Chemistry Exam Solutions

Chemistry Exam Solutions

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Question 1

A student finds the concentration of a dilute acid, HA(aq), by titration.

The student adds 25.0 cm³ of aqueous sodium hydroxide to a conical flask, adds a few drops of methyl orange indicator, slowly adds HA(aq) until the methyl orange changes colour, and records the volume of HA(aq) added.

Solution

(a)(i) The apparatus shown in Fig. 1.1 is a burette.

(a)(ii) A more suitable piece of apparatus to measure 25.0 cm³ of aqueous sodium hydroxide is a pipette.

(a)(iii) The apparatus shown in Fig. 1.2 is a burette.

(a)(iv) Volume of HA(aq) used = 25.0 cm³ (final reading 26.0 cm³ – initial reading 1.0 cm³).

(b) The substance used to wash the apparatus is distilled water.

(c) The colour change of methyl orange at the end-point is from yellow to red.

Strategy & Key Concepts

Titration Fundamentals: Titration is a quantitative analytical technique used to determine the concentration of an unknown solution by reacting it with a solution of known concentration.

Apparatus Selection:

  • Burette – Used to deliver variable, precise volumes of titrant
  • Pipette – Used to deliver fixed, precise volumes of solution
  • Conical flask – Preferred over beakers for titrations as its shape minimizes splashing and facilitates mixing

Methyl Orange Indicator: This is an acid-base indicator that changes color in the pH range of 3.1 (red) to 4.4 (yellow). It’s suitable for strong acid-strong base titrations with equivalence point around pH 7.

Washing Technique: Always rinse glassware with distilled water to prevent contamination. For volumetric apparatus like burettes and pipettes, rinse with the solution to be used to maintain concentration accuracy.

Related Problems & Solutions

Problem 1: A student uses phenolphthalein instead of methyl orange for the same titration. What would be the color change at the endpoint?

Solution: Phenolphthalein changes from colorless to pink at endpoint (pH 8.3-10.0).

Problem 2: If a student forgets to rinse the burette with the acid solution before filling it, how would this affect the calculated acid concentration?

Solution: The acid would be diluted by residual water, leading to a higher calculated concentration of the acid (more volume needed to reach endpoint).

Problem 3: Why is a conical flask preferred over a beaker for titrations?

Solution: The conical shape minimizes splashing during swirling and allows for better mixing of solutions.

Question 2

A student investigates the reaction of four metals, A, B, C and D, with aqueous copper(II) sulfate.

The student measures temperature changes and observes physical changes to determine reactivity.

Solution

(a)(i) The value recorded to an incorrect degree of precision is 69.5°C (should be recorded to 0.5°C like other measurements).

(a)(ii) Completed table:

Metal Initial Temperature (°C) Highest Temperature (°C) Temperature Increase (°C)
A 20.0 69.5 49.5
B 24.5 46.0 21.5
C 22.0 61.0 39.0
D 20.0 22.0 2.0

(a)(iii) The colorless solution indicates all Cu²⁺ ions have been reduced to copper metal. The presence of unreacted grey solid (metal B) confirms it is in excess.

(a)(iv)

  • Colorless solution: BSO₄(aq)
  • Grey solid: B(s)
  • Brown solid: Cu(s)

(b) Order of reactivity: A → C → B → D (most to least reactive)

Explanation: More reactive metals cause greater temperature increases as they release more heat energy during the displacement reaction.

(c) Suggested temperature increase for the fifth metal (second most reactive): 45.0°C (between 39.0°C for C and 49.5°C for A)

(d)

Reason: Heat loss to surroundings (beaker, air)

Improvement: Use a polystyrene cup or place a lid on the beaker to insulate the reaction mixture

(e)

Effect: The temperature increase would be smaller

Explanation: Lower concentration means fewer copper(II) sulfate particles available to react with a fixed mass of metal A. The reaction will be less extensive, releasing less heat energy.

Strategy & Key Concepts

Reactivity Series: Metals can be arranged in order of their reactivity based on their ability to displace other metals from their compounds. The more reactive a metal, the more readily it loses electrons to form positive ions.

Displacement Reactions: A more reactive metal will displace a less reactive metal from its salt solution. General equation: Metal A + Salt of Metal B → Salt of Metal A + Metal B

Temperature Change as Indicator: Displacement reactions are exothermic. The greater the temperature increase, the more vigorous the reaction, indicating higher reactivity.

Excess Reactant Concept: When a reactant is in excess, it means there is more than enough to completely react with the other reactant. Some of the excess reactant will remain unreacted.

Experimental Improvements:

  • Use insulation to minimize heat loss
  • Use similar masses of metals for fair comparison
  • Ensure constant concentration and volume of copper(II) sulfate solution
  • Stir consistently to ensure even temperature distribution

Related Problems & Solutions

Problem 1: If zinc is added to copper(II) sulfate solution, what observations would you make?

Solution: The blue color of copper(II) sulfate fades, a brown precipitate of copper forms, and the temperature increases.

Problem 2: Why doesn’t copper react with zinc sulfate solution?

Solution: Copper is less reactive than zinc, so it cannot displace zinc from its salt solution.

Problem 3: How would using powdered metal instead of solid pieces affect the temperature change?

Solution: Powdered metal has a larger surface area, leading to a faster reaction and potentially a greater maximum temperature increase.

Question 3

A student does a series of experiments to investigate solution R through various chemical tests.

Solution

(a)(i) Two possible conclusions:

  • R contains lithium ions (Li⁺)
  • R contains calcium ions (Ca²⁺)

(a)(ii) It is difficult to make a definite conclusion because the yellow Bunsen flame can mask or obscure the red flame color, making it hard to distinguish between red (lithium) and orange-red (calcium).

(b) The observation is no white precipitate forms.

(c) R contains chloride ions (Cl⁻).

(d)(i) Another observation: effervescence/bubbles of gas are produced.

(d)(ii) It is important to add dilute nitric acid to remove carbonate ions (CO₃²⁻) which would also form a precipitate with silver nitrate and interfere with the test for chloride ions.

(e) The observation is a white precipitate forms.

(f)

What the student does: Perform a flame test

Observation: Orange-red flame

(g)(i) The observation is the gas turns damp red litmus paper blue.

(g)(ii) The cation present is ammonium ions (NH₄⁺).

(h) The two compounds are calcium chloride and ammonium chloride.

Strategy & Key Concepts

Qualitative Analysis: This involves identifying ions present in an unknown solution through systematic chemical tests.

Flame Tests: Certain metal ions impart characteristic colors to flames:

  • Lithium (Li⁺) – Crimson red
  • Sodium (Na⁺) – Yellow
  • Potassium (K⁺) – Lilac
  • Calcium (Ca²⁺) – Orange-red
  • Barium (Ba²⁺) – Pale green

Precipitation Tests for Anions:

  • Sulfate (SO₄²⁻) – White precipitate with Ba²⁺ ions in acidic conditions
  • Chloride (Cl⁻) – White precipitate with Ag⁺ ions in acidic conditions
  • Carbonate (CO₃²⁻) – Effervescence with acid, producing CO₂ gas

Tests for Cations:

  • Ammonium (NH₄⁺) – Produces ammonia gas when warmed with NaOH
  • Calcium (Ca²⁺) – White precipitate with NaOH, insoluble in excess

Systematic Approach: Always follow a logical sequence when identifying unknown ions to avoid false positives and ensure accurate identification.

Related Problems & Solutions

Problem 1: How would you distinguish between solutions of sodium chloride and sodium carbonate?

Solution: Add dilute acid – sodium carbonate will effervesce (produce CO₂), while sodium chloride will not.

Problem 2: A solution gives a white precipitate with NaOH that dissolves in excess. What cation might be present?

Solution: Aluminum (Al³⁺) or zinc (Zn²⁺) ions.

Problem 3: Why is nitric acid used instead of hydrochloric acid to acidify solutions before testing for sulfate ions?

Solution: Hydrochloric acid would introduce chloride ions which could form precipitates with other tests, while nitric acid doesn’t introduce interfering anions.

Question 4

Plan an experiment to determine the percentage loss in mass when barium carbonate is heated.

Barium carbonate → barium oxide + carbon dioxide

Solution

Apparatus Needed:

  • Balance (accurate to at least 0.01 g)
  • Porcelain crucible with lid
  • Pipe-clay triangle
  • Tripod stand
  • Bunsen burner
  • Tongs
  • Desiccator (optional, for cooling)

Method:

  1. Clean and dry the crucible and lid. Heat them strongly for a few minutes to remove any moisture or impurities. Allow to cool completely (ideally in a desiccator).
  2. Weigh the empty crucible with lid accurately and record mass (M₁).
  3. Add a known mass of solid barium carbonate (2-3 g) to the crucible. Weigh the crucible, lid, and barium carbonate together and record mass (M₂).
  4. Set up apparatus: place crucible on pipe-clay triangle on tripod. Heat strongly with Bunsen burner, lifting lid occasionally to allow CO₂ to escape.
  5. Continue heating strongly for 10-15 minutes until no further change is apparent (no more fizzing).
  6. Allow crucible and lid to cool completely (ideally in desiccator).
  7. Weigh the crucible, lid and contents after cooling and record mass (M₃).
  8. Reheat for 5 minutes, cool, and reweigh. Repeat until constant mass (M₃) is obtained.

Procedures for Accuracy:

  • Heat strongly and for sufficient time to ensure complete decomposition
  • Lift lid occasionally to allow CO₂ to escape, but carefully to prevent loss of solid
  • Cool crucible completely in desiccator to prevent barium oxide from absorbing moisture
  • Heat to constant mass to confirm reaction is complete

Calculations:

  • Mass of barium carbonate before heating = M₂ – M₁
  • Mass of solid after heating (barium oxide) = M₃ – M₁
  • Mass of carbon dioxide lost = (M₂ – M₁) – (M₃ – M₁) = M₂ – M₃
  • Percentage loss in mass = (Mass of CO₂ lost / Mass of barium carbonate) × 100% = [(M₂ – M₃) / (M₂ – M₁)] × 100%

Strategy & Key Concepts

Thermal Decomposition: Some compounds break down into simpler substances when heated. Barium carbonate decomposes to barium oxide and carbon dioxide gas.

Mass Loss Measurement: By measuring mass before and after heating, we can determine the mass of gas evolved, which allows calculation of the percentage composition.

Experimental Design Principles:

  • Control variables: Use consistent heating conditions and timing
  • Minimize errors: Use accurate balance, prevent moisture absorption, ensure complete reaction
  • Safety: Use tongs to handle hot apparatus, work in well-ventilated area

Stoichiometry Application: The mass loss corresponds directly to the carbon dioxide produced, allowing calculation of the reaction extent and verification of the decomposition equation.

Common Experimental Issues:

  • Incomplete decomposition – solved by heating to constant mass
  • Absorption of moisture by product – solved by cooling in desiccator
  • Loss of solid during heating – solved by careful handling and using a lid

Related Problems & Solutions

Problem 1: How would the percentage loss in mass differ if the heating was insufficient?

Solution: The percentage loss would be lower as not all barium carbonate would have decomposed.

Problem 2: Why is it important to cool the crucible in a desiccator?

Solution: Barium oxide is hygroscopic and will absorb moisture from air, increasing the mass and giving inaccurate results.

Problem 3: Calculate the theoretical percentage loss in mass for pure barium carbonate.

Solution: Molar mass BaCO₃ = 137 + 12 + 48 = 197 g/mol
Mass of CO₂ lost = 44 g
Theoretical % loss = (44/197) × 100 = 22.34%