Dilution formula M₁V₁ = M₂V₂: Visual walkthrough

This page is the visual companion to the parent note. If any word below feels new, the prerequisites live in Molarity and concentration units and The Mole concept.

Step 1 — Draw a solution and name its three parts

WHAT. A solution is a liquid with something dissolved in it. Three things live inside the glass:

solute — the stuff dissolved (drawn as red dots),
solvent — the liquid doing the dissolving (usually water, drawn as the clear background),
volume — how much liquid there is in total (the height of the fill).

WHY. Before we can talk about a formula, we must have names for the objects it talks about. Every symbol later ( $M$ , $V$ , $n$ ) points back to one of these pictures. We earn them here.

PICTURE. In the figure, the red dots are the solute particles. Count them: there are exactly 8. Hold that number in your head — the whole derivation is a fight to keep it at 8.

Step 2 — Define "how crowded": molarity

WHAT. We invent a number that measures how crowded the red dots are. Take the number of dots (call it $n$ , for "number of moles" — one mole is just a fixed huge count of particles, see The Mole concept) and divide by the volume they float in:

crowdedness M = \frac{n how many red dots}{how much liquid V}

WHY divide, not subtract? Because "crowded" is a ratio: 8 dots in a thimble is crowded; 8 dots in a bathtub is not. Dividing dots by liquid is exactly the question "how many dots share each unit of space?" Subtraction would have no meaning here — dots and millilitres are different things and cannot be subtracted.

PICTURE. Two glasses, the same 8 red dots. Left glass: small volume → dots packed tight → high $M$ . Right glass: big volume → dots spread out → low $M$ . Same dots, different crowding.

Step 3 — Flip the definition: count dots from crowdedness

WHAT. We rearrange the definition to solve for the dot-count $n$ . Multiply both sides by $V$ :

n = crowdedness M \times how much liquid V

WHY. In dilution the thing we care about protecting is the number of dots $n$ . So we want a formula that spits out $n$ . Crowdedness alone can't do it — a crowded thimble and a spread-out bathtub can hold the same total dots. You need crowdedness × size to recover the actual count.

PICTURE. The rectangle trick: draw $M$ as the height of a box and $V$ as its width. The area of the box is $n = M \times V$ . This is our master picture — every step from here is just "the area of this box didn't change."

Step 4 — The "before" box

WHAT. Take the starting, concentrated solution. Its crowdedness is $M_{1}$ and its volume is $V_{1}$ . The dot-count before dilution is the area of this box:

n_{before} = start crowdedness M_{1} \times start volume V_{1}

WHY the subscript 1? "1" means before, first in time. It is just a label so we never confuse the start with the end.

PICTURE. A tall, thin red box: high $M_{1}$ (tall) because the solution is concentrated, small $V_{1}$ (thin) because there is little liquid. Its area is the number of dots — still our 8.

Step 5 — Pour in water: the "after" box

WHAT. Now add solvent (water). The volume grows to $V_{2}$ , and because the same dots are now spread through more liquid, the crowdedness falls to $M_{2}$ . The dot-count after is:

n_{after} = end crowdedness M_{2} \times end volume V_{2}

WHY does $M$ drop when $V$ grows? Water carries zero red dots. Pouring it in stretches the box wider ( $V_{2} > V_{1}$ ) without adding any area's worth of dots — so to keep the same area, the box must get shorter ( $M_{2} < M_{1}$ ).

PICTURE. A short, wide box next to the tall thin one. Wider ( $V_{2}$ bigger), shorter ( $M_{2}$ smaller) — but shade both: the shaded area is identical. That equal shading is the whole idea.

Step 6 — The conservation law: same dots, so same area

WHAT. Water added no solute. So the dot-count before and after must be identical:

n_{before} = n_{after}

WHY. This is the physical heart. Adding solvent is like moving the same 8 marbles from a small tray to a big tray — you did not create or vanish a single marble. The count is frozen. Notice this argument used only "solute is conserved" — it did not assume the volumes added up neatly, which is why the formula survives even non-ideal mixing.

PICTURE. Both boxes overlaid, the tall-thin one and the short-wide one, with their equal areas highlighted in red. Different shapes, one area.

Step 7 — Substitute and read off the formula

WHAT. Replace each side of $n_{before} = n_{after}$ with its box area from Steps 4 and 5:

area of before-box M_{1} V_{1} = area of after-box M_{2} V_{2}

WHY this is the finish line. The equation is the sentence "the before-box and the after-box have the same area," written in symbols. Nothing new was assumed — we only renamed the picture.

PICTURE. The two boxes with a bold red equals sign between their areas.

Step 8 — Edge cases: what if you barely dilute, or dilute infinitely?

WHAT. Push the picture to its limits.

No water added: $V_{2} = V_{1}$ . Then $M_{2} = M_{1}$ — nothing changed. The box is untouched. ✔
Huge water added: $V_{2} \to \infty$ . Then $M_{2} = M_{1} V_{1} / V_{2} \to 0$ . Spread 8 dots through an ocean and the crowding approaches zero — but it never hits exactly zero, because the 8 dots are still in there. ✔
Zero starting concentration: $M_{1} = 0$ (pure water). Then $M_{1} V_{1} = 0$ , so $M_{2} V_{2} = 0$ , so $M_{2} = 0$ . Diluting nothing gives nothing. The formula stays sane. ✔

WHY show these? A formula you trust only in the middle is a formula you don't trust. Watching the box behave correctly at the extremes is how we know it never lies.

PICTURE. A strip of three boxes: unchanged (same shape), stretched-flat (area same, height nearly 0), and empty (no red at all).

The one-picture summary

Everything above collapses into a single frame: two boxes of equal red area, one tall-thin (before), one short-wide (after), joined by an equals sign — with the arrow reminding you water only widened the box, never changed its area.

Recall Feynman retelling of the whole walkthrough

I drew a glass with 8 red sugar-dots in it. I called "how crowded the dots are" $M$ and "how much liquid" $V$ , and I noticed that crowdedness times liquid gives back the actual dot-count $n$ — like height times width gives the area of a box. So I drew the solution as a box: a tall thin box means very crowded, a short wide box means spread out. When I pour in water, the box gets wider (more liquid) but water brings no dots, so the box must get shorter to keep the same area. Same area = same dots = the law of dilution. Writing "before-box area = after-box area" in symbols is exactly $M_{1} V_{1} = M_{2} V_{2}$ — as long as I measure $V_{1}$ and $V_{2}$ in the same unit. And I never assumed the volumes add up neatly, so it holds even when water and solution don't mix to exactly the summed volume. When I stretch the box to the extremes — no water, an ocean of water, no dots at all — the box always behaves, so I trust it everywhere.

Active recall

The area of the box represents

the moles of solute

n = M \times V

(the conserved red-dot count).

Why does molarity fall as volume rises in dilution?

Water adds no solute, so the same dots spread through more liquid — the box widens but keeps its area, forcing the height (molarity) down.

V_{2} \to \infty

, what happens to

M_{2}

It approaches 0 but never reaches it, since the fixed

n

dots are always present.

Must

V_{1}

and

V_{2}

share a unit?

Yes —

V

appears on both sides so it cancels only if both are the same unit (both mL or both L); never mix.

Does the derivation assume

V_{2} = V_{1} +

water added?

No — it only assumes the solute is conserved, so it holds even for non-ideal mixtures where volumes are not strictly additive.

50 mL of 2 M HCl diluted to 250 mL gives what molarity?

M_{2} = (2) (50) /250 = 0.4

Connections

Molarity and concentration units — the box's height, $M = n / V$ , is built here.
The Mole concept — the red dots are moles $n$ , the conserved area.
Normality and N1V1 = N2V2 — same "equal-area" picture for equivalents.
Titration and neutralisation — where dots are consumed, so the area is NOT conserved.
ppm and parts-per notation — same box logic for any amount-per-volume unit.