IB Chemistry Quiz: Applying Mathematics To Chemistry
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Applying Mathematics To ChemistryQuestion 1 of 20

In a calorimetry experiment, 100.0 cm³ of 0.50 mol dm⁻³ AgNO₃(aq) is mixed with 100.0 cm³ of 0.50 mol dm⁻³ HCl(aq). The temperature increases by 3.25 K. Assuming the solution has a density of 1.00 g cm⁻³ and a specific heat capacity of 4.18 J g⁻¹ K⁻¹, what is the molar enthalpy change, in kJ mol⁻¹, for the precipitation of AgCl? AgNO₃(aq) + HCl(aq) → AgCl(s) + HNO₃(aq)

-54.3
-27.2
+54.3
-108.6
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IB Chemistry Quiz

IB Chemistry Quiz: Applying Mathematics To Chemistry

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

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

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Question 1

In a calorimetry experiment, 100.0 cm³ of 0.50 mol dm⁻³ AgNO₃(aq) is mixed with 100.0 cm³ of 0.50 mol dm⁻³ HCl(aq). The temperature increases by 3.25 K. Assuming the solution has a density of 1.00 g cm⁻³ and a specific heat capacity of 4.18 J g⁻¹ K⁻¹, what is the molar enthalpy change, in kJ mol⁻¹, for the precipitation of AgCl? AgNO₃(aq) + HCl(aq) → AgCl(s) + HNO₃(aq)

  1. -54.3 (correct answer)
  2. -27.2
  3. +54.3
  4. -108.6
Explanation: First, calculate the heat released (Q): Q = mcΔT. Total volume = 100.0 + 100.0 = 200.0 cm³. Total mass = 200.0 cm³ × 1.00 g cm⁻³ = 200.0 g. Q = 200.0 g × 4.18 J g⁻¹ K⁻¹ × 3.25 K = 2717 J. Next, find the limiting reactant moles (n). Moles AgNO₃ = 0.1000 dm³ × 0.50 mol dm⁻³ = 0.050 mol. Moles HCl = 0.1000 dm³ × 0.50 mol dm⁻³ = 0.050 mol. The reaction is 1:1, so n = 0.050 mol. Finally, calculate molar enthalpy change: ΔH = -Q/n. The negative sign is used because the temperature increased (exothermic). ΔH = -2717 J / 0.050 mol = -54340 J mol⁻¹. Converting to kJ mol⁻¹ gives -54.3 kJ mol⁻¹.

Question 2

A solution is prepared by dissolving 4.05 g of sodium hydroxide (NaOH) in deionized water to make a 200.0 cm³ stock solution. A 25.00 cm³ portion of this stock solution is then diluted with water to a final volume of 500.0 cm³. What is the final concentration of the diluted solution in mol dm⁻³, expressed to the appropriate number of significant figures?

  1. 0.0253 (correct answer)
  2. 0.025
  3. 0.02531
  4. 0.025313
Explanation: The measurement with the fewest significant figures is the mass of NaOH (4.05 g), which has three. All other measurements (200.0 cm³, 25.00 cm³, 500.0 cm³) have four. Therefore, the final answer must be given to three significant figures. Calculation: M(NaOH) = 39.997 g mol⁻¹. Moles of NaOH = 4.05 g / 39.997 g mol⁻¹ = 0.101257... mol. Stock concentration = 0.101257... mol / 0.2000 dm³ = 0.50628... mol dm⁻³. Using the dilution formula C₁V₁ = C₂V₂, C₂ = (C₁V₁) / V₂ = (0.50628... mol dm⁻³ × 25.00 cm³) / 500.0 cm³ = 0.025314... mol dm⁻³. Rounding to three significant figures gives 0.0253 mol dm⁻³.

Question 3

For the reaction A(g) → 2B(g), the concentration of reactant A was measured at various times. At t = 20 s, [A] = 0.64 mol dm⁻³. At t = 50 s, [A] = 0.28 mol dm⁻³. What is the average rate of formation of product B over this time interval, in mol dm⁻³ s⁻¹?

  1. 0.012
  2. 0.024 (correct answer)
  3. 0.006
  4. 0.018
Explanation: First, calculate the average rate of reaction with respect to A. Rate = -Δ[A]/Δt = -(0.28 - 0.64) mol dm⁻³ / (50 - 20) s = -(-0.36) / 30 = 0.012 mol dm⁻³ s⁻¹. This is the rate of consumption of A. From the stoichiometry A → 2B, the rate of formation of B is twice the rate of consumption of A. Therefore, the rate of formation of B = 2 × 0.012 mol dm⁻³ s⁻¹ = 0.024 mol dm⁻³ s⁻¹.

Question 4

A 50.0 cm³ sample of gas is collected at 20 °C and 98.0 kPa. The sample is found to have a mass of 0.088 g. Which of the following is the most likely identity of the gas?

  1. N₂ (M = 28.02 g mol⁻¹)
  2. O₂ (M = 32.00 g mol⁻¹)
  3. Ar (M = 39.95 g mol⁻¹)
  4. CO₂ (M = 44.01 g mol⁻¹) (correct answer)
Explanation: Use the ideal gas law PV=nRT to find the moles (n), then calculate molar mass (M = mass/n). Convert units: P = 98000 Pa, V = 50.0 cm³ = 5.00 × 10⁻⁵ m³, T = 20 °C = 293.15 K. n = PV/RT = (98000 × 5.00 × 10⁻⁵) / (8.31 × 293.15) = 0.00201 mol. Now calculate M: M = 0.088 g / 0.00201 mol = 43.78 g mol⁻¹. This value is very close to the molar mass of carbon dioxide, CO₂ (44.01 g mol⁻¹).

Question 5

Given the following thermochemical equations: I: C(s) + O₂(g) → CO₂(g) ΔH = -394 kJ II: 2H₂(g) + O₂(g) → 2H₂O(l) ΔH = -572 kJ III: C₂H₂(g) + ⁵/₂O₂(g) → 2CO₂(g) + H₂O(l) ΔH = -1300 kJ

What is the standard enthalpy of formation, in kJ, for ethyne (C₂H₂)? The formation reaction is 2C(s) + H₂(g) → C₂H₂(g).

  1. +334
  2. -227
  3. +227 (correct answer)
  4. -334
Explanation: To find the enthalpy of formation for C₂H₂, we need to manipulate the given equations to match the target equation 2C(s) + H₂(g) → C₂H₂(g) using Hess's Law. 1. We need 2C(s) on the reactant side, so multiply equation I by 2: 2C(s) + 2O₂(g) → 2CO₂(g), ΔH = 2(-394) = -788 kJ. 2. We need 1H₂(g) on the reactant side, so take half of equation II: H₂(g) + ¹/₂O₂(g) → H₂O(l), ΔH = (-572)/2 = -286 kJ. 3. We need 1C₂H₂(g) on the product side, so reverse equation III: 2CO₂(g) + H₂O(l) → C₂H₂(g) + ⁵/₂O₂(g), ΔH = +1300 kJ. 4. Add the three modified equations and their ΔH values: (-788) + (-286) + 1300 = +226 kJ. The closest answer is +227 kJ.

Question 6

A student dissolves 10.0 g of a hydrated salt with the formula MgSO₄·xH₂O in water and adds an excess of barium chloride solution, BaCl₂. A precipitate of barium sulfate, BaSO₄, is formed, which is filtered, dried, and weighed. The mass of the dry precipitate is 9.68 g. What is the value of x? (M(MgSO₄)=120.37 g mol⁻¹, M(BaSO₄)=233.39 g mol⁻¹)

  1. 3
  2. 5
  3. 7 (correct answer)
  4. 10
Explanation: The reaction is MgSO₄ + BaCl₂ → BaSO₄ + MgCl₂. Moles of BaSO₄ precipitated = 9.68 g / 233.39 g mol⁻¹ = 0.04147 mol. From the 1:1 stoichiometry, the moles of MgSO₄ in the hydrated salt sample is also 0.04147 mol. The mass of anhydrous MgSO₄ is 0.04147 mol × 120.37 g mol⁻¹ = 4.992 g. The total mass of the hydrated salt was 10.0 g, so the mass of water of hydration is 10.0 g - 4.992 g = 5.008 g. Moles of water = 5.008 g / 18.02 g mol⁻¹ = 0.2779 mol. The mole ratio x = moles of H₂O / moles of MgSO₄ = 0.2779 / 0.04147 = 6.70, which rounds to 7.

Question 7

The Haber process for synthesizing ammonia has the following reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). The process is typically run at high pressure. If 1.0 m³ of N₂ is reacted with 3.0 m³ of H₂ at constant temperature and pressure, what is the maximum volume of NH₃ that can be produced, assuming the reaction goes to completion?

  1. 1.0 m³
  2. 2.0 m³ (correct answer)
  3. 3.0 m³
  4. 4.0 m³
Explanation: According to Avogadro's law, at constant temperature and pressure, the ratio of volumes of reacting gases is equal to the ratio of their moles in the balanced chemical equation. The stoichiometric ratio is 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃. Therefore, the volume ratio is 1 volume of N₂ reacts with 3 volumes of H₂ to produce 2 volumes of NH₃. Since the reactants are provided in the exact stoichiometric ratio (1.0 m³ of N₂ and 3.0 m³ of H₂), neither is in excess. The volume of NH₃ produced will be twice the volume of N₂ reacted. Volume of NH₃ = 2 × Volume of N₂ = 2 × 1.0 m³ = 2.0 m³.

Question 8

Given the following bond enthalpies in kJ mol⁻¹: H-H: 436 O=O: 498 O-H: 463

The reaction for the formation of hydrogen peroxide is: H₂(g) + O₂(g) → H₂O₂(g). The structure of hydrogen peroxide is H-O-O-H. The enthalpy change for this reaction is -136 kJ mol⁻¹. Using this information and the values provided, what is the bond enthalpy of the O-O single bond in hydrogen peroxide?

  1. 146 kJ mol⁻¹ (correct answer)
  2. 204 kJ mol⁻¹
  3. 282 kJ mol⁻¹
  4. 358 kJ mol⁻¹
Explanation: The enthalpy change of reaction is ΔH = Σ(bonds broken) - Σ(bonds formed). Bonds broken: 1 × H-H (436 kJ) and 1 × O=O (498 kJ). Total = 436 + 498 = 934 kJ. Bonds formed in H-O-O-H: 2 × O-H and 1 × O-O. Let the O-O bond enthalpy be x. Total formed = 2 × 463 + x = 926 + x. We are given ΔH = -136 kJ. So, -136 = 934 - (926 + x). -136 = 8 - x. Therefore, x = 8 - (-136) = 144 kJ mol⁻¹. The closest answer is 146 kJ mol⁻¹, the accepted average value.

Question 9

A 2.50 g sample of impure anhydrous sodium carbonate (Na₂CO₃) was added to 100.0 cm³ of 0.500 mol dm⁻³ hydrochloric acid, an excess amount. The resulting solution was titrated with 0.250 mol dm⁻³ sodium hydroxide, requiring 24.80 cm³ to neutralize the remaining acid. What is the percentage purity of the sodium carbonate in the original sample? (M(Na₂CO₃) = 105.99 g mol⁻¹)

  1. 52.5%
  2. 92.6% (correct answer)
  3. 46.3%
  4. 83.4%
Explanation: This is a back-titration problem. First, calculate the initial moles of HCl: 0.1000 dm³ × 0.500 mol dm⁻³ = 0.0500 mol HCl. Second, calculate the moles of NaOH used to neutralize the excess acid: 0.02480 dm³ × 0.250 mol dm⁻³ = 0.00620 mol NaOH. Since NaOH and HCl react in a 1:1 ratio, this is also the moles of excess HCl. Third, calculate the moles of HCl that reacted with Na₂CO₃: 0.0500 mol (initial) - 0.00620 mol (excess) = 0.0438 mol HCl. The reaction is Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂. The mole ratio of Na₂CO₃ to HCl is 1:2. Therefore, moles of Na₂CO₃ = 0.0438 mol HCl / 2 = 0.0219 mol. Fourth, calculate the mass of pure Na₂CO₃: 0.0219 mol × 105.99 g mol⁻¹ = 2.321 g. Finally, calculate the percentage purity: (2.321 g / 2.50 g) × 100% = 92.84%, which is approximately 92.6% (slight rounding difference).

Question 10

A student measures the concentration of a solution using Beer's law, where A=εbcA = \varepsilon bc. The measured absorbance is 0.245±0.0080.245 \pm 0.008, the path length is 1.00±0.011.00 \pm 0.01 cm, and the molar absorptivity is 1250±501250 \pm 50 M−1^{-1}cm−1^{-1}. What is the concentration with its absolute uncertainty?

  1. 1.96×10−4±8×10−61.96 \times 10^{-4} \pm 8 \times 10^{-6} M
  2. 1.96×10−4±1.5×10−51.96 \times 10^{-4} \pm 1.5 \times 10^{-5} M (correct answer)
  3. 1.96×10−4±2.1×10−51.96 \times 10^{-4} \pm 2.1 \times 10^{-5} M
  4. 1.96×10−4±3.2×10−51.96 \times 10^{-4} \pm 3.2 \times 10^{-5} M
Explanation: First calculate concentration: c = A/(εb) = 0.245/(1250 × 1.00) = 1.96 × 10⁻⁴ M. For uncertainty propagation in division/multiplication, use relative uncertainties: (Δc/c) = √[(ΔA/A)² + (Δε/ε)² + (Δb/b)²] = √[(0.008/0.245)² + (50/1250)² + (0.01/1.00)²] = √[0.00107 + 0.0016 + 0.0001] = 0.076. Absolute uncertainty = 0.076 × 1.96 × 10⁻⁴ = 1.5 × 10⁻⁵ M. A uses only absorbance uncertainty, C adds uncertainties linearly instead of quadratically, D incorrectly uses absolute uncertainties in the formula.

Question 11

A buffer solution contains 0.12 M CH₃COOH and 0.08 M CH₃COONa. When 15.0 mL of 0.20 M NaOH is added to 100.0 mL of this buffer, the pH changes from 4.57 to a new value. Given that Ka for acetic acid is 1.8×10−51.8 \times 10^{-5}, what is the final pH?

  1. 4.72
  2. 4.89 (correct answer)
  3. 5.06
  4. 5.23
Explanation: Initial moles: CH₃COOH = 0.12 × 0.100 = 0.012 mol, CH₃COO⁻ = 0.08 × 0.100 = 0.008 mol. Added OH⁻ = 0.20 × 0.015 = 0.003 mol. The reaction CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O consumes 0.003 mol acid and produces 0.003 mol conjugate base. Final moles: CH₃COOH = 0.012 - 0.003 = 0.009 mol, CH₃COO⁻ = 0.008 + 0.003 = 0.011 mol. Final volume = 115.0 mL. Using Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]) = 4.74 + log(0.011/0.009) = 4.89. A neglects volume change, C uses initial concentrations, D incorrectly calculates the effect of added base.

Question 12

The equilibrium constant for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is Kp = 6.2 × 10⁻⁴ at 500°C. If the reaction is initiated with partial pressures of 2.0 atm N₂, 3.0 atm H₂, and 0.5 atm NH₃, what is the reaction quotient Qp, and which direction will the reaction proceed?

  1. Qp = 4.6 × 10⁻³; reaction proceeds left since Qp > Kp (correct answer)
  2. Qp = 9.3 × 10⁻³; reaction proceeds left since Qp > Kp
  3. Qp = 4.6 × 10⁻³; reaction proceeds right since Qp < Kp
  4. Qp = 9.3 × 10⁻³; reaction proceeds right since Qp < Kp
Explanation: Calculate Qp using: Qp = [P(NH₃)]²/([P(N₂)][P(H₂)]³) = (0.5)²/((2.0)(3.0)³) = 0.25/(2.0 × 27) = 0.25/54 = 4.6×10⁻³. Since Qp (4.6×10⁻³) > Kp (6.2×10⁻⁴), the reaction will proceed left (toward reactants) to reach equilibrium. B uses incorrect denominator calculation, C and D have wrong direction predictions.

Question 13

How many oxygen atoms are present in 1.84 g of potassium dichromate(VI), K₂Cr₂O₇? (NA = 6.02 × 10²³ mol⁻¹, M(K₂Cr₂O₇) = 294.18 g mol⁻¹)

  1. 3.76 × 10²¹
  2. 2.63 × 10²² (correct answer)
  3. 5.27 × 10²¹
  4. 1.84 × 10²³
Explanation: First, calculate the moles of K₂Cr₂O₇: n = mass / M = 1.84 g / 294.18 g mol⁻¹ = 0.006254... mol. Second, calculate the number of formula units of K₂Cr₂O₇: Number = n × NA = 0.006254... mol × 6.02 × 10²³ mol⁻¹ = 3.765... × 10²¹ formula units. Third, since each formula unit of K₂Cr₂O₇ contains 7 oxygen atoms, multiply the number of formula units by 7: Number of O atoms = 7 × (3.765... × 10²¹) = 2.635... × 10²² atoms. This rounds to 2.63 × 10²².

Question 14

The first five successive ionization energies for an element are 786, 1577, 3232, 4356, and 16091 kJ mol⁻¹. What is the charge on the most common ion formed by this element?

  1. 1+
  2. 2+
  3. 3+
  4. 4+ (correct answer)
Explanation: Successive ionization energies increase as more electrons are removed. A very large jump in ionization energy indicates the removal of an electron from a new, inner electron shell (a core electron), which is held much more tightly by the nucleus. The energies are approximately 786, 1577, 3232, 4356, and 16091. The jump from the 4th to the 5th ionization energy (4356 to 16091) is significantly larger than the previous jumps. This suggests that the element has four valence electrons. After removing these four electrons, the fifth electron is removed from a stable inner shell. Therefore, the element is in Group 14 and is most likely to form an ion by losing its four valence electrons, resulting in a 4+ charge.

Question 15

An aqueous solution has a hydroxide ion concentration, [OH⁻], of 2.5 × 10⁻⁴ mol dm⁻³. What is the pH of this solution at 298 K? (K_w = 1.0 × 10⁻¹⁴)

  1. 3.60
  2. 4.40
  3. 9.60
  4. 10.40 (correct answer)
Explanation: There are two common methods. Method 1: Calculate pOH first. pOH = -log₁₀[OH⁻] = -log₁₀(2.5 × 10⁻⁴) = 3.60. Then use the relationship pH + pOH = 14 at 298 K. pH = 14 - pOH = 14 - 3.60 = 10.40. Method 2: Calculate [H⁺] first. [H⁺] = K_w / [OH⁻] = (1.0 × 10⁻¹⁴) / (2.5 × 10⁻⁴) = 4.0 × 10⁻¹¹ mol dm⁻³. Then calculate pH. pH = -log₁₀[H⁺] = -log₁₀(4.0 × 10⁻¹¹) = 10.40.

Question 16

What volume of 0.125 mol dm⁻³ phosphoric(V) acid, H₃PO₄, is required to completely neutralize 45.0 cm³ of 0.100 mol dm⁻³ potassium hydroxide, KOH?

  1. 12.0 cm³ (correct answer)
  2. 18.0 cm³
  3. 36.0 cm³
  4. 108 cm³
Explanation: First, write the balanced neutralization equation: H₃PO₄(aq) + 3KOH(aq) → K₃PO₄(aq) + 3H₂O(l). The mole ratio of H₃PO₄ to KOH is 1:3. Next, calculate the moles of KOH: n(KOH) = C × V = 0.100 mol dm⁻³ × 0.0450 dm³ = 0.00450 mol. Using the mole ratio, calculate the moles of H₃PO₄ needed: n(H₃PO₄) = 0.00450 mol KOH / 3 = 0.00150 mol. Finally, calculate the volume of H₃PO₄ solution: V = n / C = 0.00150 mol / 0.125 mol dm⁻³ = 0.0120 dm³. Convert this to cm³: 0.0120 dm³ × 1000 = 12.0 cm³.

Question 17

A compound contains 49.3% carbon, 6.9% hydrogen, and 43.8% oxygen by mass. Its molar mass is 146 g mol⁻¹. What is the molecular formula of the compound?

  1. C₃H₅O₂
  2. C₆H₁₀O₄ (correct answer)
  3. C₇H₁₄O₃
  4. C₆H₁₄O₃
Explanation: Assume a 100 g sample. Moles of each element are: C = 49.3 g / 12.01 g mol⁻¹ = 4.105 mol; H = 6.9 g / 1.01 g mol⁻¹ = 6.83 mol; O = 43.8 g / 16.00 g mol⁻¹ = 2.738 mol. Divide by the smallest value (2.738) to find the simplest ratio: C = 1.5, H = 2.5, O = 1. Multiply by 2 to get whole numbers: C₃H₅O₂. This is the empirical formula. The molar mass of the empirical formula is 3(12.01) + 5(1.01) + 2(16.00) = 73.08 g mol⁻¹. The ratio of the molecular mass to the empirical formula mass is 146 / 73.08 ≈ 2. Therefore, the molecular formula is (C₃H₅O₂)₂, which is C₆H₁₀O₄.

Question 18

What is the final concentration of chloride ions, [Cl⁻], when 50.0 cm³ of 0.200 mol dm⁻³ MgCl₂ solution is mixed with 150.0 cm³ of 0.100 mol dm⁻³ NaCl solution? Assume volumes are additive.

  1. 0.125 mol dm⁻³
  2. 0.150 mol dm⁻³
  3. 0.175 mol dm⁻³ (correct answer)
  4. 0.200 mol dm⁻³
Explanation: First, calculate the moles of Cl⁻ ions from each solution. From MgCl₂, each formula unit provides 2 Cl⁻ ions. Moles of MgCl₂ = 0.0500 dm³ × 0.200 mol dm⁻³ = 0.0100 mol. Moles of Cl⁻ from MgCl₂ = 2 × 0.0100 mol = 0.0200 mol. From NaCl, each formula unit provides 1 Cl⁻ ion. Moles of NaCl = 0.1500 dm³ × 0.100 mol dm⁻³ = 0.0150 mol. Moles of Cl⁻ from NaCl = 0.0150 mol. The total moles of Cl⁻ is 0.0200 + 0.0150 = 0.0350 mol. The total volume is 50.0 + 150.0 = 200.0 cm³ = 0.2000 dm³. The final concentration [Cl⁻] = total moles / total volume = 0.0350 mol / 0.2000 dm³ = 0.175 mol dm⁻³.

Question 19

What mass of water, in grams, must be added to 50.0 g of a 12.0% by mass solution of NaOH to dilute it to a 4.00% by mass solution?

  1. 50.0
  2. 150
  3. 100 (correct answer)
  4. 200
Explanation: First, find the mass of NaOH in the initial solution: 0.120 × 50.0 g = 6.00 g NaOH. In the final solution, this same mass of NaOH will constitute 4.00% of the total mass. Let M_final be the total mass of the final solution. Then, 0.0400 × M_final = 6.00 g. Solving for M_final gives M_final = 6.00 / 0.0400 = 150 g. The initial mass of the solution was 50.0 g. Therefore, the mass of water that must be added is the difference: 150 g - 50.0 g = 100 g.

Question 20

The solubility of CaF₂ in pure water is 2.1×10−42.1 \times 10^{-4} M at 25°C. What is the solubility of CaF₂ in a solution that is already 0.015 M in Ca(NO₃)₂?

  1. 1.4×10−61.4 \times 10^{-6} M
  2. 2.8×10−62.8 \times 10^{-6} M
  3. 5.6×10−65.6 \times 10^{-6} M (correct answer)
  4. 1.1×10−51.1 \times 10^{-5} M
Explanation: First calculate Ksp from pure water solubility: CaF₂ ⇌ Ca²⁺ + 2F⁻. If solubility = s, then [Ca²⁺] = s and [F⁻] = 2s. Ksp = [Ca²⁺][F⁻]² = s(2s)² = 4s³ = 4(2.1×10⁻⁴)³ = 3.7×10⁻¹¹. In Ca(NO₃)₂ solution: [Ca²⁺]total = 0.015 + s ≈ 0.015 M (since s << 0.015). Therefore: 3.7×10⁻¹¹ = (0.015)(2s)² = 0.015 × 4s² = 0.06s². Solving: s² = 6.2×10⁻¹⁰, s = 5.6×10⁻⁶ M. A uses wrong Ksp expression, B neglects the coefficient 2 for fluoride, D doesn't account for common ion effect properly.