IB Chemistry IA calibration curves: catch the dilution error before your conclusion

The IB uses a shared scientific-investigation model for Biology, Chemistry and Physics. A calibration curve is just one possible analytical method; its suitability depends on the chemical system, instrument response, and your actual research question. The numbers below are invented educational data, not an approved lab protocol, safety instruction or measured IB example.

Work through an entire result

Suppose absorbance follows the fitted relationship:

A = 0.010 + 0.050c

where A is unitless absorbance and c is concentration in mg/L. The synthetic standards are:

The standards deliberately sit on an exact line so the arithmetic is easy to inspect. Real calibration data have noise, and a perfect line should not be expected or fabricated. A diluted unknown gives an absorbance of 0.185. Inverting the line gives:

c_diluted = (0.185 - 0.010) / (0.050) = 3.50\ mg/L

Now comes the part students sometimes miss. The unknown was prepared by transferring 10.0 mL of original sample into a 50.0 mL volumetric flask and making it up to the mark. The dilution factor is 50.0/10.0 = 5.00. Therefore:

c_original = 3.50 × 5.00 = 17.5\ mg/L

Reporting 3.50 mg/L would describe the diluted mixture, not the original sample. Reporting 17.5 mg/mL would introduce a factor-of-1,000 unit error. A one-line unit check is worth doing before discussing chemical mechanisms.

Three checks that change your interpretation

Blank correction. Here the nonzero intercept, 0.010, represents the calibration's modeled baseline. Simply dividing 0.185 by the slope without accounting for the intercept would overestimate the concentration. In a real analysis, document whether a blank was measured and whether the fitted intercept should be retained.

Range. The diluted estimate, 3.50 mg/L, falls inside the 0–6 mg/L standard range, so this calculation interpolates. The undiluted concentration is higher, but you do not compare it directly with the diluted calibration range. If the measured diluted sample had yielded 8 mg/L, you should prepare an appropriate further dilution and remeasure rather than quietly extrapolating.

Uncertainty. Imagine absorbance repeats of 0.180, 0.185 and 0.190. Their mean is 0.185; the spread shows some reading variability, but it is not the total uncertainty. The standards, slope and intercept, flask and pipette, matrix effects, instrument drift and measurement replicates may all matter. Do not claim a confidence interval from those three readings alone.

A result paragraph that can be checked

"The calibrated mean absorbance of the diluted unknown was 0.185. Using the fitted relation A=0.010+0.050c, its estimated concentration was 3.50 mg/L, within the measured standard range. After applying the documented five-fold volumetric dilution, the estimated concentration in the original sample was 17.5 mg/L. This value remains subject to calibration, volume-measurement and possible matrix uncertainties; the present synthetic example does not quantify their combined effect."

This paragraph earns its specificity because the arithmetic and assumptions are visible. A vague claim that the line had a "very high R²" would not replace these checks. In real work, retain the individual calibration points and inspect residuals as well as the fitted regression.

Primary sources and limits

• IB 2026 Chemistry examiner instructions, scientific investigation (2026 French-language public original): https://ibpublishing.ibo.org/exinst/apps/exinst/index.html?chapter=2&doc=EX_instructions_2026_f&part=10
• IB public curriculum updates for checking which Chemistry assessment framework applies: https://ibo.org/university-admission/latest-curriculum-updates

The chemistry and numerical example are original teaching illustrations, not real experimental measurements. Actual safety, chemical preparation and lab procedures require school supervision. See ibacalao.com for independent student-controlled draft review.