Dilution稀释

Big Idea · Scale, Proportion, and Quantity (SPQ)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 3~35 min

Learning goals

  • Use M₁V₁ = M₂V₂ to find the molarity after dilution or the volume of stock solution needed
  • Convert a stock solution’s mass percent and density into molarity
  • Describe how to make a dilute solution from a stock solution, and how to dilute concentrated acid safely

Before you start

  • Mass percent of a solute (1.7.6)
  • Molarity (2.9.2)

Stock solutions

Acids and ammonia arrive in the lab as concentrated stock solutions浓溶液（储备液）, and you dilute them before use. In 1.7.6 you diluted by mass: the mass of solute stays the same. But liquids are measured by volume, so it’s easier to work with molarity.

The dilution equation

Adding water makes the volume larger, but the moles of solute stay the same:

M1V1=M2V2M_1V_1 = M_2V_2

Here 1 is the concentrated solution and 2 is the dilute one. The two volumes just need the same unit, so mL on both sides is fine. V₂ is the total volume after dilution, not the volume of water you add.

The ratio V₂/V₁ is the dilution factor稀释倍数: if the volume grows 10 times, the molarity drops to 1/10.

When you mix two solutions of the same solute, the moles add: M₁V₁ + M₂V₂ = MmixVmix. Unless you’re told otherwise, you can treat the volumes of dilute solutions as adding.

On the left is the stock solution you measured out; on the right is the solution after dilution (drawn as blue CuSO₄). Count the dots on each side:

Try it: dilution and stock solutions
Solve for

    Molarity of a stock solution

    A stock bottle’s label usually gives the mass percent and the density. Work with 1 L (1000 mL) of solution: its mass is 1000 × density, the mass of solute is that times the mass fraction, and dividing by the molar mass gives the moles of solute in that liter. So

    M=1000×density×mass fractionmolar massM = \frac{1000 \times \text{density} \times \text{mass fraction}}{\text{molar mass}}

    with density in g/mL and molar mass in g/mol. For example, concentrated sulfuric acid is about 96% H₂SO₄ by mass with a density of 1.84 g/mL, so its molarity is 1000 × 1.84 × 0.96 ÷ 98.08 = 18.0 M.

    Making a dilute solution from a stock

    The steps match 2.9.2, except you measure a volume instead of weighing: calculate, measure the stock with a volumetric pipet, add it to water in a beaker and let it cool, transfer to a volumetric flask, rinse, fill to the mark, and mix.

    When diluting concentrated acid, always pour the acid slowly into the water while stirring: “Always add acid to water.” Concentrated sulfuric acid releases a lot of heat as it dissolves. If you pour water into the acid, the less dense water floats on top, boils instantly, and spatters acid.

    This lesson uses the atomic masses on the AP periodic table (H 1.008, N 14.01, O 16.00, Na 22.99, S 32.06, Cl 35.45).

    How Chinese textbooks differ

    • Chinese textbooks write the dilution equation as c(浓)·V(浓) = c(稀)·V(稀), “concentrated” and “dilute,” with c for molarity and units of mol/L.
    • Chinese school labs usually measure the stock with a graduated cylinder (to 0.1 mL). US labs more often use a volumetric pipet (to 0.01 mL).
    • Chinese problems often quote the classic label for concentrated sulfuric acid, 98% and 1.84 g/cm³, which gives 18.4 mol/L, and write the conversion as c = 1000ρw/M.
    • US labs often do a serial dilution逐级稀释: dilute 10.00 mL to 100.0 mL, then repeat with the new solution, so each step lowers the molarity by a factor of 10. For very dilute solutions this is far more accurate than one big dilution.

    Example 1: diluting sulfuric acid

    How many milliliters of 18.0 M H₂SO₄ do you need to make 250.0 mL of 1.00 M H₂SO₄? How should you do it?

    Step-by-step solution

    1. The rule. Adding water doesn’t change the moles of H₂SO₄: M₁V₁ = M₂V₂.

    2. Solve for V₁. Both volumes are in mL, so there is no need to convert to liters:

      V1=M2V2M1=1.00×250.018.0=13.9 mL\begin{aligned} V_1 &= \frac{M_2 V_2}{M_1} \\ &= \frac{1.00 \times 250.0}{18.0} \\ &= \boxed{13.9\ \text{mL}} \end{aligned}

    3. Measure. Measure 13.9 mL of the concentrated acid with a pipet or a small graduated cylinder; a 100-mL cylinder would be far less precise.

    4. Dilute: acid into water. Put about 100 mL of water in a beaker and pour the acid in slowly down the side, stirring all the time. Dissolving concentrated H₂SO₄ releases a lot of heat; if you poured water into the acid, the water would float on top, boil instantly, and spatter acid.

    5. Cool, then transfer. Wait until the solution is at room temperature before transferring it to a 250 mL volumetric flask. A warm solution shrinks as it cools, so filling to the mark while warm makes the final solution too concentrated.

    6. Rinse and fill to the mark. Rinse the beaker and stirring rod into the flask two or three times so all the acid gets there. Add water until the bottom of the meniscus touches the mark, using a dropper for the last few milliliters, then stopper and mix.

    Example 2: the molarity of concentrated hydrochloric acid

    Concentrated hydrochloric acid is 36.0% HCl by mass, with a density of 1.18 g/mL. (a) What is its molarity? (b) If you dilute 10.0 mL of it to 100.0 mL, what is the new molarity?

    Step-by-step solution

    1. The idea. Mass percent is about mass; molarity is moles per liter of solution. Working with exactly 1 L (1000 mL) of solution makes it easy.

    2. Mass of HCl in 1 L.

      m(soln)=1000×1.18=1180 gm(HCl)=1180×0.360=424.8 g\begin{aligned} m(\text{soln}) &= 1000 \times 1.18 \\ &= 1180\ \text{g} \\ m(\ce{HCl}) &= 1180 \times 0.360 \\ &= 424.8\ \text{g} \end{aligned}

    3. Molarity.

      n(HCl)=424.8/36.46=11.65 molM=11.65/1 L≈11.7 M\begin{aligned} n(\ce{HCl}) &= 424.8 / 36.46 \\ &= 11.65\ \text{mol} \\ M &= 11.65 / 1\ \text{L} \approx \boxed{11.7\ \text{M}} \end{aligned}

      In one step: 1000 × 1.18 × 0.360 ÷ 36.46 = 11.7 M.

    4. Dilute. The moles of HCl don’t change:

      M2=11.65×10.0100.0=1.17 M\begin{aligned} M_2 &= \frac{11.65 \times 10.0}{100.0} \\ &= \boxed{1.17\ \text{M}} \end{aligned}

    5. A common mistake. Reading 36.0% as “36.0 g of HCl per 100 mL.” Mass percent is a mass ratio, so use the density to turn the volume of solution into a mass.

    Practice: what is the molarity after dilution, or the volume of stock needed?
    1. 10.0 mL of 1.00 M solution diluted to 100.0 mL: new molarity
    2. 50.0 mL of 2.00 M solution diluted to 500.0 mL: new molarity
    3. 100.0 mL of 1.00 M solution diluted to 200.0 mL: new molarity
    4. Volume of 1.00 M stock needed for 100.0 mL of 0.100 M
    5. Volume of 2.00 M stock needed for 250.0 mL of 0.200 M
    6. Volume of 5.00 M stock needed for 500.0 mL of 0.500 M
    7. A 0.200 M solution diluted by a factor of 2: new molarity
    8. 100.0 mL of 0.100 M NaCl mixed with 100.0 mL of 0.300 M NaCl (volumes add): molarity of the mixture
    9. 10.0 mL of 5.00 M solution diluted to 100.0 mL: new molarity

    Common mistakes

    Mistake 1: “V₂ is the volume of water you add.” V₂ is the total volume after dilution. To dilute 10.0 mL to 100.0 mL, you don’t add 100.0 mL of water; you add water up to the 100.0 mL mark.

    Mistake 2: “It doesn’t matter whether you pour water into acid or acid into water.” Concentrated sulfuric acid releases so much heat that water poured onto it boils and spatters acid. Always add the acid slowly to water, stirring.

    Mistake 3: “You can put mass percent straight into M₁V₁ = M₂V₂.” Mass percent goes with mass of solution, not volume. If you’re given mass percent and need volumes, first convert to molarity using the density.

    Quick check

    You dilute 20.0 mL of 5.00 M HCl to 500.0 mL, then pour out 100.0 mL. What is the molarity of the portion you poured out?

    Concentrated aqueous ammonia is 28% NH₃ by mass with a density of 0.90 g/mL. What is its molarity?

    Summary

    • Diluting doesn’t change the moles of solute: M₁V₁ = M₂V₂, where V₂ is the total volume after dilution.
    • The molarity of a stock solution comes from its density and mass percent: take 1 L, find the mass of solute, and convert to moles.
    • To dilute a stock: calculate, measure, dilute and cool, transfer, rinse, fill to the mark, and mix. Always add acid to water.
    On to AP (for later)

    The AP equation sheet gives only M = nsolute/Lsolution, not a separate dilution formula, so you derive M₁V₁ = M₂V₂ from the fact that moles of solute stay the same. A common AP lab question: dilute a stock solution into a series of standards with a pipet and volumetric flasks, measure their absorbance with a spectrophotometer, and use Beer’s law (A = abc, on the equation sheet) to build a calibration curve.

    Data sources: the second mode of the tool uses nominal values for commercial stock solutions from OpenStax Chemistry 2e, Appendix F.

    Practice this chapter: 2.9 · Solutions · 10 questions

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