5 Steps to a 5 AP Chemistry 2019

334 ❯ STEP 5. Build Your Test-Taking Confidence

❯ Answers and Explanations for Exam 1, Section II

(Free Response)

Question 1

(a) (i) Ni(IO 3 ) 2 (s) ^ Ni^2 +(aq) + 2 IO 3 – (aq)

You get 1 point for this answer. The equilibrium arrow and the ionic charges must be present.

(ii) Ksp = [Ni^2 +] [IO 3 – ]^2

You get 1 point for this answer. The charges must be present and the iodate ion concentration

must be squared.

(iii) Ksp = [Ni^2 +] [IO 3 – ]^2 = 1.4 × 10 –8

Setting [Ni^2 +] = x and [IO 3 – ] = 2 x and inserting into the mass-action expression gives:

(x) (2x)^2 = 4 x^3 = 1.4 × 10 –8.

Solving for x gives x = 1.5 × 10 –3, and [IO 3 – ] = 2 x = 3.0 × 10 –3 M.

You get 1 point for the correct [IO 3 – ].

(b) Copper(II) nitrate is soluble in water (obviously because you are given a solution), which separates into

copper(II) ions and nitrate ions. The copper(II) ion concentration is 0.10 M and it is a common ion

affecting the equilibrium.

Ksp = [Cu^2 +] [IO 3 – ]^2 = 7.4 × 10 –8

Setting [Cu^2 +] = 0.10 + x and [IO 3 – ] = 2 x and inserting into the mass-action expression gives:

(0.10 + x) (2x)^2 = 7.4 × 10 –8. Assuming x is much smaller than 0.10 allows a simplification of the

calculation to (0.10) (2x)^2 = 0.4 x^2 = 7.4 × 10 –8.

Solving for x gives x = 4.3 × 10 –4 M, which leads to [IO 3 – ] = 2 x = 8.6 × 10 –4 M.

You get 1 point for the correct [IO 3 – ].

(c) (i) Silver iodate is more soluble in nitric acid than in pure water.

You get 1 point for this answer.

(ii) Iodic acid is a weak acid (Ka = 0.16); therefore, according to Le Châtelier’s principle, the solubil-

ity equilibrium will be displaced to the right as iodate ion combines with hydrogen ions from the

nitric acid to form unionized iodic acid. The reactions are: AgIO 3 (s) ^ Ag+(aq) + IO 3 – (aq) and

H+(aq) + IO 3 – (aq) → HIO 3 (aq)

You get 1 point for this answer.

(d) If the two compounds have the same stoichiometry, the larger Ksp would generate the higher iodate ion

concentration. However, this is not the case, so it is necessary to calculate the iodate ion concentration

for each.

Ksp = [Ba^2 +] [IO 3 – ]^2 = 1.5 × 10 –9

Setting [Ba^2 +] = x and [IO 3 – ] = 2 x, and inserting into the mass-action expression gives:

(x) (2x)^2 = 4 x^3 = 1.5 × 10 –9.

Solving for x gives x = 7.2 × 10 –4, and [IO 3 – ] = 2 x = 1.4 × 10 –3 M.

Ksp = [La^3 +] [IO 3 – ]^3 = 6.2 × 10 –12

20-Moore_PE01_p307-340.indd 334 31/05/18 1:58 pm