar mass — the grams in one mole, our exchange rate between grams and moles.

Forecast: $5.30/106 = 0.05$ mol in a quarter-litre → roughly $0.2$ M. Guess.

Grams → moles using $n = m / M_{r}$ : $n = \frac{5.30}{106} = 0.0500 mol$ Why this step? Molarity needs moles, but we weighed grams; the molar mass $M_{r}$ is the exchange rate between them.
Moles → concentration ( $250 mL = 0.250 L$ ): $c = \frac{n}{V} = \frac{0.0500}{0.250} = 0.200 M$ Why litres? Molarity is per litre by definition.

Verify: $0.200 M \times 0.250 L = 0.0500$ mol $= 5.30/106$ . Round-trip back to the weighed mass. ✓

Example 7 — Cell H: mixing two solutions of the SAME solute

Worked example Blending two acid batches

You pour together $100 mL$ of $0.500 M$ HCl and $300 mL$ of $0.100 M$ HCl. What is the concentration of the mixture?

Forecast: It's a weighted average, pulled toward the bigger volume ( $0.100$ M batch). Expect a result closer to $0.100$ than to $0.500$ — somewhere around $0.2$ M.

Moles from each batch, then ADD them (same solute, so moles simply pile up): $n_{1} = 0.500 \times 0.100 = 0.0500 mol, n_{2} = 0.100 \times 0.300 = 0.0300 mol$ $n_{total} = 0.0500 + 0.0300 = 0.0800 mol$ Why add moles? No reaction occurs (same substance); particles just join the same pool.
Add the volumes (they merge into one liquid): $V = 0.100 + 0.300 = 0.400 L$ .
Concentration of blend: $c = \frac{0.0800}{0.400} = 0.200 M$ .

Verify: $0.200$ M sits between $0.100$ and $0.500$ , nearer $0.100$ (the larger volume), as forecast. ✓

Common mistake "But the parent said combined volume is irrelevant!"

Where the confusion comes from: in a titration the parent note warned you to use the delivered/pipetted volume of each solution, not the combined volume in the flask. Why the rule flips here: the two situations ask different questions.

In a titration you want to count reacted moles, and each solution's moles are fixed by its own $c$ and its own delivered $V$ — the moment they mix and react, the "concentration of the mixture" is a distraction, so combined volume is irrelevant.
Here nothing reacts, and the very thing you are asked for is the concentration of the final mixture. To get a concentration you must divide total moles by the total (combined) volume. The one-line test: ask "am I counting moles that react (use own volumes), or describing the pooled liquid itself (use combined volume)?" This example is the second kind.

Example 8 — Cell I: back-titration

Worked example Purity of an insoluble antacid (CaCO₃)

A $0.500 g$ chalk tablet (mostly CaCO₃, $M_{r} = 100 g mol^{- 1}$ ) is stirred into $50.0 mL$ of $0.500 M$ HCl (a known excess). The leftover, unreacted acid needs $32.0 mL$ of $0.500 M$ NaOH to neutralise. How many moles of CaCO₃ were in the tablet?

Reactions: $CaCO_{3} + 2 HCl \to CaCl_{2} + H_{2} O + CO_{2}$ (ratio 1 CaCO₃ : 2 HCl) $HCl + NaOH \to NaCl + H_{2} O$ (leftover acid, 1:1)

Forecast: Some acid was eaten by chalk, some survived to meet the NaOH. Reacted = added − survived. Guess whether more than half the acid was consumed before reading the steps.

Total HCl added: $n_{HCl,added} = 0.500 \times 0.0500 = 0.0250 mol$ . Why this step? This is all the acid we put in — the ceiling. Some of it reacts with chalk, the rest survives; we tally the ceiling first.
Leftover HCl (= moles NaOH, 1:1): $n_{HCl,left} = 0.500 \times 0.0320 = 0.0160 mol$ . Why does NaOH volume tell us leftover acid? The NaOH only meets acid that the chalk didn't touch; the leftover HCl + NaOH reaction is 1:1, so leftover acid moles equal the NaOH moles delivered.
Acid that reacted with chalk (conservation of moles): $n_{HCl,reacted} = n_{added} - n_{left} = 0.0250 - 0.0160 = 9.00 \times 1 0^{- 3} mol$ Why subtract? What vanished into the chalk = what we put in − what remained. This is the whole point of the excess method: you can't watch the chalk react, but you can measure what's left over.
Cross the 1:2 ratio bridge to CaCO₃: $\frac{n _{CaCO_{3}}}{1} = \frac{n _{HCl,reacted}}{2} \Rightarrow n_{CaCO_{3}} = \frac{1}{2} (9.00 \times 1 0^{- 3}) = 4.50 \times 1 0^{- 3} mol$ Why halve? From the first reaction, each CaCO₃ consumes two HCl, so the carbonate moles are half the reacted-acid moles.

Verify: Reacted acid $9.00 \times 1 0^{- 3}$ mol is more than half of the $0.0250$ mol added — most acid was consumed, matching a chalky tablet. This same tablet's purity is finished in Ex 10. ✓ See Limiting reagent and excess for why the excess must be known.

Example 9 — Cell J: real-world word problem

Worked example Vinegar strength on a food label

A vinegar is labelled as acetic acid (CH₃COOH). You pipette $10.0 mL$ of it and it requires $16.0 mL$ of $0.500 M$ NaOH to reach the phenolphthalein end point. Reaction is 1:1. What is the molarity of the acetic acid?

Forecast: More NaOH volume than acid volume, at a higher-ish base concentration, 1:1 ⇒ the acid should come out stronger than the base. Do a quick guess before computing.

Strip the words to $c, V, a, b$ : known = NaOH ( $c = 0.500$ M, $V = 16.0$ mL); unknown = acid ( $V = 10.0$ mL); coefficients $a = b = 1$ . Why first? Word problems bury the same four numbers; find them and the machinery is identical to Ex 3.
Moles of NaOH (the known): $n_{NaOH} = c V = 0.500 \times 0.0160 = 8.00 \times 1 0^{- 3} mol$ . Why this step? NaOH is the fully known solution, so we can turn its $c$ and $V$ straight into a mole count with $n = c V$ — the entry onto the moles hub of the map.
Cross the 1:1 ratio bridge: $n_{acid} = n_{NaOH} = 8.00 \times 1 0^{- 3} mol$ . Why equal? The balanced reaction has coefficients $a = b = 1$ , so one acetic acid is consumed per NaOH — the bridge doesn't change the number.
Back out to concentration: $c_{acid} = \frac{n}{V} = \frac{8.00 \times 1 0 ^{- 3}}{0.0100} = 0.800 M$ . Why divide by $0.0100$ L? We now have the acid's moles and its pipetted volume; molarity is moles per litre, so we divide by the acid's own volume in litres.

Verify: $0.800$ M acetic acid — plausible for a table vinegar. It is higher than the $0.500$ M base because the titre volume ( $16$ mL) exceeds the acid volume ( $10$ mL) at a 1:1 ratio, exactly as forecast. ✓

Example 10 — Cell K: exam twist (percent purity)

Worked example Finishing Example 8 — how pure was the tablet?

Using $n_{CaCO_{3}} = 4.50 \times 1 0^{- 3} mol$ from Ex 8, $M_{r} (CaCO_{3}) = 100 g mol^{- 1}$ , and the tablet mass $0.500 g$ , find the percent purity of the tablet.

Forecast: Not all $0.500$ g is CaCO₃ (there's binder holding the tablet together). Expect purity below $100%$ — guess how far below before reading.

Moles → mass of pure CaCO₃ using $m = n M_{r}$ : $m_{CaCO_{3}} = n M_{r} = 4.50 \times 1 0^{- 3} \times 100 = 0.450 g$ Why this step? Purity compares masses, so we convert the carbonate mole count back into grams using the molar mass $M_{r}$ (the exact reverse of Ex 6's grams→moles step).
Percent purity = fraction of the tablet that is genuinely CaCO₃, ×100: $purity = \frac{m _{pure}}{m _{tablet}} \times 100% = \frac{0.450}{0.500} \times 100% = 90.0%$ Why divide by tablet mass? Purity asks: of the whole tablet, what fraction was the real active ingredient? That is (useful mass) ÷ (total mass).

Verify: $0.450 g < 0.500 g$ ⇒ purity $90.0%$ , below $100%$ as forecast, and $0.450$ g of pure carbonate is physically less than the $0.500$ g tablet — consistent, no impossibility. ✓

Recall Quick self-test across the matrix

Dilution: which quantity is conserved? ::: Moles $n = c V$ ; water adds no solute. Which cell needs you to divide by stoichiometric coefficients? ::: Cell D (and any $a \neq = b$ ), e.g. $H_{2} SO_{4} + 2 NaOH$ . As you dilute toward infinite volume, what does $c$ approach and why? ::: $c \to 0$ ; fixed moles spread through unbounded volume. When you mix two batches of the same solute, what do you add? ::: Add the moles, add the volumes, then divide. In a back-titration, reacted moles = ? ::: Moles added (excess) − moles left over (titrated). Percent purity formula? ::: (mass of pure substance ÷ mass of sample) × 100%.

Connections

Parent: Solution stoichiometry — the tools these examples exercise.
Mole concept and Avogadro's number — what $n$ counts.
Balancing chemical equations — where the coefficients $a : b$ come from.
Limiting reagent and excess — the logic behind back-titration.
Acid–base neutralisation — the reactions in Ex 3, 4, 9.
Indicators and pH — spotting the end point.
Standard solutions and primary standards — Ex 6's preparation.