IB Chemistry Quiz: Using Experimental Techniques
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Using Experimental TechniquesQuestion 1 of 20

In a paper chromatography experiment to separate a mixture of amino acids, a polar stationary phase (cellulose) is used with a non-polar solvent. Amino acid X has an R_f value of 0.20 and amino acid Y has an R_f value of 0.80. Which deduction is correct?

Amino acid X is more soluble in the non-polar solvent than amino acid Y.
Amino acid Y has stronger intermolecular forces of attraction to the stationary phase than amino acid X.
Amino acid Y is less polar than amino acid X.
The mobile phase is polar and the stationary phase is non-polar.
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IB Chemistry Quiz

IB Chemistry Quiz: Using Experimental Techniques

Practice Using Experimental Techniques in IB Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Using Experimental Techniques, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

In a paper chromatography experiment to separate a mixture of amino acids, a polar stationary phase (cellulose) is used with a non-polar solvent. Amino acid X has an R_f value of 0.20 and amino acid Y has an R_f value of 0.80. Which deduction is correct?

  1. Amino acid X is more soluble in the non-polar solvent than amino acid Y.
  2. Amino acid Y has stronger intermolecular forces of attraction to the stationary phase than amino acid X.
  3. Amino acid Y is less polar than amino acid X. (correct answer)
  4. The mobile phase is polar and the stationary phase is non-polar.
Explanation: C is correct. The R_f value is a ratio of the distance travelled by the substance to the distance travelled by the solvent front. A higher R_f value (Y = 0.80) indicates the substance travels further with the mobile phase. Since the mobile phase is non-polar, this implies that substance Y is more soluble in it and less attracted to the polar stationary phase. Substances that are more soluble in a non-polar solvent are themselves less polar. Therefore, Y is less polar than X. A is incorrect because Y has the higher R_f and is thus more soluble in the non-polar solvent. B is incorrect because Y has a higher R_f, meaning it has weaker, not stronger, attraction to the stationary phase. D incorrectly describes the phases.

Question 2

During the purification of an impure solid by recrystallization, a minimum amount of hot solvent is used to dissolve the crude product. What is the primary reason for using the minimum amount of solvent?

  1. To ensure that any insoluble impurities are left undissolved.
  2. To minimize the amount of energy required to heat the solution to boiling.
  3. To maximize the yield of crystals upon cooling the solution. (correct answer)
  4. To speed up the rate of filtration when collecting the crystals.
Explanation: C is correct. Recrystallization relies on the desired compound being less soluble in the cold solvent. After the compound is dissolved in hot solvent, the solution is cooled to induce crystallization. If an excess of solvent is used, the solution may not become saturated upon cooling, and a significant amount of the desired product will remain dissolved, leading to a low yield. Using the minimum amount of solvent ensures the solution is saturated or near-saturated with respect to the product when it cools, maximizing the amount that crystallizes out. A is related to the hot filtration step, not the amount of solvent. B and D are practical considerations but not the primary chemical reason for this step.

Question 3

Which experimental technique would be most suitable for monitoring the rate of the following reaction? Zn(s)+H2SO4(aq)→ZnSO4(aq)+H2(g)Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

  1. Measuring the change in pH over time.
  2. Measuring the change in electrical conductivity over time.
  3. Measuring the change in the total mass of the flask and contents over time. (correct answer)
  4. Measuring the change in colour intensity over time.
Explanation: C is correct. The reaction produces hydrogen gas (H₂), which escapes from the open flask. This leads to a decrease in the total mass of the flask and its contents. By placing the flask on a balance, this change in mass can be measured over time and used to determine the rate of reaction. A is not ideal because the change in [H⁺] is what drives the reaction, but measuring pH during a vigorous reaction with a solid is difficult. B is also possible as 2H⁺ ions are replaced by one Zn²⁺ ion, changing the conductivity, but C is a more direct and common method. D is incorrect as all reactants and products in solution are colorless.

Question 4

A student synthesizes a solid product via precipitation, collects it by filtration, and washes it with deionized water. The product is then dried in an oven. If the student does not dry the product sufficiently, how will this affect the calculated percentage yield?

  1. The percentage yield will be lower than the true value.
  2. The percentage yield will be higher than the true value. (correct answer)
  3. The percentage yield will be unaffected if the same balance is used for all weighings.
  4. The effect cannot be determined without knowing the molar mass of the product.
Explanation: B is correct. The percentage yield is calculated as (actual mass / theoretical mass) × 100%. If the product is not dried completely, its measured mass will include the mass of the residual solvent (water). This makes the measured 'actual mass' artificially high. A value that is higher than the true mass of the product will lead to a calculated percentage yield that is also higher than the true value. It is common for this error to lead to calculated yields greater than 100%.

Question 5

An aqueous solution contains an organic compound X which is much more soluble in ethyl acetate than in water. Ethyl acetate is immiscible with water and has a density of 0.90 g cm⁻³. To extract X, a student adds ethyl acetate to the aqueous solution in a separating funnel, shakes the mixture, and allows the layers to separate. Which statement is correct?

  1. The lower layer is the ethyl acetate layer containing compound X.
  2. Shaking is avoided to prevent the two immiscible liquids from mixing.
  3. The top layer is run out of the funnel by opening the stopcock.
  4. The bottom aqueous layer is run out through the stopcock first. (correct answer)
Explanation: D is correct. Since ethyl acetate has a density (0.90 g cm⁻³) less than water (~1.0 g cm⁻³), it will form the top layer. Compound X will move into this top layer. In a separating funnel, the lower layer is always drained out through the stopcock at the bottom first. The top layer is then either drained through the stopcock or poured out of the top opening. A is incorrect because the ethyl acetate is the top layer. B is incorrect because vigorous shaking is required for efficient extraction. C is incorrect because the bottom layer must be removed before the top layer can be run out through the stopcock.

Question 6

A student heats a sample of hydrated magnesium sulfate, MgSO₄·xH₂O, to determine the value of x. After heating, the crucible is allowed to cool on the benchtop before weighing. The anhydrous MgSO₄ is hygroscopic. How would this error affect the calculated value of x?

  1. The calculated value of x will be too high.
  2. The calculated value of x will be too low. (correct answer)
  3. The calculated value of x will be unaffected.
  4. The effect will be random, so x could be too high or too low.
Explanation: B is correct. The value of x is calculated from the ratio of moles of water lost to moles of anhydrous salt remaining. Moles of water = (initial mass - final mass) / M_r(H₂O). Moles of salt = final mass / M_r(MgSO₄). If the anhydrous salt is cooled on the bench, it will absorb moisture from the air because it is hygroscopic. This will make the measured 'final mass' of the anhydrous salt higher than its true value. A higher final mass means the calculated mass loss ('mass of water') will be lower. A lower mass of water and a higher mass of salt will both contribute to making the calculated ratio of moles (and thus the value of x) too low.

Question 7

Which measurement in an acid-base titration typically has the lowest percentage uncertainty, assuming standard class A glassware is used?

  1. The mass of the primary standard used to make the standard solution. (correct answer)
  2. The volume of the aliquot measured with a 25.00 cm³ pipette.
  3. The final volume of the 250.0 cm³ standard solution in the volumetric flask.
  4. The titre volume of approximately 20 cm³ delivered from a 50.00 cm³ burette.
Explanation: A is correct. A modern analytical balance can measure mass to ±0.0001 g. For a typical mass of a primary standard (e.g., ~2 g), the percentage uncertainty is extremely small: (0.0001 / 2) × 100% ≈ 0.005%. The uncertainties for glassware are larger. A 250.0 cm³ volumetric flask (C) has an uncertainty of about ±0.12 cm³, giving (0.12 / 250) × 100% ≈ 0.05%. A 25.00 cm³ pipette (B) is about ±0.03 cm³, giving (0.03 / 25) × 100% ≈ 0.12%. A burette reading (D) has two uncertainties (initial and final reading), each about ±0.02 cm³, for a total uncertainty of about ±0.04 cm³ in the titre. For a 20 cm³ titre, this is (0.04 / 20) × 100% ≈ 0.2%. Therefore, the mass measurement is the most precise.

Question 8

In a titration to determine the concentration of an unknown acid, a student uses a pipette to transfer 25.00 cm³ of the acid into a conical flask and then titrates it with a standard solution of NaOH from a burette. Which procedural error would lead to a calculated concentration of the acid that is consistently lower than its true value?

  1. Rinsing the pipette with deionized water immediately before filling it with the acid solution. (correct answer)
  2. Rinsing the burette with the standard NaOH solution before filling it.
  3. Adding approximately 50 cm³ of deionized water to the conical flask after adding the acid.
  4. Reading the initial burette volume from the top of the meniscus and the final volume from the bottom.
Explanation: A is correct because rinsing the pipette with water and not the acid solution will leave water droplets inside, which dilutes the acid as it is being measured. This means fewer moles of acid are transferred to the conical flask than the 25.00 cm³ volume would suggest. Consequently, a smaller volume of NaOH titrant will be required to reach the endpoint, leading to a calculated acid concentration that is erroneously low. B is a correct procedure, not an error. C is a correct procedure; adding water to the conical flask does not change the number of moles of acid being titrated. D describes a random error, not a systematic error that would lead to a consistently lower value.

Question 9

A student is titrating 25.0 cm³ of 0.100 mol dm⁻³ ammonia (NH₃, a weak base) with 0.100 mol dm⁻³ hydrochloric acid (HCl, a strong acid). An indicator's pKₐ should be close to the pH at the equivalence point. Which indicator would be most appropriate?

  1. Phenolphthalein (pKₐ ≈ 9.6), which changes colour in the pH range 8.3–10.0.
  2. Bromothymol blue (pKₐ ≈ 7.1), which changes colour in the pH range 6.0–7.6.
  3. Methyl orange (pKₐ ≈ 3.7), which changes colour in the pH range 3.1–4.4. (correct answer)
  4. Any of these indicators can be used as it is a strong acid-weak base titration.
Explanation: C is correct. This is a weak base-strong acid titration. At the equivalence point, all the ammonia has been converted to its conjugate acid, the ammonium ion (NH₄⁺). The ammonium ion is a weak acid and will hydrolyze water: NH₄⁺(aq) ⇌ NH₃(aq) + H⁺(aq). This production of H⁺ ions means the pH at the equivalence point will be less than 7 (i.e., acidic), typically around 5-6. The best indicator is one whose color change range brackets this pH. Methyl orange changes color in the acidic range of 3.1-4.4, which is the closest and most appropriate choice as the pH changes very rapidly in this region around the equivalence point. Phenolphthalein (A) would change too late, and bromothymol blue (B) would change slightly after the equivalence point on the steep part of the curve, but methyl orange is a better fit. D is incorrect because the choice of indicator is critical.

Question 10

A student performs an experiment to determine the enthalpy of neutralization of HCl and NaOH in a polystyrene cup. The calculated enthalpy change is found to be less exothermic than the data booklet value. Which of the following is the most likely significant reason for this discrepancy, assuming the concentrations and volumes of solutions are accurate?

  1. The heat capacity of the polystyrene cup was assumed to be negligible.
  2. The final mass of the solution was calculated by adding the masses of the two initial solutions.
  3. The density of the final solution was assumed to be the same as that of pure water.
  4. Heat was lost to the surroundings through the unlidded top of the cup. (correct answer)
Explanation: D is correct because heat loss to the surroundings is the most significant source of systematic error in simple calorimetry. This loss prevents the solution from reaching its theoretical maximum temperature, resulting in a smaller measured ΔT. A smaller ΔT leads to a smaller calculated heat change (Q = mcΔT), and thus a calculated enthalpy change (ΔH = -Q/n) that is less exothermic (less negative) than the true value. While A and C are also sources of error, they are typically less significant in magnitude than the heat lost from an open container over the course of the experiment.

Question 11

A student needs to separate a mixture of propan-1-ol (boiling point 97 °C) and butan-1-ol (boiling point 117 °C). Which statement best justifies the use of fractional distillation instead of simple distillation?

  1. The difference in boiling points (20 °C) is too small for effective separation by simple distillation. (correct answer)
  2. Simple distillation cannot separate miscible liquids, only a solid dissolved in a liquid.
  3. Fractional distillation is faster because it does not require a condenser.
  4. The large surface area in a simple distillation flask ensures a more complete separation.
Explanation: A is correct. Fractional distillation is specifically designed to separate liquid mixtures where the boiling points of the components are close (typically < 25 °C difference). The fractionating column provides a large surface area for repeated vaporization-condensation cycles, which enriches the vapor in the more volatile component (propan-1-ol), allowing for an effective separation. Simple distillation would only achieve a partial separation, with the distillate being only slightly enriched in the lower-boiling point component. B is incorrect; simple distillation can separate miscible liquids, just not effectively if their boiling points are close. C is incorrect; fractional distillation is generally slower and requires a fractionating column in addition to a condenser. D is incorrect; the large surface area for separation is a feature of the fractionating column, not the distillation flask.

Question 12

A student collects hydrogen gas from the reaction of a metal with acid by displacing water in a measuring cylinder. Before reading the volume, the student ensures the water level inside the cylinder is equal to the water level outside. What is the reason for this step?

  1. To ensure the temperature of the gas is equal to the temperature of the water.
  2. To ensure the pressure of the collected gas is equal to the atmospheric pressure. (correct answer)
  3. To correct for the volume of water vapour mixed with the hydrogen gas.
  4. To make it easier to read the meniscus on the measuring cylinder scale.
Explanation: B is correct. When the water levels inside and outside the collection vessel are equal, the pressures at that level are also equal. The pressure on the outside is atmospheric pressure. The pressure on the inside is the total pressure of the collected gas (hydrogen + water vapour). By equating the levels, the total pressure inside is made equal to the atmospheric pressure. This simplifies the subsequent calculation using the ideal gas law, as the atmospheric pressure can be easily measured with a barometer. A and C relate to other aspects of the experiment (temperature and partial pressure of water) that are not addressed by this specific action. D is incorrect as the ease of reading is not the primary scientific reason for this crucial step.

Question 13

To prepare 100.0 cm³ of a 0.0200 mol dm⁻³ solution from a 0.100 mol dm⁻³ stock solution, a student requires 20.00 cm³ of the stock solution. Which set of glassware provides the highest accuracy for this dilution?

  1. A 20 cm³ measuring cylinder and a 100 cm³ beaker.
  2. A 50 cm³ burette and a 100 cm³ measuring cylinder.
  3. A 20.00 cm³ volumetric pipette and a 100.0 cm³ volumetric flask. (correct answer)
  4. A 25 cm³ graduated pipette and a 100.0 cm³ conical flask.
Explanation: C is correct. For preparing accurate solutions (indicated by the number of significant figures), the most precise glassware should be used. A volumetric pipette is calibrated to deliver a single, highly accurate volume (e.g., 20.00 cm³). A volumetric flask is calibrated to contain a single, highly accurate volume (e.g., 100.0 cm³). This combination is the standard and most accurate method for performing a dilution. Measuring cylinders (A, B) and conical flasks (D) are not designed for high-precision volume measurements.

Question 14

A student is preparing a buffer solution by mixing 50.0 mL of 0.100 M acetic acid with 50.0 mL of 0.100 M sodium acetate. They accidentally add 2.0 mL of 0.10 M HCl to the mixture. To restore the original buffer capacity most effectively, what should the student add?

  1. 2.0 mL of 0.10 M NaOH to neutralize the excess acid and restore the original pH
  2. 0.20 g of solid sodium acetate to replace the acetate consumed by the added HCl (correct answer)
  3. 2.0 mL of 0.10 M sodium acetate to maintain the buffer ratio and total volume
  4. 1.0 mL of 0.20 M NaOH to neutralize the HCl while minimizing volume change
Explanation: The HCl converts acetate ions to acetic acid (CH₃COO⁻ + HCl → CH₃COOH + Cl⁻), reducing the buffer's base component. Adding solid sodium acetate restores the lost acetate without significantly changing the volume or adding excess base. NaOH would create excess base, while adding sodium acetate solution would dilute the buffer unnecessarily.

Question 15

A student is extracting caffeine from tea using a separatory funnel with dichloromethane and water. After shaking and allowing the layers to separate, they observe three distinct layers instead of two. The middle layer appears cloudy and does not settle completely even after 10 minutes of standing. What is the most appropriate action to take?

  1. Add anhydrous sodium sulfate to the funnel to absorb excess water and clarify the layers
  2. Filter the entire mixture through filter paper to remove suspended particles causing the cloudiness
  3. Drain each layer separately and dry the organic phases individually over anhydrous magnesium sulfate
  4. Add a few drops of saturated salt solution to break the emulsion and improve layer separation (correct answer)
Explanation: When you encounter multiple layers in a liquid-liquid extraction, you're likely dealing with an emulsion—a stabilized mixture where tiny droplets of one phase are suspended in another. This commonly occurs when extracting organic compounds like caffeine because vigorous shaking can create persistent droplets that resist separation. The cloudy middle layer indicates an emulsion has formed between your aqueous and organic phases. To break this emulsion, you need to disrupt the forces stabilizing these tiny droplets and encourage them to coalesce into their respective phases. Adding saturated salt solution (answer D) is the correct approach because the high ionic strength destabilizes the emulsion. The salt ions interact with water molecules and any surface-active compounds that might be stabilizing the droplets, allowing the phases to separate cleanly into two distinct layers. Answer A is incorrect because anhydrous sodium sulfate is a drying agent used to remove dissolved water from organic solvents—it won't break an emulsion or clarify the layers. Answer B won't work because you can't effectively filter an emulsion; the droplets are too small and the mixture will likely clog your filter paper. Answer C assumes you can already separate the layers cleanly, but the whole problem is that the emulsion prevents proper separation—you'd just be transferring the emulsified mixture between containers. Remember: when you see unexpected multiple layers or cloudiness in extractions, think "emulsion first." Salt solutions, gentle swirling (not shaking), or sometimes just patience can resolve most emulsion problems without losing your product.

Question 16

A student is performing a redox titration using potassium permanganate (KMnO₄) as the titrant against an unknown concentration of iron(II) sulfate solution. The reaction is: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

During the titration, the student observes that the endpoint color change (from colorless to pale pink) appears gradually over 10-15 seconds rather than sharply. After adding one additional drop, the solution becomes deep purple. What is the most likely cause of this observation and how should the student proceed?

  1. Manganese dioxide precipitation is interfering; filter the solution and continue the titration from that point
  2. The solution pH is too low causing side reactions; add buffer to maintain optimal pH for the reaction
  3. The titration rate is too fast; repeat the titration adding drops more slowly near the expected endpoint (correct answer)
  4. The permanganate solution is too concentrated; dilute it by half and repeat the entire titration procedure
Explanation: When performing redox titrations, the sharpness of the endpoint depends on how quickly you add titrant near the equivalence point. In permanganate titrations, you're looking for the first persistent pale pink color that indicates a slight excess of MnO4−\text{MnO}_4^- ions. The gradual color change over 10-15 seconds followed by a deep purple after one more drop is a classic sign that you're adding titrant too quickly. Near the endpoint, the concentration of Fe2+\text{Fe}^{2+} becomes very low, so each drop of permanganate creates a temporary local excess that takes time to react completely. When you add drops too fast, you can't distinguish between this temporary color and the true endpoint. Option C correctly identifies that slower addition near the expected endpoint will give you a sharper, more accurate endpoint. You should repeat the titration, adding drops rapidly at first, then switching to dropwise addition when you approach the expected volume. Option A is wrong because manganese dioxide (brown precipitate) isn't mentioned in the observations - the student sees purple, which indicates excess permanganate, not MnO2\text{MnO}_2 formation. Option B misses the point since this reaction actually requires acidic conditions (note the 8H+8\text{H}^+ in the equation), and the gradual endpoint isn't due to pH issues. Option D unnecessarily complicates the procedure when the problem is simply technique-related. Study tip: In any titration, when you observe a gradual rather than sharp endpoint, your first thought should be titration technique - specifically whether you're adding titrant too quickly near the equivalence point.

Question 17

During a crystallization experiment, a student dissolves 5.0 g of impure benzoic acid in 100 mL of hot water, then allows it to cool slowly. After filtration, they recover 3.8 g of crystals. Analysis shows the crystals are 98% pure benzoic acid. What can be concluded about the original sample's purity?

  1. The original sample was 74% pure, calculated from the mass recovery ratio
  2. The original sample was 76% pure, accounting for both recovery and final purity
  3. The original sample was 98% pure, since crystallization doesn't change the purity significantly
  4. The purity cannot be determined without knowing the solubility of the impurities in water (correct answer)
Explanation: The purity cannot be calculated without knowing whether the impurities are soluble or insoluble in water. If impurities are highly soluble, they remain in solution and the recovery loss is mainly pure benzoic acid. If impurities are insoluble, they may be retained in the crystals or removed during filtration. The mass recovery alone doesn't indicate original purity without this information.

Question 18

A student performs a calorimetry experiment to determine the enthalpy of combustion of ethanol. They burn 0.250 g of ethanol and observe a temperature increase of 12.4°C in 200.0 g of water. However, they forgot to account for the heat capacity of the calorimeter itself, which is 45.2 J/°C. What is the corrected value for the heat released?

  1. 10.4 kJ, calculated using only the water's heat capacity and temperature change
  2. 10.9 kJ, calculated by including both water and calorimeter heat capacities (correct answer)
  3. 10.1 kJ, calculated by subtracting the calorimeter's heat absorption from water heating
  4. 11.5 kJ, calculated by adding the heat lost to the calorimeter walls separately
Explanation: Total heat released = (mass_water × c_water + C_calorimeter) × ΔT = (200.0 g × 4.184 J/g°C + 45.2 J/°C) × 12.4°C = (836.8 + 45.2) J/°C × 12.4°C = 882.0 × 12.4 J = 10,937 J = 10.9 kJ. Both the water and calorimeter absorb heat, so their heat capacities must be added.

Question 19

During a gravimetric analysis experiment, a student obtains a mass of 1.2456 g for their dried precipitate using an analytical balance. However, they realize they forgot to pre-dry the filter paper, which typically retains 0.0008 g of moisture per square centimeter. If the filter paper has an area of 78.5 cm², what is the corrected mass of the precipitate?

  1. 1.1828 g, by subtracting the total moisture content from the measured mass (correct answer)
  2. 1.1834 g, by subtracting moisture and accounting for measurement uncertainty
  3. 1.3085 g, by adding the moisture that should have been removed by pre-drying
  4. 1.2456 g, since the moisture content is within the balance's uncertainty range
Explanation: The measured mass includes both precipitate and moisture from the undried filter paper. Moisture mass = 0.0008 g/cm² × 78.5 cm² = 0.0628 g. Corrected precipitate mass = 1.2456 g - 0.0628 g = 1.1828 g. The moisture must be subtracted because it was included in the measurement but is not part of the actual precipitate.

Question 20

A student is conducting a kinetics experiment by monitoring the disappearance of a colored reactant using a spectrophotometer. They notice that their absorbance readings are gradually decreasing even in a control sample that contains no catalyst. The decrease follows a consistent pattern over 30 minutes. What is the most likely explanation and appropriate correction?

  1. Photochemical decomposition is occurring; use a reference cell and subtract its absorbance change from sample readings (correct answer)
  2. The instrument baseline is drifting; re-zero the spectrophotometer every 5 minutes during the experiment
  3. Solvent evaporation is concentrating the solution; cover all samples and use sealed cuvettes for future runs
  4. Temperature fluctuations are affecting the electronic components; allow longer equilibration time before measurements
Explanation: A consistent decrease in absorbance in a control sample (without catalyst) suggests photochemical decomposition of the colored compound due to exposure to the spectrometer's light source. Using a reference cell with the same solution and subtracting its absorbance change accounts for this effect. Baseline drift would be random, evaporation would increase concentration (increase absorbance), and temperature effects would be irregular.