5 Steps to a 5 AP Chemistry 2019

AP Chemistry Practice Exam 1 ❮ 335

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

(x) (3x)^3 = 27 x^4 = 6.2 × 10 –12.

Solving for x gives x = 6.9 × 10 –4, and [IO 3 – ] = 3 x = 2.0 × 10 –3 M.

You get 1 point for doing the calculations and showing La(IO 3 ) 3 is more soluble and 1 additional point

for the correct answers.

(e) It is easier to begin with the balanced molecular equation:

2 NaIO 3 (aq) + Zn(NO 3 ) 2 (aq) → Zn(IO 3 ) 2 (s) + 2 NaNO 3 (aq)

Converting the molecular equation to a complete ionic equation:

2 Na+(aq) + 2 NO 3 – (aq) + Zn^2 +(aq) + 2 NO 3 – (aq) → Zn(IO 3 ) 2 (s) + 2 Na+(aq) + 2 NO 3 – (aq)

The ionic compounds NaIO 3 and Zn(IO 3 ) 2 should be separated into their ions since you are supplied

with solutions and the sodium nitrate is also shown in the ionic form since you are told it is soluble in

water. There is a Ksp for Zn(IO 3 ) 2 given in the table; therefore, it does not separate into ions.

Removing the spectator ions (Na+ and NO 3 – ) from the complete ionic equation leaves the net ionic

equation:

Zn^2 +(aq) + 2 IO 3 – (aq) → Zn(IO 3 ) 2 (s)

You get 1 point for this answer.

Total your points for the different parts. There is a maximum of 10 points possible. Subtract one point if

all answers did not have the correct number of significant figures.

Question 2

(a) The order for H 2 SeO 3 is 2. The order for H 2 O 2 is 1. The order for H+ is –1.

If you get all three of these correct, you get 1 point.

Comparing experiments 2 and 1 (placing the larger over the smaller makes the calculations a little

easier): The concentration of H 2 SeO 3 is doubled, and the rate is quadrupled;

×

×

−

−

2.9 10

7.4 10

7

8 ≈ 4. This leads

to H 2 SeO 3 having an order of 2.

You get 1 point for this reasoning.

Comparing experiments 3 and 1: The concentration of H 2 O 2 is doubled, and the rate is doubled;

×

×

−

−

1.5  10

7.4  10

7

8 ≈ 2. This leads to H^2 O^2 having an order of 1.

You get 1 point for this reasoning.

Comparing experiments 4 and 1: The concentration of H+ is doubled, and the rate is halved;

×

×

−

−

3.7  10

7.4  10

8

8 ≈ 0.5 =^2

–1. This leads to H+ having an order of –1.

You get 1 point for this reasoning.

(b) Rate = k[H 2 SeO 3 ]^2 [H 2 O 2 ][H+]–1

This answer is worth 1 point. You will still get 1 point if you use incorrect orders from part (a).

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