E. Chemical Systems and Equilibrium · Chemical EquilibriumInvestigation: Determining an Equilibrium Constant
E2.3 E2.4 A1.2 A1.5 A1.6 A1.8 A1.10 A1.12 A1.13About 75 minutes
Learning goals- I can explain how a colorimeter and a calibration line give the concentration of a coloured species.
- I can calculate initial concentrations after mixing solutions.
- I can use ICE tables on experimental data to calculate K for several trials and judge whether K is constant.
- I can identify sources of error in an equilibrium-constant experiment and suggest improvements.
Words to know
colorimeter (spectrophotometer)
An instrument that shines light of a chosen wavelength through a sample and measures how much is absorbed.
absorbance (A)
A measure of how much light a sample absorbs; for dilute solutions it is directly proportional to concentration.
Beer-Lambert law
A = εbc, where ε is a constant for the substance and wavelength, b is the path length and c is the concentration.
calibration line (standard curve)
A graph of absorbance against concentration made from standard solutions of known concentration, used to find unknown concentrations.
standard solution
A solution whose concentration is known accurately.
Measuring a constant you cannot see
In lesson 2 you calculated K from equilibrium concentrations that were simply given to you. In a real lab, someone had to measure them. One of the classic Grade 12 investigations does exactly that for the blood-red iron(III) thiocyanate ion, using light: the redder the solution, the more FeSCN²⁺ it contains. If K really is constant, every trial should give about the same value, no matter how much of each reactant you mix.
The plan
Equilibrium: Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) K = [FeSCN²⁺] ÷ ([Fe³⁺][SCN⁻])
Question: Is K constant when the starting concentration of SCN⁻ changes?
Prediction: If the equilibrium law is valid, then K will be the same (within experimental error) for every trial, because K depends only on temperature.
Method in three parts:
- Standards and calibration. Make solutions with a known [FeSCN²⁺] by mixing a small amount of SCN⁻ with a large excess of Fe³⁺ (0.200 mol/L iron(III) nitrate in dilute nitric acid). The excess pushes the equilibrium so far right that essentially all the SCN⁻ becomes FeSCN²⁺, so [FeSCN²⁺] = [SCN⁻] added. Measure the absorbance of each standard at about 450 nm (blue light, which the red ion absorbs strongly) and plot a calibration line.
- Trials. Mix 5.00 mL of 2.00 × 10⁻³ mol/L Fe(NO₃)₃ with different volumes of 2.00 × 10⁻³ mol/L KSCN and enough water to make 10.00 mL. Measure each absorbance.
- Analysis. Use the calibration line to get [FeSCN²⁺] at equilibrium, then an ICE table to find [Fe³⁺] and [SCN⁻], and calculate K for each trial.
Safety: iron(III) nitrate solution is an irritant and is made up in dilute nitric acid (corrosive); potassium thiocyanate is harmful. Wear goggles and gloves, and use the labelled waste container. School lab only.
Step 1: The calibration line
Five standards gave these results (path length 1.00 cm):
| [FeSCN²⁺] (mol/L) | 2.00 × 10⁻⁵ | 4.00 × 10⁻⁵ | 6.00 × 10⁻⁵ | 8.00 × 10⁻⁵ | 1.00 × 10⁻⁴ |
|---|
| absorbance | 0.097 | 0.190 | 0.289 | 0.386 | 0.478 |
The points lie on a straight line through the origin with slope 4.80 × 10³ L/mol (the value of εb). So for any trial,
[FeSCN²⁺] = A ÷ (4.80 × 10³ L/mol)
Worked exampleExample 1: Initial concentrations after mixing
Trial 3 mixes 5.00 mL of 2.00 × 10⁻³ mol/L Fe(NO₃)₃, 3.00 mL of 2.00 × 10⁻³ mol/L KSCN and 2.00 mL of water. Find the initial [Fe³⁺] and [SCN⁻] in the 10.00 mL mixture.
- Use the dilution formula c₁V₁ = c₂V₂ from Grade 11.
- [Fe³⁺]ᵢ = (2.00 × 10⁻³)(5.00 mL) ÷ 10.00 mL = 1.00 × 10⁻³ mol/L.
- [SCN⁻]ᵢ = (2.00 × 10⁻³)(3.00 mL) ÷ 10.00 mL = 6.00 × 10⁻⁴ mol/L.
[Fe³⁺]ᵢ = 1.00 × 10⁻³ mol/L and [SCN⁻]ᵢ = 6.00 × 10⁻⁴ mol/L. Forgetting to account for mixing is one of the most common errors in this lab.
Worked exampleExample 2: K for trial 3
Trial 3 has an absorbance of 0.355. Calculate K.
- [FeSCN²⁺]ₑ = 0.355 ÷ 4800 = 7.396 × 10⁻⁵ mol/L = x.
- ICE: I: 1.00 × 10⁻³, 6.00 × 10⁻⁴, 0. C: −x, −x, +x.
- [Fe³⁺]ₑ = 1.00 × 10⁻³ − 7.396 × 10⁻⁵ = 9.260 × 10⁻⁴ mol/L.
- [SCN⁻]ₑ = 6.00 × 10⁻⁴ − 7.396 × 10⁻⁵ = 5.260 × 10⁻⁴ mol/L.
- K = 7.396 × 10⁻⁵ ÷ [(9.260 × 10⁻⁴)(5.260 × 10⁻⁴)] = 7.396 × 10⁻⁵ ÷ 4.871 × 10⁻⁷ = 152.
| Trial | [SCN⁻]ᵢ (mol/L) | Absorbance | [FeSCN²⁺]ₑ (mol/L) | [Fe³⁺]ₑ (mol/L) | [SCN⁻]ₑ (mol/L) | K |
|---|
| 1 | 2.00 × 10⁻⁴ | 0.124 | 2.58 × 10⁻⁵ | 9.74 × 10⁻⁴ | 1.74 × 10⁻⁴ | 152 |
| 2 | 4.00 × 10⁻⁴ | 0.236 | 4.92 × 10⁻⁵ | 9.51 × 10⁻⁴ | 3.51 × 10⁻⁴ | 147 |
| 3 | 6.00 × 10⁻⁴ | 0.355 | 7.40 × 10⁻⁵ | 9.26 × 10⁻⁴ | 5.26 × 10⁻⁴ | 152 |
| 4 | 8.00 × 10⁻⁴ | 0.455 | 9.48 × 10⁻⁵ | 9.05 × 10⁻⁴ | 7.05 × 10⁻⁴ | 148 |
| 5 | 1.00 × 10⁻³ | 0.566 | 1.18 × 10⁻⁴ | 8.82 × 10⁻⁴ | 8.82 × 10⁻⁴ | 152 |
Analysing the results
The five K values range from 147 to 152, with a mean of about 150. The spread is only about ±2 %, while the starting SCN⁻ concentration changed by a factor of five. The evidence supports the prediction: K is constant at constant temperature.
(Published values of K for this reaction vary, because K depends on temperature and on the other ions in the solution. Your class's value is valid for your conditions.)
Sources of error and improvements:
- Volume measurements: use volumetric pipettes and burettes rather than graduated cylinders.
- Temperature: K depends on temperature; keep all solutions at room temperature and record it.
- Cuvettes: fingerprints, scratches or bubbles scatter light; handle by the frosted sides and wipe clean.
- Calibration assumption: if the standards' Fe³⁺ excess were not large enough, not all SCN⁻ would be converted, and the slope would be too low. Using a larger excess tests this.
- Timing: measure each trial a few minutes after mixing, once readings are steady.
Other K investigations: Ksp of a slightly soluble salt such as calcium hydroxide can be found by titrating a saturated solution. That is part of the unit Acid-Base and Solubility Equilibria.
Just for funThe iron(III) thiocyanate ion is so intensely red that chemists use it to detect very small amounts of iron. It is the lab's best blusher.
Hands-on activityKitchen colorimetry: a calibration series by eye
You need: red or blue food colouring, five identical clear glasses, a measuring cup, a measuring spoon or medicine syringe, white paper, a partner. Safety: food colouring stains; cover the table.
- Make a stock solution: 10 drops of food colouring in 500 mL of water. Call its concentration 100 units.
- Make standards in the glasses, each 100 mL total: 100 units (100 mL stock), 75 (75 mL stock + 25 mL water), 50, 25 and 10 units.
- Line them up on white paper and look down through each glass from above (this makes the path length the same for all).
- Ask your partner to secretly make a "mystery" solution between 10 and 100 units. Match it to your standards and estimate its concentration.
What to look for: you should be able to place the mystery between two standards (for example "between 50 and 75, nearer 50"). This is the same idea as a calibration line, but your eye is a much less precise detector than a colorimeter, and judging colour is subjective. Looking down through the glasses works because colour depth depends on both concentration and path length (A = εbc).
Your turn
1Why must Fe³⁺ be in large excess when making the standards?
So that the equilibrium lies so far to the right that essentially all the SCN⁻ is converted to FeSCN²⁺. Then [FeSCN²⁺] in the standard equals the [SCN⁻] added, which is known.
25.00 mL of 2.00 × 10⁻³ mol/L Fe(NO₃)₃ is mixed with 1.00 mL of 2.00 × 10⁻³ mol/L KSCN and 4.00 mL of water. Find the initial [Fe³⁺] and [SCN⁻].
[Fe³⁺]ᵢ = (2.00 × 10⁻³)(5.00) ÷ 10.00 = 1.00 × 10⁻³ mol/L; [SCN⁻]ᵢ = (2.00 × 10⁻³)(1.00) ÷ 10.00 = 2.00 × 10⁻⁴ mol/L.
3Using the calibration slope 4.80 × 10³ L/mol, find [FeSCN²⁺] for an absorbance of 0.240.
0.240 ÷ 4800 = 5.00 × 10⁻⁵ mol/L
4Verify K for trial 1 from the results table.
x = 0.124 ÷ 4800 = 2.583 × 10⁻⁵. [Fe³⁺] = 1.00 × 10⁻³ − 2.583 × 10⁻⁵ = 9.742 × 10⁻⁴; [SCN⁻] = 2.00 × 10⁻⁴ − 2.583 × 10⁻⁵ = 1.742 × 10⁻⁴. K = 2.583 × 10⁻⁵ ÷ (9.742 × 10⁻⁴ × 1.742 × 10⁻⁴) = 2.583 × 10⁻⁵ ÷ 1.697 × 10⁻⁷ = 152.
5What evidence in the table shows that K is a constant?
The starting [SCN⁻] changed fivefold and the equilibrium concentrations all changed, yet K stayed between 147 and 152 (mean about 150).
6Why is blue light (about 450 nm) used instead of red light?
A red solution looks red because it absorbs blue-green light and lets red pass. Measuring at a wavelength the ion absorbs strongly gives the biggest, most sensitive change in absorbance.
More practice: warm-up
1What does a colorimeter measure?
How much light of a chosen wavelength a sample absorbs (absorbance).
2If absorbance doubles, what happens to the concentration (Beer-Lambert law)?
It doubles (A is proportional to c).
3State the dilution formula.
4Which variable was changed between trials?
The initial concentration of SCN⁻ (the volume of KSCN solution added).
5Name one way to reduce error in measuring volumes.
Use volumetric pipettes or burettes instead of graduated cylinders.
More practice: on your own
1Verify K for trial 5 in the table.
x = 0.566 ÷ 4800 = 1.179 × 10⁻⁴. [Fe³⁺] = [SCN⁻] = 1.00 × 10⁻³ − 1.179 × 10⁻⁴ = 8.821 × 10⁻⁴. K = 1.179 × 10⁻⁴ ÷ (8.821 × 10⁻⁴)² = 1.179 × 10⁻⁴ ÷ 7.781 × 10⁻⁷ = 152.
2A student forgets to account for mixing and uses [Fe³⁺]ᵢ = 2.00 × 10⁻³ and [SCN⁻]ᵢ = 1.20 × 10⁻³ mol/L for trial 3 (A = 0.355). What K would they get, and is it too high or too low?
x = 7.396 × 10⁻⁵. [Fe³⁺] = 1.926 × 10⁻³; [SCN⁻] = 1.126 × 10⁻³. K = 7.396 × 10⁻⁵ ÷ (2.169 × 10⁻⁶) = 34.1, far too low.
3A cuvette has a fingerprint on it. Will the measured K be too high or too low? Explain.
The fingerprint scatters or absorbs extra light, so the absorbance reads too high. That makes [FeSCN²⁺] too high and the reactant concentrations too low, so K is too high.
4The temperature of the lab rose by 5 °C between the standards and the trials. Why does this matter?
K depends on temperature, so the trials may not be at the same equilibrium conditions as assumed. (ε can also vary slightly.) Keep all solutions at the same temperature and record it.
5Calculate the percent deviation of each K from the mean of 150 for trials 2 and 4.
Trial 2: (147 − 150) ÷ 150 × 100 % = −2 %. Trial 4: (148 − 150) ÷ 150 × 100 % = −1.3 %.
Challenge
1Predict the absorbance of a sixth trial with [Fe³⁺]ᵢ = 1.00 × 10⁻³ and [SCN⁻]ᵢ = 1.00 × 10⁻⁴ mol/L, using K = 150.
x ÷ [(1.00 × 10⁻³ − x)(1.00 × 10⁻⁴ − x)] = 150. Expanding: 150x² − 1.165x + 1.50 × 10⁻⁵ = 0 → x = [1.165 − √(1.3572 − 0.0090)] ÷ 300 = (1.165 − 1.16113) ÷ 300 = 1.29 × 10⁻⁵ mol/L. A = 4800 × 1.29 × 10⁻⁵ = 0.062.
2Explain why plotting A against [FeSCN²⁺] for the standards should give a straight line through the origin, and what it would mean if the line curved downward at high concentration.
Beer-Lambert says A = εbc, so A is proportional to c: a straight line through (0, 0). A downward curve at high concentration means the law is breaking down (too concentrated, or the detector cannot measure very low light levels); use more dilute solutions.
3Design a way to test whether temperature changes K for this equilibrium, using the same equipment.
Prepare identical trial mixtures, bring them to different temperatures (for example 10 °C, 25 °C and 40 °C water baths), measure absorbance quickly at each temperature, calculate K for each, and compare. (A calibration at each temperature would make the results more reliable.)
Quiz
1. In the standards, why is [FeSCN²⁺] taken to equal the [SCN⁻] added?
2. 2.00 mL of 5.00 × 10⁻³ mol/L KSCN is diluted to 10.00 mL. The new [SCN⁻] is:
3. A calibration line has slope 4.80 × 10³ L/mol. A sample has A = 0.192. Its [FeSCN²⁺] is:
4. If the five trials give K values of 147 to 152, the best conclusion is:
5. Which error would make the calculated K too HIGH?
6. Why should all trials be done at the same temperature?
Big idea (E2.3, E2.4): K can be measured. A colorimeter and a calibration line give the equilibrium concentration of a coloured ion; ICE tables give the rest; trials with different starting amounts give the same K, supporting the equilibrium law.
How to help: the full lab needs a school colorimeter, but the kitchen colorimetry activity shows the idea of a calibration series safely. Ask your teen to explain why every K in the table is about the same even though the mixtures differ.
Common difficulties:
- Forgetting to account for dilution when solutions are mixed.
- Using initial instead of equilibrium concentrations in K.
- Confusing the standards (all SCN⁻ converted) with the trials (equilibrium mixtures).
- Vague error analysis: encourage students to say which way each error pushes K and why.
Safety: Fe(NO₃)₃ in nitric acid and KSCN are school-lab chemicals only; food colouring is a safe substitute at home.