Aim for FiveChemistry

Unit 3 · Topic 3.7

3.7 Solutions and Mixtures

A solution is a mixture that is the same all the way through, while a heterogeneous mixture varies from place to place. You'll describe how concentrated a solution is with molarity, the moles of solute per liter of solution, and use it to find moles, volumes and numbers of particles.

Key terms

  • solution
  • homogeneous mixture
  • heterogeneous mixture
  • solute
  • solvent
  • molarity

Homogeneous and heterogeneous mixtures

A solution, also called a homogeneous mixture, has the same composition and properties everywhere in the sample. Sugar water tastes equally sweet at the top and the bottom. Solutions can be liquids (salt water), gases (air) or solids (brass, an alloy of copper and zinc).

A heterogeneous mixture has properties that depend on where you sample it. Sand in water, oil floating on vinegar and granite are heterogeneous.

In a solution, the solvent is the substance present in the largest amount (often water), and the solute is what's dissolved in it.

Molarity

Molarity (M) = moles of solute ÷ liters of solution. A 0.50 M NaCl solution contains 0.50 mol NaCl in every liter of solution.

Notice the denominator is liters of solution, not liters of solvent. To make 1.00 L of 0.50 M NaCl, you dissolve 0.50 mol of NaCl in some water, then add water until the total volume is 1.00 L. A volumetric flask is used for this.

Rearranging gives moles = M × V (in liters). That's the most-used form in stoichiometry and titration problems.

Using molarity as a conversion factor

Treat molarity like any other conversion factor in dimensional analysis. 0.250 M means 0.250 mol per 1 L, so you can multiply by (0.250 mol / 1 L) to go from liters to moles, or by (1 L / 0.250 mol) to go from moles to liters.

Preparing a solution in the lab follows the same logic: calculate the mass of solute needed, weigh it, dissolve it in less water than the final volume, transfer it to a volumetric flask, and add water up to the line on the neck. Dissolving the solute in exactly 1 L of water doesn't work, because the final volume of solution generally isn't exactly 1 L, so the concentration would be off.

Dilution

Adding solvent doesn't change the number of moles of solute, only the volume. So moles before = moles after, or M₁V₁ = M₂V₂. You can use any volume unit as long as both volumes use the same one.

Concentrations of ions

When a soluble ionic compound dissolves, it separates into ions. The concentration of each ion depends on the formula. In 0.20 M Al₂(SO₄)₃, [Al³⁺] = 2 × 0.20 = 0.40 M and [SO₄²⁻] = 3 × 0.20 = 0.60 M. Square brackets mean 'concentration of, in mol/L'. Molecular solutes like sugar don't break apart, so 0.20 M sugar is just 0.20 M sugar molecules.

You won't be tested on

Molality, percent by mass and percent by volume of solutions aren't assessed. Molarity is the unit you need.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1Calculator allowed

    Making a solution

    What mass of NaOH is needed to make 250.0 mL of 0.150 M NaOH solution?

    Show the solution
    1. Step 1: Convert volume to liters: 250.0 mL = 0.2500 L.
    2. Step 2: Moles = M × V = 0.150 mol/L × 0.2500 L = 0.0375 mol NaOH.
    3. Step 3: Mass = 0.0375 mol × 40.00 g/mol = 1.50 g.
    4. Step 4: Dissolve 1.50 g NaOH in some water, then add water to a total volume of 250.0 mL.

    Answer: 1.50 g NaOH

  2. Example 2Calculator allowed

    Dilution

    What volume of 6.0 M HCl is needed to make 500. mL of 0.30 M HCl?

    Show the solution
    1. Step 1: M₁V₁ = M₂V₂, so V₁ = M₂V₂ ÷ M₁.
    2. Step 2: V₁ = (0.30 M)(500. mL) ÷ 6.0 M = 25 mL.
    3. Step 3: Measure 25 mL of the 6.0 M acid and add it to water, then dilute to 500. mL.

    Answer: 25 mL of 6.0 M HCl

  3. Example 3Calculator allowed

    Counting ions (classic trap)

    How many ions in total are present in 50.0 mL of 0.20 M Al₂(SO₄)₃?

    Show the solution
    1. Step 1: Moles of Al₂(SO₄)₃ = 0.20 mol/L × 0.0500 L = 0.010 mol.
    2. Step 2: Each formula unit gives 2 Al³⁺ and 3 SO₄²⁻, so 5 ions in all.
    3. Step 3: Moles of ions = 5 × 0.010 = 0.050 mol.
    4. Step 4: Number of ions = 0.050 × 6.022 × 10²³ = 3.0 × 10²² ions. Forgetting the factor of 5 gives 6.0 × 10²¹, five times too few.

    Answer: 3.0 × 10²² ions

Common mistakes

  • Using milliliters in M = n/V. Convert to liters.
  • Dividing by the volume of solvent added instead of the final volume of solution.
  • Forgetting that ionic solutes split into several ions, so ion concentrations can be larger than the compound's molarity.
  • Thinking dilution changes the number of moles of solute. Only the volume changes.

On the exam

  • Molarity shows up in almost every quantitative free-response question: titrations, stoichiometry, equilibrium and Beer's law. Write 'mol = M × L' explicitly and keep track of units.
  • Lab-based questions may ask which glassware gives an accurate final volume. A volumetric flask is the precise choice for making a solution of known molarity.

Connected topics

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Check yourself

4 questions on 3.7 Solutions and Mixtures. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A student dissolves 5.85 g of NaCl (molar mass 58.44 g/mol) in water and adds enough water to make 250.0 mL of solution. What is the molarity of NaCl in the solution?

Question 2 of 4Calculator allowed

A 25.0 mL sample of 2.00 M HCl is diluted with water to a total volume of 100.0 mL. What is the molarity of HCl in the diluted solution?

Question 3 of 4Calculator allowed

A student mixes 50.0 mL of 0.100 M NaCl with 150.0 mL of 0.300 M NaCl. Assuming the volumes add, what is the concentration of NaCl in the final solution?

Question 4 of 4

A student needs exactly 250.0 mL of 0.100 M CuSO₄ solution. Which procedure will give the most accurate concentration?

0 of 4 answered