A solution of copper sulfate is blue because it absorbs orange-red light and lets blue through, the same kind of electron jump you met in What Light Does to Molecules. The more copper ions in the light's path, the more light they absorb. Measuring that absorption is one of the fastest ways to find a Molarity, and it rests on a straight-line model.
Key idea
$$A = \varepsilon b c$$ The absorbance $A$ is proportional to the concentration $c$ (mol/L) and the path length $b$ (cm); the molar absorptivity $\varepsilon$ (L/(mol·cm)) depends on the substance and the wavelength. Absorbance is found from the fraction of light transmitted, $T = I/I_0$: $A = -\log_{10} T$. A calibration curve (A against c for known standards) is a straight line through the origin; read an unknown's concentration from it.
Why it works
Think of the light passing through thin slices of solution. Each slice absorbs the same fraction of the light reaching it, so the light fades by equal factors: after $n$ slices only $(1 - f)^n$ is left. Fractions that multiply turn into logarithms that add, so $-\log T$ grows in proportion to the number of absorbing particles in the beam: the concentration times the path length. That is why absorbance, not percent transmittance, is the quantity that makes a straight line. Measurements are made at the wavelength the substance absorbs most strongly ($\lambda_\text{max}$), where the line is steepest and the reading most sensitive.
Worked example: Absorbance from concentration
A dye has $\varepsilon = 5.0 \times 10^{3}\ \text{L/(mol·cm)}$ at its $\lambda_\text{max}$. Find the absorbance of a $1.2 \times 10^{-4}\ \text{M}$ solution in a $1.00\ \text{cm}$ cell.
- $A = \varepsilon bc = (5.0 \times 10^{3})(1.00)(1.2 \times 10^{-4})$.
- $A = 0.60$ (no units).
Answer: $0.60$
Worked example: Using a calibration line
Standards give a line $A = (2.50 \times 10^{3}\ \text{L/mol})\,c$ in a 1 cm cell. An unknown reads $A = 0.450$. Find its concentration.
- The line is $A = (\text{slope})\,c$, so $c = \dfrac{A}{\text{slope}}$.
- $c = \dfrac{0.450}{2.50 \times 10^{3}\ \text{L/mol}} = 1.80 \times 10^{-4}\ \text{M}$. Check: 0.450 lies between the standards' absorbances, so we are reading inside the calibrated range.
Answer: $1.80 \times 10^{-4}\ \text{M}$
Worked example: From percent transmittance
A sample transmits 25.0% of the light. What is its absorbance?
- $T = 0.250$.
- $A = -\log_{10}(0.250) = 0.602$.
Answer: $0.602$
Common mistake
Using percent transmittance as if it were absorbance, or thinking a darker solution absorbs less. Transmittance falls as concentration rises, and not in a straight line; absorbance rises in proportion. A darker solution absorbs more. Check: $A = 0$ means all light gets through; $A = 1$ means only 10% does; $A = 2$, only 1%.