Collecting & Processing Data - IB Chemistry

1. To ensure a reliable mean, a student performed five titrations to determine the concentration of an acid. The volumes of base required were: 21.55 cm³, 21.60 cm³, 21.50 cm³, 22.10 cm³, and 21.58 cm³. Which statement represents the best method for processing this data?
   Answer: The result of 22.10 cm³ should be discarded as an outlier, and the mean of the other four values should be calculated. In a set of repeated measurements, outliers should be identified and excluded before calculating the mean. The value 22.10 cm³ is significantly different from the other four values, which are clustered between 21.50 cm³ and 21.60 cm³. Including the outlier would skew the mean and reduce the accuracy of the final result. The best practice is to discard 22.10 cm³ and calculate the average of the four concordant titres.
2. In a titration to determine the concentration of a sample of ethanoic acid, a standard solution of NaOH is used. The student consistently overshoots the endpoint, adding more NaOH than required to reach the equivalence point. How will this systematic error affect the calculated concentration of the ethanoic acid?
   Answer: The calculated concentration will be consistently higher than the true value. The calculation for the concentration of the acid is based on the mole ratio from the balanced equation and the volumes and concentrations of the reactants (C_acid × V_acid = C_base × V_base). If the volume of the base (V_base) recorded from the titration is consistently too high due to overshooting the endpoint, the calculated moles of base will be too high. This will lead to a calculated number of moles of acid, and thus a calculated concentration of the acid, that is consistently higher than the true value.
3. A student performs UV-Vis spectroscopy to determine concentration using Beer's Law. The absorbance readings for standards are: 0.1 M (A = 0.245), 0.2 M (A = 0.502), 0.3 M (A = 0.751), 0.4 M (A = 1.008). The unknown sample gives A = 0.628. When calculating concentration, the student realizes the spectrophotometer was set to 98% transmittance instead of 100% when blanked. How should this systematic error be addressed?
   Answer: The error cancels out since both standards and unknown were measured under identical conditions Since all measurements (standards and unknown) were made under identical conditions with the same systematic error, the error cancels when using the calibration curve. The relative relationships between absorbance and concentration remain valid. Choice A incorrectly applies a logarithmic correction. Choice B uses an incorrect linear correction factor. Choice D arbitrarily subtracts 0.02 without theoretical basis.
4. In a calorimetry experiment, the enthalpy change is calculated using ΔH = -mcΔT/n. The following data with uncertainties were collected: m = 100.0 ± 0.5 g, ΔT = 20.0 ± 0.5 K, n = 0.050 ± 0.001 mol. Which measurement contributes the largest percentage uncertainty to the calculated value of ΔH?
   Answer: The temperature change, ΔT To find the largest contribution, calculate the percentage uncertainty for each measurement. % uncertainty(m) = (0.5 / 100.0) × 100% = 0.5%. % uncertainty(ΔT) = (0.5 / 20.0) × 100% = 2.5%. % uncertainty(n) = (0.001 / 0.050) × 100% = 2.0%. Comparing the three values (0.5%, 2.5%, 2.0%), the temperature change (ΔT) has the largest percentage uncertainty and therefore contributes most to the uncertainty in the final result.
5. A student performs a titration and obtains the following concordant titres in cm³: 24.55, 24.65, and 24.60. What is the mean volume that should be used for subsequent calculations, reported to the appropriate precision?
   Answer: 24.60 cm³ The mean of the three titres is (24.55 + 24.65 + 24.60) / 3 = 73.80 / 3 = 24.60 cm³. When adding measurements, the result should have the same number of decimal places as the measurement with the fewest decimal places. All titres have two decimal places, so their sum (73.80) does as well. Dividing by an exact number (3) does not change the number of significant figures or precision. Therefore, the mean should also be reported to two decimal places.
6. A student monitors the decomposition of hydrogen peroxide by measuring oxygen evolution over time. The gas collection apparatus shows volumes of 12.3 mL at 2 min, 23.1 mL at 4 min, 32.8 mL at 6 min, and 41.2 mL at 8 min. However, the student realizes that the collection tube's graduated markings are 5% larger than actual due to thermal expansion. Additionally, the reaction temperature rose from 18°C to 23°C during the experiment. Which data processing approach provides the most accurate kinetic information?
   Answer: Correct volumes by multiplying by 0.95, then apply individual temperature corrections to each data point using linear interpolation When analyzing kinetic data from gas evolution experiments, you must carefully consider how multiple systematic errors compound and how temperature changes affect gas volumes throughout the reaction. The key principle here is that temperature corrections must be applied individually to each data point because the temperature changed during the experiment. Using the ideal gas law, $V_1/T_1 = V_2/T_2$, you need to account for the specific temperature at each measurement time, not an average temperature. Option B is correct because it follows the proper sequence: first correct the systematic volume error (multiply by 0.95), then apply temperature corrections individually to each data point using linear interpolation to determine the exact temperature at each measurement time. This approach preserves the true kinetic behavior by maintaining the correct relative changes between data points. Option A fails because using an average temperature (20.5°C) for all points incorrectly assumes constant temperature throughout the reaction, which distorts the kinetic profile. Option C is unnecessary and potentially introduces additional error—normalizing to 0°C isn't required for kinetic analysis, and working with such large temperature corrections can amplify uncertainties. Option D ignores correctable systematic errors that would significantly affect rate constant calculations, making this approach scientifically inadequate. Remember for IB Chemistry: when multiple corrections are needed for kinetic data, always correct systematic instrumental errors first, then apply time-dependent corrections (like temperature) individually to each data point to preserve the true reaction profile.
7. A student performs a series of dilutions to prepare standards for atomic absorption spectroscopy. Starting with 1000 ppm stock solution, they prepare: 100 ppm (10-fold dilution), 10 ppm (10-fold dilution of 100 ppm), and 1 ppm (10-fold dilution of 10 ppm). The absorption readings are 0.089, 0.087, 0.084, and 0.081 respectively. What is the most likely explanation for this trend?
   Answer: Container adsorption losses become proportionally more significant at lower concentrations, reducing effective analyte concentration The decreasing absorption with decreasing concentration suggests non-linear behavior due to adsorption losses. At low concentrations, surface adsorption to container walls becomes a significant fraction of total analyte, causing greater proportional losses. Choice A would show random variation, not systematic decrease. Choice B would affect slope uniformly. Choice C would show the opposite trend (increasing absorption with dilution).
8. In an experiment to measure the rate of reaction between magnesium and hydrochloric acid by collecting the gas produced, there are uncontrolled fluctuations in the room temperature. How would this factor most likely influence the collected data?
   Answer: It would introduce random errors, decreasing the precision of the rate measurements. Uncontrolled fluctuations in an environmental variable like temperature will cause the reaction rate to vary unpredictably during the experiment. Sometimes the temperature might be slightly higher, increasing the rate, and sometimes slightly lower, decreasing it. This lack of consistency introduces random errors, which manifest as scatter in the data points and reduce the precision (reproducibility) of the measurements.
9. During the synthesis of an ester, a student records the smell of the product as "fruity, similar to bananas". Which statement correctly classifies this observation and a potential limitation?
   Answer: This is qualitative data; its main limitation is subjectivity, as different observers may describe the smell differently. Observations based on senses, such as smell, are classified as qualitative data because they are descriptive and not numerical. A significant limitation of such data is its subjective nature. What one person describes as 'fruity' another might describe differently, and perceptions can vary. While useful, it is not as objective as quantitative data.
10. A student measures the mass of a metal block as 27.2 ± 0.1 g and its volume as 3.0 ± 0.1 cm³. What is the calculated density with its absolute uncertainty?
    Answer: 9.1 ± 0.3 g cm⁻³ Density (ρ) = mass/volume = 27.2 g / 3.0 cm³ = 9.066... g cm⁻³. The volume has 2 significant figures, so the density should be reported as 9.1 g cm⁻³. To find the uncertainty, first calculate the percentage uncertainties and add them: %unc(m) = (0.1/27.2)×100% ≈ 0.37%. %unc(V) = (0.1/3.0)×100% ≈ 3.33%. Total %unc = 0.37% + 3.33% ≈ 3.7%. The absolute uncertainty is this percentage of the calculated density: 3.7% of 9.066... g cm⁻³ = 0.037 × 9.066... ≈ 0.335... g cm⁻³. This should be rounded to one significant figure, giving ±0.3 g cm⁻³. The final result is 9.1 ± 0.3 g cm⁻³.
11. The combustion of 0.025 mol of a fuel in a calorimeter causes the temperature of 100.0 g of water to rise by 15.5 K. Given the specific heat capacity of water is 4.18 J g⁻¹ K⁻¹, what is the enthalpy of combustion (ΔH) of the fuel in kJ mol⁻¹, reported with the correct sign and significant figures?
    Answer: -260 kJ mol⁻¹ First, calculate the heat absorbed by the water: Q = mcΔT = (100.0 g) × (4.18 J g⁻¹ K⁻¹) × (15.5 K) = 6479 J. The number of significant figures in the result is limited by the measurement with the fewest significant figures, which is 4.18 and 15.5 (both 3 s.f.). So, Q = 6.48 kJ. The moles of fuel (0.025 mol) has 2 s.f. Enthalpy change ΔH = -Q/n = -6.48 kJ / 0.025 mol = -259.2 kJ mol⁻¹. The final answer must be given to 2 significant figures, limited by the moles of fuel. Thus, ΔH = -260 kJ mol⁻¹.
12. Which statement provides the best comparison of the two sets of results?
    Answer: Student 1's results are precise but not accurate, while Student 2's results are, on average, accurate but not precise. Student 1's results (94.5%, 94.6%, 94.4%) are very close to each other, so they are precise. However, their average (94.5%) is far from the accepted value of 97.2%, so they are not accurate. Student 2's results (97.1%, 95.5%, 98.9%) have a wide range, so they are not precise. However, their average (97.17%) is very close to the accepted value, so the results are accurate on average.
13. During a gravimetric analysis, a student precipitates BaSO₄ from a solution and obtains the following mass data after each washing step: Initial precipitate: 0.8432 g, After wash 1: 0.8401 g, After wash 2: 0.8396 g, After wash 3: 0.8394 g, After wash 4: 0.8394 g. The student also analyzes the wash solutions and finds the fourth wash contains 0.3 ppm dissolved BaSO₄ while the third wash contained 0.8 ppm. What is the most appropriate conclusion?
    Answer: Mass measurements indicate equilibrium, but the decreasing dissolved BaSO₄ levels suggest the washing endpoint has been reached The constant mass (0.8394 g) indicates equilibrium between the solid and solution phases. The decreasing dissolved BaSO₄ concentration (0.8 → 0.3 ppm) shows the washing endpoint is being approached as equilibrium is established. Choice A ignores the dissolution data. Choice B doesn't recognize that some dissolution at equilibrium is expected. Choice C overestimates dissolution risk at these low concentrations.
14. A student determines the concentration of a standard solution with a known concentration of 0.100 mol dm⁻³. Three trials yielded concentrations of 0.125 mol dm⁻³, 0.126 mol dm⁻³, and 0.124 mol dm⁻³. How should these results be described?
    Answer: High precision and low accuracy. Precision refers to how close the measured values are to each other. The three results (0.125, 0.126, 0.124) are very close to one another, indicating high precision. Accuracy refers to how close the measured values are to the true or accepted value. The average result (0.125 mol dm⁻³) is significantly different from the true value (0.100 mol dm⁻³), indicating low accuracy. This suggests a systematic error in the experiment.
15. A student investigates the relationship between pressure (P) and volume (V) of a fixed mass of gas at constant temperature. To obtain a linear graph from which the relationship can be determined, which quantities should be plotted?
    Answer: Pressure on the y-axis and the inverse of volume (1/V) on the x-axis. Boyle's Law states that for a fixed mass of gas at constant temperature, pressure is inversely proportional to volume (P ∝ 1/V), or PV = k. This can be rewritten as P = k(1/V). This equation is in the form of a straight line, y = mx, where y = P and x = 1/V. Therefore, plotting pressure versus the inverse of volume will produce a straight line passing through the origin, confirming the relationship.
16. A student calculates the mass of a product using the following data: Concentration of reactant A = 0.50 mol dm⁻³ Volume of reactant A = 25.55 cm³ Molar mass of product B = 158.03 g mol⁻¹ The reaction stoichiometry is A → 2B. What is the maximum theoretical mass of B, reported to the correct number of significant figures?
    Answer: 4.0 g First, calculate the moles of A: n(A) = C × V = (0.50 mol dm⁻³) × (0.02555 dm³) = 0.012775 mol. The result of a multiplication is limited by the term with the fewest significant figures, which is the concentration (0.50 mol dm⁻³) with two significant figures. The moles of B is 2 × n(A) = 0.02555 mol. The mass of B is n(B) × M(B) = (0.02555 mol) × (158.03 g mol⁻¹) = 4.0376... g. The final answer must be rounded to two significant figures, which gives 4.0 g.
17. A student measures the mass of a hydrated salt sample before and after heating to determine water content. The initial mass is recorded as 2.847 g, and after complete dehydration, the mass is 1.923 g. However, the student realizes the balance was not zeroed properly and consistently read 0.015 g high throughout the experiment. What is the correct percentage of water by mass in the original hydrated salt?
    Answer: 32.5% Since both measurements were 0.015 g high, the corrected masses are: initial = 2.847 - 0.015 = 2.832 g, final = 1.923 - 0.015 = 1.908 g. Water lost = 2.832 - 1.908 = 0.924 g. Percentage water = (0.924/2.832) × 100% = 32.5%. Choice B uses uncorrected masses. Choice C incorrectly subtracts 0.015 from the water mass. Choice D incorrectly adds 0.015 to both corrected masses before calculating.
18. A student records a burette reading as 15.4 cm³. A typical burette is marked to the nearest 0.1 cm³. What is the implied uncertainty of this single measurement?
    Answer: ±0.05 cm³ For an analogue instrument like a burette, the uncertainty of a single reading is typically taken as half of the smallest graduation or marking. Since the burette is marked to the nearest 0.1 cm³, the uncertainty is ±(0.1 / 2) = ±0.05 cm³. The student correctly records the reading to two decimal places (15.40 cm³ would be better practice), with the second decimal place being an estimated digit, and the uncertainty applies to this estimated value.
19. In a calorimetry experiment, the initial temperature of a solution was recorded as 22.5 °C and the final temperature as 27.5 °C. The thermometer has an uncertainty of ±0.2 °C for each reading. What is the temperature change and its absolute uncertainty?
    Answer: 5.0 ± 0.4 °C The temperature change (ΔT) is the difference between the final and initial temperatures: 27.5 °C - 22.5 °C = 5.0 °C. When subtracting two measurements, their absolute uncertainties are added. Therefore, the uncertainty in ΔT is the sum of the uncertainties of the initial and final readings: 0.2 °C + 0.2 °C = 0.4 °C. The result is reported as 5.0 ± 0.4 °C.
20. During a calorimetry experiment measuring enthalpy of neutralization, a student records temperature data every 30 seconds. The data shows: 0 min (22.1°C), 0.5 min (22.0°C), 1.0 min (21.9°C), 1.5 min (28.4°C), 2.0 min (28.1°C), 2.5 min (27.8°C), 3.0 min (27.6°C). The reaction was initiated at t = 1.0 min. What is the most appropriate method to determine ΔT for enthalpy calculation?
    Answer: Use extrapolated initial temperature of 21.8°C and extrapolated final temperature of 28.6°C: ΔT = 6.8°C Proper calorimetry analysis requires extrapolation to account for heat loss. Pre-reaction trend: 22.1°C → 21.9°C gives slope of -0.2°C/min, extrapolated to t=1.5 min gives 21.8°C. Post-reaction cooling from 28.4°C → 27.6°C gives slope of -0.4°C/min, extrapolated back to t=1.5 min gives 28.6°C. ΔT = 28.6 - 21.8 = 6.8°C. Other choices use incorrect extrapolation methods or ignore heat loss corrections.

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