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MCAT Sample Questions : Physical Sciences

Passage I

A series of chemical reactions was carried out to study the chemistry of lead.

Reaction 1

Initially, 15.0 mL of 0.300 M Pb(NO3)2(aq) was mixed with 15.0 mL of 0.300 M Na2SO4(aq).All the Pb(NO3)2 reacted to form Compound A, a white precipitate.Compound A was removed by filtration.

Reaction 2

Next, 15.0 mL of 0.300 M KI(aq) was added to Compound A. The mixture was agitated and some of Compound A dissolved. In addition, a yellow precipitate of PbI2(s) was formed.

Reaction 3

The PbI2(s) was separated and mixed with 15.0 mL of 0.300 M Na2CO3(aq). A white precipitate of PbCO3(s) formed. All of the PbI2(s) was converted into PbCO3(s).

Reaction 4

The PbCO3(s) was removed by filtration and a small sample gave off a gas when treated with dilute HCl.

Some sample questions on this passage are as follows:

  1. Which of the following reactions depicts the formation of the gas in Reaction 4?

    1. PbCO3(s) + 2 HCl(aq) = PbCl2(aq) + CO2(g) + H2O(l)
    2. Na2CO3(aq) + 2 HCl(aq) = 2 NaCl(aq) + CO2(g) + H2O(l)
    3. PbCO3(s) + 2 HCl(aq) = PbC2(s) + Cl2(g) + H2O(l)
    4. PbI 2(s) + HCl(aq) = PbCl2(aq) + HI(g)

    Answer: A

    Explanation: Reaction 4 is shown in the following equation, which is answer choice A.

    PbCO3(s) + 2 HCl(aq) = PbCl2(aq) + CO2(g) + H2O(l)

    Answers B and D do not show a reaction involving PbCO3(s), as required by Reaction 4. Answer C shows an implausible and unbalanced equation. Thus, answer choice A is the best answer.

  2. The identity of Compound A is:

    1. Pb(NO3)2.
    2. PbI2.
    3. NaNO3.
    4. PbSO4.

    Answer: D

    Explanation: Reaction 1 is shown in the following equation.

    Pb(NO3)2(aq) + Na2SO4(aq) = PbSO4(s) + 2 NaNO3(aq)

    Compound A, the white solid, is PbSO4(s). Neither the reactant Pb(NO3)2 nor the product NaNO3 can precipitate because all nitrates and sodium salts are water soluble. PbI2 cannot precipitate because iodide is not present. Thus, answer choice D is the best answer.

  3. Pb(OH)2(s) is slightly soluble in water. How would the amount of Pb(OH)2(s) that normally dissolves in 1 L of water be affected if the pH were 9.0?

    1. Less would dissolve.
    2. The same amount would dissolve.
    3. More would dissolve.
    4. There is no way to predict the effect of the change in pH of the water.

    Answer: A

    Explanation: The dissolution of Pb(OH)2(s) is represented by the following equation.

    Pb(OH)2(s)=Pb2(aq) + 2 OH-(aq)

    At pH 9, the concentration of OH-(aq) is greater than the concentration of OH-(aq) at pH 7. According to Le Châtelier's principle, the additional common ion, OH-(aq), will shift the position of equilibrium to the left, and less Pb(OH)2 will dissolve. Thus, answer choice A is the best answer.

  4. A soluble form of Pb2+can be carefully added to a solution tosequentially precipitate and separate anions present in the solution. When Pb2+is added, in what order will the following anions be precipitated?

    1. SO42- then I-
    2. CO32- then I-
    3. SO42- then CO32-
    4. I- then CO32-

    Answer: B

    Explanation: The reactions described in the passage show that lead(II) is successively precipitated as PbSO4, PbI2, and PbCO3. This sequence shows (assuming equal anion concentrations, as must be done here) that PbCO3 is less soluble than PbI2, and PbI2 is less soluble than PbSO4. The order in which the anions precipitate Pb2- is: CO32- then I- then SO42-. When this sequence is applied to the question, answer choice B is in the correct order, and answers A, C, and D are all in the opposite order. Thus, answer choice B is the best answer.

  5. How many moles of Na+ ions are there in the initial Na2SO4(aq) solution used in Reaction 1?

    1. 0.0018 mole
    2. 0.009 mole
    3. 0.045 mole
    4. 0.090 mole

    Answer: B

    Explanation: The initial Na2SO4(aq) solution in Reaction 1 is 15 mL of 0.300 MNa2SO4(aq).

    (15.0 mL) ( 1 L ) (0.300 mol Na2SO4) (2 mol Na+)
    (1000 mL) (1 L Na2SO4(aq) ) (1 mol Na2SO4)

    = 0.00900 mol = Answer B

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