← Back to course

Solutions & Spectroscopy

You’ll be able to

Preparing solutions and diluting them

To prepare a solution of known molarity (M = mol solute ÷ L solution), you weigh out a calculated mass of solute and add solvent up to a marked volume. To make a dilute solution from a concentrated stock, you take a measured volume of stock and add solvent. Dilution changes the concentration but not the moles of solute already measured out, so moles are conserved: M₁V₁ = M₂V₂. Solving for the stock volume needed, V₁ = M₂V₂ ÷ M₁, tells you exactly how much concentrated solution to pipette before topping up with water.

Solution stoichiometry

Molarity is the bridge between a volume you can measure and the moles a reaction consumes. Multiply molarity × volume (in L) to get moles, then use the mole ratio from the balanced equation to find moles of another species, and convert back to mass or volume as needed. This is the backbone of titration: at the equivalence point the moles of titrant delivered exactly satisfy the mole ratio to the analyte, letting you back-calculate an unknown concentration.

Beer–Lambert law: color into concentration

A colored solution absorbs light, and the more concentrated it is, the more light it absorbs. A spectrophotometer measures absorbance (A), a unitless number, at a chosen wavelength. The Beer–Lambert law states that absorbance is directly proportional to concentration: A = εbc, where ε is the molar absorptivity (a constant for the substance at that wavelength, in L·mol⁻¹·cm⁻¹), b is the path length of the cuvette (usually 1.00 cm), and c is the molar concentration. Because A rises linearly with c, you can build a calibration line and read an unknown concentration straight off a measured absorbance.

Separating mixtures: chromatography and distillation

Chromatography separates components by how strongly each is drawn to a stationary phase versus a moving mobile phase. On paper, a component’s travel is reported as a retention factor, Rf = distance moved by the spot ÷ distance moved by the solvent front — a value between 0 and 1. A larger Rf means weaker attraction to the stationary phase (it rides farther with the solvent). Distillation instead separates by boiling point: the more volatile component vaporizes first, then recondenses in a cooler tube, leaving the less volatile component behind.

Beer–Lambert law
A = εbc
A is absorbance (unitless); ε is molar absorptivity (L·mol⁻¹·cm⁻¹); b is path length (cm); c is molar concentration (mol·L⁻¹). Rearranged, c = A ÷ (εb).
Dilution relationship
M₁V₁ = M₂V₂
Moles of solute are conserved on dilution. Subscript 1 is the concentrated stock, 2 is the diluted solution; any consistent volume unit works since it cancels.
Worked example

A solution of a colored complex is measured in a 1.00 cm cuvette and gives an absorbance of A = 0.42. The molar absorptivity at that wavelength is ε = 1.5 × 10³ L·mol⁻¹·cm⁻¹. What is the concentration?

  1. 1.Start from the Beer–Lambert law, A = εbc, and solve for concentration: c = A ÷ (εb).
  2. 2.Substitute the values: c = 0.42 ÷ (1.5 × 10³ L·mol⁻¹·cm⁻¹ × 1.00 cm).
  3. 3.Evaluate the denominator: 1.5 × 10³ × 1.00 = 1.5 × 10³, then divide: 0.42 ÷ 1500.
  4. 4.c = 2.8 × 10⁻⁴ mol·L⁻¹. Check: 1500 × 2.8 × 10⁻⁴ = 0.42, matching the measured absorbance.
Answer: c = 2.8 × 10⁻⁴ M
Worked example

How many milliliters of 2.00 M NaCl stock are needed to prepare 500. mL of 0.250 M NaCl?

  1. 1.Dilution conserves moles: M₁V₁ = M₂V₂, with stock as 1 and the target as 2.
  2. 2.Solve for the stock volume: V₁ = M₂V₂ ÷ M₁.
  3. 3.Substitute: V₁ = (0.250 M × 500. mL) ÷ 2.00 M = 125 ÷ 2.00.
  4. 4.V₁ = 62.5 mL. Measure 62.5 mL of the 2.00 M stock, then add water up to 500. mL total.
Answer: 62.5 mL of stock, diluted to 500. mL
Worked example

How many moles of NaOH are required to completely neutralize 25.0 mL of 0.100 M H₂SO₄? The reaction is H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O.

  1. 1.Find moles of acid: mol H₂SO₄ = M × L = 0.100 mol·L⁻¹ × 0.0250 L = 2.50 × 10⁻³ mol.
  2. 2.Read the mole ratio from the balanced equation: 2 mol NaOH per 1 mol H₂SO₄.
  3. 3.Apply the ratio: mol NaOH = 2.50 × 10⁻³ mol × (2 ÷ 1).
  4. 4.mol NaOH = 5.00 × 10⁻³ mol.
Answer: 5.00 × 10⁻³ mol NaOH
Tip

Beer’s law is linear, so absorbance and concentration scale together: if a solution’s absorbance doubles, its concentration doubled (at fixed path length and wavelength). A quick way to check a Beer’s-law answer is to confirm εbc reproduces the given A.

Checkpoint

A solution measured in a 1.0 cm cuvette has an absorbance of A = 0.60. Its molar absorptivity is ε = 2.0 × 10³ L·mol⁻¹·cm⁻¹. What is its concentration?

Checkpoint

What volume of 2.0 M HCl stock is needed to prepare 250 mL of 0.10 M HCl?

Checkpoint

On a paper chromatogram, a pigment spot travels 3.0 cm while the solvent front travels 12.0 cm. What is the retention factor (Rf) of the pigment?

On the exam

For a Beer’s-law free-response, expect a calibration graph of absorbance vs concentration: its slope is εb, so a steeper line means a larger molar absorptivity. Read an unknown by finding its absorbance on the line and dropping to the concentration axis — and always confirm the path length is 1.00 cm before using ε directly.

Answer the 3 checkpoints as you read.

Sign in to save your progress