Dilution formula M₁V₁ = M₂V₂: Worked examples

Before line one: recall the only fact we ever use. Molarity $M$ means "moles of solute divided by litres of solution", written $M = \frac{n}{V}$ , so $n = M \cdot V$ . The single unbreakable idea is: ==adding or removing pure solvent (water) never changes the number of solute particles $n$ ==. Everything below is that sentence, rearranged.

The scenario matrix

Every dilution question is one cell of this grid. The columns are which unknown you solve for; the rows are what physical situation creates the two states.

Situation ↓ \ Unknown →	Solve for $M_{2}$ (final conc.)	Solve for $V_{1}$ (start vol.)	Solve for volume of water
Add water (dilute)	Ex 1	Ex 4	Ex 2
Remove water (concentrate ↑)	Ex 3	—	Ex 3
Mix two solutions	Ex 5 (weighted avg)	—	—
Degenerate / limiting	Ex 6 (add nothing; add ∞ water)	—	—
Trap: reaction consumes solute	Ex 7 (formula FAILS)	—	—
Non-molarity unit (ppm)	Ex 8 (same logic, different unit)	—	—

We hit every cell (the two dashes are combinations that are just Ex 4 / Ex 2 re-labelled). Each example says which cell it fills.

Example 1 — Add water, find $M_{2}$ · (Dilute → $M_{2}$ )

Worked example Statement

You take $25 mL$ of $4 M$ sulfuric acid and add water until the total is $200 mL$ . What is the final molarity?

Forecast: The volume grew from $25$ to $200$ mL — that is an $8 \times$ jump. So guess the concentration falls by $8 \times$ : about $4/8 = 0.5 M$ . Let's confirm.

Step 1. List what we know: $M_{1} = 4$ , $V_{1} = 25 mL$ , $V_{2} = 200 mL$ , unknown $M_{2}$ . Why this step? Three of the four slots in $M_{1} V_{1} = M_{2} V_{2}$ are known, so the fourth is forced.

Step 2. Rearrange the formula for the unknown: divide both sides by $V_{2}$ . $M_{2} = \frac{M _{1} V _{1}}{V _{2}} = \frac{4 \times 25}{200}$ Why this step? We want $M_{2}$ alone; dividing by $V_{2}$ peels it off the right side.

Step 3. Compute: $M_{2} = \frac{100}{200} = 0.5 M$ . Why this step? Just arithmetic on the numbers from Step 2.

Verify: Volume ratio $\frac{V _{2}}{V _{1}} = \frac{200}{25} = 8$ , and concentration ratio $\frac{M _{1}}{M _{2}} = \frac{4}{0.5} = 8$ . They match — concentration dropped by exactly the factor the volume rose. Units: mL cancelled mL, leaving M. ✔

Example 2 — Add water, find water added · (Dilute → water)

Worked example Statement

How many millilitres of water must be added to $120 mL$ of $2.5 M$ KCl to bring it down to $0.75 M$ ?

Forecast: Concentration must fall by $2.5/0.75 \approx 3.33 \times$ , so the final volume is about $120 \times 3.33 \approx 400 mL$ , meaning roughly $280 mL$ of water. Careful — the formula gives the total, not the water.

Step 1. Find the final total volume $V_{2}$ : $V_{2} = \frac{M _{1} V _{1}}{M _{2}} = \frac{2.5 \times 120}{0.75}$ . Why this step? The formula relates the two total volumes. Water added is a separate, later subtraction.

Step 2. Compute: $V_{2} = \frac{300}{0.75} = 400 mL$ . Why this step? Arithmetic; this is the whole solution after topping up.

Step 3. Water added $= V_{2} - V_{1} = 400 - 120 = 280 mL$ . Why this step? You already had $120$ mL of liquid; only the extra $280$ mL is new water.

Verify: Check the diluted solution really is $0.75$ M: $M_{2} = \frac{2.5 \times 120}{400} = \frac{300}{400} = 0.75$ M. ✔ And $280$ mL sits between "some water" and "less than the full $400$ ", as expected.

Example 3 — Remove water (concentrate up) · (Concentrate → $M_{2}$ and water removed)

Worked example Statement

$500 mL$ of $0.20 M$ glucose is left in an open dish and $300 mL$ of water evaporates. What is the new concentration, and did we add or remove solvent?

Forecast: We removed water, so volume shrinks $500 \to 200 mL$ and concentration should rise. Volume fell by $2.5 \times$ , so guess $\approx 0.20 \times 2.5 = 0.5 M$ .

Step 1. New total volume $V_{2} = 500 - 300 = 200 mL$ . Why this step? Evaporation removes solvent, so the total volume goes down by the amount lost — but the glucose molecules stay in the dish.

Step 2. Apply $M_{1} V_{1} = M_{2} V_{2}$ (still valid — glucose was neither made nor destroyed, just less water): $M_{2} = \frac{M _{1} V _{1}}{V _{2}} = \frac{0.20 \times 500}{200}$ Why this step? Conservation of moles works for removing solvent exactly as for adding it — the moles are frozen either way.

Step 3. Compute: $M_{2} = \frac{100}{200} = 0.5 M$ . Why this step? Arithmetic; concentration rose, as forecast.

Verify: Moles before $= 0.20 \times 0.500 L = 0.10 mol$ . Moles after $= 0.50 \times 0.200 L = 0.10 mol$ . Identical — the same $0.10$ mol of glucose, now crowded into less water. ✔

Example 4 — Add water, find starting stock volume · (Dilute → $V_{1}$ )

Worked example Statement

You need $250 mL$ of $0.05 M$ HNO₃. Your stock bottle is $15.8 M$ . What volume of stock do you measure out (then top up to $250 mL$ )?

Forecast: Stock is enormously stronger ( $15.8$ vs $0.05$ ) — a ratio of about $316 \times$ . So we need a tiny sip of stock: about $250/316 \approx 0.8 mL$ .

Step 1. Identify the unknown as $V_{1}$ : here state 1 is the stock, state 2 is the final dilute solution. Known: $M_{1} = 15.8$ , $M_{2} = 0.05$ , $V_{2} = 250 mL$ . Why this step? "Before" always means the more concentrated stock; "after" the diluted target.

Step 2. Rearrange for $V_{1}$ : $V_{1} = \frac{M _{2} V _{2}}{M _{1}} = \frac{0.05 \times 250}{15.8}$ . Why this step? Divide both sides by $M_{1}$ to isolate the stock volume.

Step 3. Compute: $V_{1} = \frac{12.5}{15.8} \approx 0.79 mL$ of stock, then add water up to $250$ mL. Why this step? Arithmetic; the answer is tiny, matching the forecast.

Verify: Plug back: $M_{2} = \frac{15.8 \times 0.79}{250} = \frac{12.48}{250} \approx 0.05 M$ . ✔ Sanity: a very strong acid needs only a droplet to make a weak one — never trust an answer of "hundreds of mL of stock" here.

Example 5 — Mix two solutions · (Mix → weighted average)

Worked example Statement

You pour together $200 mL$ of $3 M$ NaCl and $300 mL$ of $1 M$ NaCl. What is the concentration of the mixture? (Assume volumes just add.)

Forecast: The answer must land between $1$ and $3$ M — never below, never above. Since there is more of the weak solution, expect it closer to $1$ than to $3$ : maybe about $1.8 M$ .

Step 1. This is NOT a single $M_{1} V_{1} = M_{2} V_{2}$ . Instead, conserve total moles by adding each solution's moles: $n_{total} = M_{a} V_{a} + M_{b} V_{b} = (3) (200) + (1) (300)$ Why this step? Each solution carries its own moles; mixing pools them. You add moles, never molarities.

Step 2. Compute total "moles" (using mL, we'll divide by mL too so units are consistent): $n_{total} = 600 + 300 = 900$ (mmol). Why this step? Arithmetic; keeping mL throughout means the answer still comes out in mol/L.

Step 3. Total volume $V_{total} = 200 + 300 = 500 mL$ . Why this step? Volumes are additive; that is the "spread into" step.

Step 4. Final concentration $M_{mix} = \frac{n _{total}}{V _{total}} = \frac{900}{500} = 1.8 M$ . Why this step? Molarity is total moles ÷ total volume — the definition, applied to the pooled system.

Verify: $1.8$ lies between $1$ and $3$ , and closer to $1$ (there was more of it). ✔ If you had wrongly added the molarities you'd get $4$ M — impossible, since no part was stronger than $3$ M. That is the trap the parent note's mistake box warns about.

Example 6 — Degenerate & limiting cases · (Add nothing; add infinite water)

Worked example Statement

Two edge cases on $100 mL$ of $2 M$ solution: (a) Add zero water. (b) Add enormously much water — say dilute to $100 L$ . What are the concentrations, and what do these limits tell us?

Forecast: (a) Adding nothing changes nothing → still $2 M$ . (b) A huge volume should crush the concentration toward almost zero.

Step 1 (case a). $V_{2} = V_{1} = 100 mL$ , so $M_{2} = \frac{M _{1} V _{1}}{V _{2}} = \frac{2 \times 100}{100} = 2 M$ . Why this step? When $V_{2} = V_{1}$ the ratio $V_{1} / V_{2} = 1$ , so $M_{2} = M_{1}$ . The formula correctly does nothing when you do nothing — a good self-check that it isn't magic.

Step 2 (case b). Convert: $100 L = 100000 mL$ . Then $M_{2} = \frac{2 \times 100}{100000} = \frac{200}{100000} = 0.002 M$ . Why this step? We must keep the same unit on both sides, so express both volumes in mL before dividing.

Step 3 (the limit). As $V_{2} \to \infty$ , $M_{2} = \frac{M _{1} V _{1}}{V _{2}} \to 0$ . Why this step? The numerator $M_{1} V_{1}$ (the fixed moles) is constant while the denominator explodes — a constant over a growing number heads to zero. Physically: the same solute smeared through an ocean is essentially undetectable, but never exactly zero (the solute never disappears).

Verify: Case (a): moles before $= 2 \times 0.1 = 0.2$ mol, after $= 2 \times 0.1 = 0.2$ mol — unchanged. ✔ Case (b): moles $= 0.002 \times 100 = 0.2$ mol still. ✔ The conserved $0.2$ mol survives every scenario — that is the whole point.

Example 7 — The trap: a reaction happens · (Formula FAILS)

Worked example Statement

You mix $100 mL$ of $1 M$ HCl with $100 mL$ of $1 M$ NaOH. A student writes $M_{1} V_{1} = M_{2} V_{2}$ and claims the HCl is now " $0.5$ M". Is this right?

Forecast: Something smells wrong — HCl and NaOH react: $HCl + NaOH \to NaCl + H_{2} O$ . Solute is being consumed, not merely spread out.

Step 1. Count moles: HCl $= 1 \times 0.100 = 0.10 mol$ ; NaOH $= 1 \times 0.100 = 0.10 mol$ . Why this step? We must check whether the "solute" still exists after mixing before any dilution formula can apply.

Step 2. They react $1 : 1$ and are equal, so all $0.10$ mol of HCl is neutralised. Remaining HCl $= 0.10 - 0.10 = 0 mol$ . Why this step? $M_{1} V_{1} = M_{2} V_{2}$ assumes the moles of that species are conserved. Here they drop to zero — the assumption is broken.

Step 3. So the true HCl concentration afterward is $0 M$ , not $0.5$ M. The correct tool for "who used up whom" is Titration and neutralisation / Normality and N1V1 = N2V2 at the equivalence point. Why this step? When solute is destroyed by reaction, you switch from conservation-of-moles-of-one-species to stoichiometry.

Verify: Equivalence check via $N_{1} V_{1} = N_{2} V_{2}$ (both are $1$ N here): $1 \times 100 = 1 \times 100$ — they exactly cancel, confirming zero excess of either. ✔ The naive $0.5$ M answer would have been the dilution result if nothing reacted — it's the wrong model.

Example 8 — Same logic, different unit · (Non-molarity: ppm)

Worked example Statement

A tank of pool water tests at $6 ppm$ chlorine. You pump in fresh water, changing the volume from $30000 L$ to $50000 L$ (no chlorine added). New ppm?

Forecast: More water, same chlorine → ppm falls. Volume rose by $50/30 \approx 1.67 \times$ , so guess $6/1.67 \approx 3.6 ppm$ .

Step 1. Recognise ppm is also an "amount ÷ volume" measure (see ppm and parts-per notation), so the dilution formula works with ppm in place of $M$ : $C_{1} V_{1} = C_{2} V_{2}$ Why this step? The derivation only ever needed "amount is conserved". Any concentration-per-volume unit obeys it.

Step 2. Solve for $C_{2}$ : $C_{2} = \frac{C _{1} V _{1}}{V _{2}} = \frac{6 \times 30000}{50000}$ . Why this step? Isolate the unknown final concentration.

Step 3. Compute: $C_{2} = \frac{180000}{50000} = 3.6 ppm$ . Why this step? Arithmetic; matches the forecast.

Verify: Chlorine "amount" before $\propto 6 \times 30000 = 180000$ ; after $\propto 3.6 \times 50000 = 180000$ . Equal — no chlorine created or lost. ✔

Active recall

Which volume is the total, not the water added?

V_{2}

, the final total volume; water added

= V_{2} - V_{1}

Mixing 200 mL of 3 M with 300 mL of 1 M gives what concentration?

\frac{600 + 300}{500} = 1.8

Does evaporation break

M_{1} V_{1} = M_{2} V_{2}

No — removing solvent still conserves solute moles, so the formula holds.

When does the dilution formula fail?

When the solute is chemically consumed (e.g. HCl + NaOH neutralisation) — use stoichiometry instead.

V_{2} \to \infty

, what happens to

M_{2}

It tends to

0

but never reaches it (fixed moles over growing volume).

Can you use the formula with ppm?

Yes — any amount-per-volume unit obeys

C_{1} V_{1} = C_{2} V_{2}

Connections

Parent: Dilution formula — the derivation these examples exercise.
Molarity and concentration units — the $M = n / V$ definition behind every step.
The Mole concept — moles $n$ are the conserved quantity across all cases.
Normality and N1V1 = N2V2 — the tool for Example 7's reacting case.
Titration and neutralisation — where solute is consumed, not diluted.
ppm and parts-per notation — Example 8's alternative unit.