Dilution formula M₁V₁ = M₂V₂: Question bank

Before we start, three words we lean on constantly, each earned in the parent note:

Solute = the stuff dissolved (the sugar, the acid). Solvent = what dissolves it (usually water). Moles = a fixed count of particles — the thing dilution never changes.
Molarity $M = n / V$ = "amount per litre", a ratio, not an amount.

True or false — justify

TRUE / FALSE: Diluting a solution reduces the number of moles of solute.

FALSE. Adding solvent adds zero solute — the mole count is frozen; only the volume grows, so concentration (moles ÷ volume) falls.

TRUE / FALSE: In

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

you may plug

V_{1}

in mL and

V_{2}

in litres.

FALSE. Volume appears on both sides and cancels only if the unit matches; mixing mL and L breaks the cancellation and gives an answer off by 1000×.

TRUE / FALSE: If you double the volume by adding water, the molarity is exactly halved.

TRUE. Since

M V

is constant,

M

and

V

are inversely proportional — double

V

forces

M

to fall to one half.

TRUE / FALSE: The formula works for ppm and normality too, not just molarity.

TRUE. Any "amount ÷ volume" unit conserves its amount on dilution, so

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

holds for ppm, normality, etc. — see ppm and parts-per notation and Normality and N1V1 = N2V2.

TRUE / FALSE:

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

still holds if the acid neutralises a base you added.

FALSE. Neutralisation consumes solute, so moles are no longer conserved; you need reaction stoichiometry — see Titration and neutralisation.

TRUE / FALSE: When you dilute, the volume of water you add equals

V_{2}

FALSE.

V_{2}

is the total final volume. Water added

= V_{2} - V_{1}

, since the original

V_{1}

is already part of the final mixture.

TRUE / FALSE: Making

500 mL

of dilute acid from concentrated stock is NOT a dilution because you're "building up" a solution.

FALSE. You still take a fixed amount of solute and spread it into more solvent — that's dilution, just read backwards to find the stock volume.

TRUE / FALSE: If two solutions have equal molarity, mixing them gives that same molarity.

TRUE. Mixing equal concentrations keeps the ratio unchanged — the pooled moles and pooled volume scale together, so concentration is untouched.

Why questions

WHY do the volume units cancel, letting you skip converting to litres?

V

sits on both sides of

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

; any common factor (like mL→L, a factor of 1000) divides out, so only the ratio of volumes matters.

WHY is it moles — and not molarity or mass of water — that stays constant?

Adding pure solvent introduces no new solute particles and removes none, so the count (moles) is the untouched quantity; molarity changes precisely because volume does.

WHY does a tiny volume of

12 M

stock make a large volume of weak acid?

A few mL of very concentrated acid already carries a lot of moles; spreading those moles over a big volume yields a low concentration — strength traded for volume.

WHY can the same equation appear as

N_{1} V_{1} = N_{2} V_{2}

Normality is also "amount ÷ volume" (equivalents instead of moles); dilution conserves equivalents just as it conserves moles, so the identical algebra applies.

WHY must you NOT use this formula at a titration endpoint?

At the endpoint acid and base react and are consumed — moles are destroyed, breaking the conservation the formula depends on; use stoichiometry instead.

Edge cases

EDGE: What does

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

say if you add no water at all?

Then

V_{2} = V_{1}

, forcing

M_{2} = M_{1}

— a "zero dilution" leaves concentration unchanged, exactly as expected.

EDGE: Can

V_{2}

ever be smaller than

V_{1}

in a genuine dilution?

No. Dilution only adds solvent, so

V_{2} \geq V_{1}

; a smaller

V_{2}

would mean concentrating (removing solvent), a different process the formula still balances but not by "adding water".

EDGE: If you keep adding water forever, what does

M_{2}

approach?

V_{2} \to \infty

M_{2} = M_{1} V_{1} / V_{2} \to 0

. Concentration approaches (but never reaches) zero — you can dilute indefinitely but never fully to nothing.

EDGE: You start with

0 M

(pure water) and add more water — what happens?

M_{1} = 0

gives

M_{2} = 0

for any

V_{2}

: diluting nothing yields nothing. The formula gracefully handles the degenerate "no solute" case.

EDGE: Two different solutes (say salt and sugar) are both present — does one formula cover both?

No single equation covers both; apply

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

separately to each solute, since each conserves its own moles independently.

EDGE: What if some solvent evaporates after dilution?

Then

V_{2}

shrinks while moles stay fixed, so

M

rises — the formula still holds with the new smaller

V_{2}

; evaporation is just "reverse dilution".

Recall One-line summary to carry away

Conserve moles, never molarity. $M_{1} V_{1} = M_{2} V_{2}$ balances the amount on both sides; every trap above is really someone forgetting that concentration is a ratio and the solute count is what's frozen.

Connections

Molarity and concentration units — why molarity is a ratio, not an amount.
The Mole concept — moles are the conserved bookkeeping quantity.
Normality and N1V1 = N2V2 — same conservation, equivalents instead of moles.
Titration and neutralisation — the case where the formula must NOT be used.
ppm and parts-per notation — dilution logic for any amount/volume unit.

Question bank — Dilution formula M₁V₁ = M₂V₂