5 Steps to a 5 AP Chemistry

AP Chemistry Practice Exam 1  311

59. Acid Ka, acid dissociation constant

H 3 PO 4 7.2 × 10 −^3

H 2 PO 4

−

6.3 × 10 −^8

HPO 4

2 −

4.2 × 10 −^13

Using the above information, choose the best

answer for preparing a pH =7.9 buffer.

(A) K 2 HPO 4

(B) K 3 PO 4

(C) K 2 HPO 4 +KH 2 PO 4

(D) K 2 HPO 4 +K 3 PO 4

(E) H 3 PO 4 +KH 2 PO 4

60.What is the ionization constant, Ka, for a weak
monoprotic acid if a 0.6-molar solution has a pH
of 2.0?

(A) 1.7 × 10 –4

(B) 1.7 × 10 –2

(C) 6 × 10 –6

(D) 2.7 × 10 –3

(E) 3.7 ×l0–4

Questions 61–64 refer to the following aqueous
solutions. All concentrations are 1 M.

(A) CH 3 NH 2 (methylamine) and LiOH (lithium

hydroxide)

(B) C 2 H 5 NH 2 (ethylamine) and C 2 H 5 NH 3 NO 3

(ethylammonium nitrate)

(C) CH 3 NH 2 (methylamine) and HC 3 H 5 O 2

(propionic acid)

(D) KClO 4 (potassium perchlorate) and HClO 4

(perchloric acid)

(E) H 2 C 2 O 4 (oxalic acid) and KHC 2 O 4 (potas-

sium hydrogen oxalate)

61.The most basic solution (highest pH)

62.The solution with a pH nearest 7

63.A buffer with a pH > 7

64.A buffer with a pH < 7

65.At constant temperature, a change in volume will

NOT affect the moles of the substances present

in which of the following?

(A)

(B)

(C)

(D)

(E)

66.

Which species, in the above equilibrium, behave

as bases?

I. CO 3

2–

II. H 2 O

III. HCO 3

• 
  

(A) I and III

(B) II only

(C) I and II

(D) I only

(E) II and III

67.

A 1.00-L flask is filled with 0.70 mol of H 2 and

0.60 mol of CO, and allowed to come to equilib-

rium. At equilibrium, there are 0.40 mol of CO

in the flask. What is the value of Kc, the equilib-

rium constant, for the reaction?

(A) 0.74

(B) 3.2

(C) 0.0050

(D) 5.6

(E) 1.2

68.H 2 O(l) +CrO 4

2 −

(aq) +HSnO 2

−

(aq) →

CrO 2

• (aq) +OH–(aq) +HSnO 3

−

(aq)

What is the coef f icient of OH–when the above

reaction is balanced?

(A) 10

(B) 2

(C) 5

(D) 4

(E) 1

CO(g) + 2H g 23 ()→CH OH(g)

HCO aq−− 32 ()++H O e()H O aq 3 +()CO^23 ()aq

23 NHgNgHg 322 ()→+() ()

CO(g) + H O(g) 222 →+CO g()H g()

PCl g 32 ()+→Cl g() PCl g 5 ()

CO(g) + Cl g 2 ()→COCl g 2 ()

6 CN aq− +→Fe3+aq Fe CN 63 aq

−

() () [( )]()

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