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Mixing Two Solutions (C₁V₁ + C₂V₂ = C_f V_f)

Also known as blending two solutions · combining two stock solutions · mixed concentration · C1V1 + C2V2

Cf=C1V1+C2V2V1+V2C_f = \frac{C_1 V_1 + C_2 V_2}{V_1 + V_2}

Worked example: 2.00 L of 0.500 M plus 3.00 L of 1.50 M → 1.10 M — press Try an example to run it live, then adjust anything.

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See also
Dilution Equation (C1V1 = C2V2) — The special case where the second stream is pure water: C₂ = 0 and this collapses to C₁V₁ = C₂V₂.Molarity (C = n/V) — Where each C comes from, if you have masses and volumes rather than concentrations.

Learning zone

The dilution equation, C1V1=C2V2C_1V_1 = C_2V_2, assumes the thing you add carries no solute. Most of the time that is water and the assumption is fine. But an operator with a half-drum of 30% inhibitor and a full drum of 12% is not diluting anything — they are blending, and both streams bring solute to the tank. Conserve the moles rather than the concentration and the answer falls out: total solute in equals total solute out, so C1V1+C2V2=Cf(V1+V2)C_1V_1 + C_2V_2 = C_f(V_1+V_2).

Worked: pour 2.00 L of 0.500 M sodium chloride into 3.00 L of 1.50 M. Solute in stream one, 2.00 × 0.500 = 1.00 mol. Stream two, 3.00 × 1.50 = 4.50 mol. Total 5.50 mol in 5.00 L, so the blend is 1.10 M. Notice where that lands — between 0.500 and 1.50, closer to the stronger one because there was more of it. A blend never leaves the interval between its two ingredients. If your answer does, you have made an arithmetic error, and the solver will say so rather than hand it to you.

Run backwards, this is the recipe question: how much of the strong stock do I add to what I already have? Solving for V₁ gives V₂(C_f − C₂)/(C₁ − C_f), which is the algebraic twin of the Pearson square that feed and fertiliser blenders draw by hand. The two differences from the square are that this one carries units and that it will refuse an impossible target instead of quietly producing a negative volume.

The assumption to watch is that the volumes add. V_f = V₁ + V₂ is very nearly true for dilute aqueous solutions and quite false for concentrated ones. Mix 50 mL of ethanol with 50 mL of water and you get about 96 mL, not 100 — the small water molecules pack into gaps in the hydrogen-bonded ethanol network. Concentrated sulfuric acid contracts similarly, and it heats violently while doing it. Where the contraction matters, make the batch up to a measured final volume in a flask and use the dilution equation on the result, rather than trusting the sum.

Mixing Two Solutions (C₁V₁ + C₂V₂ = C_f V_f)
Cf=C1V1+C2V2V1+V2C_f = \frac{C_1 V_1 + C_2 V_2}{V_1 + V_2}
Where
  • CfC_f= Final concentration of the blend (M)
  • C1C_1= Concentration of solution 1 (M)
  • V1V_1= Volume of solution 1 (L)
  • C2C_2= Concentration of solution 2 (M)
  • V2V_2= Volume of solution 2 (L)
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