5 Steps to a 5 AP Chemistry 2019

(Marvins-Underground-K-12) #1

238 ❯ STEP 4. Review the Knowledge You Need to Score High



  1. D—If there are equal numbers of moles of gas
    on each side of the equilibrium arrow, then
    volume or pressure changes will not affect the
    equilibrium. The presence of solids, liquids, or
    aqueous phases does not make any difference as
    long as some of the phase is present.

  2. D—Perbromic acid, HBrO 4 , is expected to
    be the strongest acid in this group because the
    greater number of oxygen atoms pulls more elec-
    tron density from the hydrogen atom, making it
    easier to be lost as a hydrogen ion.

  3. B—The two substances (NH 4 Cl and
    CH 3 NH 3 Cl) do not contain a conjugate acid–
    base pair, so this is not a buffer, which elimi-
    nates A and C. Both compounds are salts of
    a weak base (NH 3 or CH 3 NH 2 ) and a strong
    acid (HCl); such salts are acidic (pH < 7). The
    Kb values shown in the table indicate that all
    three compounds are weak bases.
    24. B—All the ions can serve as Brønsted bases
    (accept a hydrogen ion to form H 3 PO 4 , HCO 3 - ,
    H 2 SeO 4 , and H 2 CO 3 ). All but B can behave as
    Brønsted acids (donate a hydrogen ion to form
    HPO 42 - , SeO 42 - , and CO 32 ).
    25. A—Aqueous ammonia contains primarily NH 3 ,
    which eliminates B and C. NH 3 Cl does not exist,
    which eliminates D. The reaction produces the
    silver–ammonia complex, [Ag(NH 3 ) 2 ]+.
    26. A—A Ksp is never homogeneous; all the others
    are normally always homogenous.
    27. A—Kc = Kp occurs whenever there are equal
    moles of gaseous molecules on each side of the
    equilibrium arrow.
    28. D—This is a heterogeneous equilibrium; there-
    fore, the solids (CaCO 3 and CaO) will not
    appear in the equilibrium expression. B is the Kc
    expression, not the Kp expression.


❯ Free-Response Question


You have 20 minutes to answer the following long question. You may use a calculator and
the tables in the back of the book.

Question
An aqueous solution is prepared that is initially 0.100 M in CdI 42 -. After equilibrium is
established, the solution is found to be 0.013 M in Cd2+. The products of the equilibrium
are Cd^2 +(aq) and I-(aq).
(a) Derive the expression for the dissociation equilibrium constant, Kd, for the equilibrium,
and determine the value of the constant.
(b) What will be the cadmium ion concentration arising when 0.400 mol of KI is added
to 1.00 L of the solution in part a?
(c) A solution is prepared by mixing 0.500 L of the solution from part b and 0.500 L of
2.0 × 10 -^5 M NaOH. Will cadmium hydroxide, Cd(OH) 2 , precipitate? The Ksp for
cadmium hydroxide is 2.2 × 10 -^14.
(d) When the initial solution is heated, the cadmium ion concentration increases. Is the
equilibrium an exothermic or an endothermic process? Explain how you arrived at your
conclusion.
(e) Complete the Lewis structure below for the cyanate ion, and determine the formal
charge for each atom. If resonance is possible, you only need to show one resonance
form.

[O C N]−
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