Name:_______Answer Key____________________________ AP Chemistry Review Acids and Bases
__B__1. Which of the following best describes the pH of a 0.01-molar solution of HBrO?
(For HBrO, Ka = 2 × 10-9)
(A) Less than or equal to 2
(D) Between 7 and 11
(B) Between 2 and 7
(C) 7
(E) Greater than or equal to 11
__A__2. A 1-molar solution of a very weak monoprotic acid has a pH of 5. What is the value of Ka for
the acid?
(A) Ka = 1 × 10-10
(B) Ka = 1 × 10-7
(C) Ka = 1 × 10-5
(D) Ka = 1 × 10-2
(E) Ka = 1 × 10-1
__C__3. The concentrations of which of the following species will be increased when HCl is added to a
solution of HC2H3O2 in water? HC2H3O2 (aq) + H2O (l) ⇄ C2H3O2- + H3O+
Adding strong acid will shift the eq to the left
I. H+
II. C2H3O2III. HC2H3O2
(A) I only
(D) II and III only
(B) I and II only
(E) I, II and III
(C) I and III only
__B__4. The first acid dissociation constant for tartaric acid, H2C4H4O6, is 1.0 × 10-3. What is the base
dissociation constant for HC4H4O6-?
(A) Kb= 1.0 × 10-13 (B) Kb= 1.0 × 10-11 (C) Kb= 1.0 × 10-7 (D) Kb= 1.0 × 10-4 (E) Kb= 1.0 × 10-1
__D__5. If 0.630 grams of HNO3 (molecular weight 63.0) are placed in 1 liter of distilled water at 25°C,
what will be the pH of the solution? (Assume that the volume of the solution is unchanged by
the addition of HNO3.)
(A) 0.01
(B) 0.1
(C) 1
(D) 2
(E) 3
Questions 6-8
A solution of a weak monoprotic acid is titrated with a solution of a strong base, KOH. Consider
the points labeled (A) through (E) on the titration curve that results as shown below.
__C__6. The point at which the moles of the added strong base are equal to the moles of the weak acid.
__E__7. The point at which the pH is closest to that of the strong base being added
__B__8. The point at which the pH of the mixture is equal to the pKa of the acid.
9. a. A 0.100M solution of NH3 is prepared in a 125mL flask. What is the pH of the solution if Kb of
ammonia is 1.79 x 10-5?
R NH3 (aq) +
I 0.100M
C -X
E 0.100M
Kb<<[NH3]i
[NH3]E = [NH3]I
Kb =
H2O (l) ⇄
-------------------------------------
[𝑁𝐻4+ ][𝑂𝐻 − ]
[𝑁𝐻3 ]
Kb =
𝑋2
0.100𝑀
pOH = - log[OH-]
14 – pOH = pH
NH4+ + OH0M
0M
+X
+X
X
X
= 1.79 x 10-5
X = 0.00134M = [OH-]
pOH = - log(0.00134)
pOH = 2.87
14 – 2.87 = 11.13
b. A different solution is prepared containing 0.100-molar NH4Cl with 0.200-molar solution of NH3.
What is the pH of this new solution ? The value of Kb for ammonia is 1.79 × 10-5.
BUFFER!!!
pH = pka + log ([A-]/[HA])
Ka • Kb = Kw
Ka (1.79 x 10-5) = 1.0 x 10-14
pH = -log(5.59 x 10-10) + log (0.200M/0.100M)
Ka = 5.59 x 10-10
pH = 9.55
10. A sample of 40.0 milliliters of a 0.100 molar HC2H3O2 solution is titrated with a 0.150 molar
NaOH solution. The Ka value for acetic acid is1.8×10-5.
(a) What volume of NaOH is used in the titration in order to reach the equivalence point?
MAVA = MBVB
MA = 0.100M
VA = 40.0 mL
MB = 0.150M
VB = ?
(0.100M)(40.0mL) = (0.150M)(VB)
VB = 26.67mL
(b) What is the molar concentration of C2H3O2- at the equivalence point?
At equivalence all moles of acid have been converted to conjugate base.
mol = M(V)
mol = (0.100M) ( 0.040L)
mol = 0.004 mol HC2H3O2
at equvilance 0.004 mol HC2H3O2 are converted to 0.004 mol C2H3O2[C2H3O2-] = 0.004 mol/(total volume)
[C2H3O2-] = 0.004 mol/(0.040L + 0.02667L) = 0.060M
(c) What is the pH of the solution at the equivalence point?
Ka • Kb = Kw
(1.8 x 10-5)Kb = 1.0 x 10-14
R C2H3O2- (aq) + H2O (l) ⇄
I 0.060M
------------C -X
------------E 0.060M
------------Kb<<[C2H3O2 ]i
[C2H3O2-]E = [C2H3O2-]I
Kb =
[HC2H3O2][𝑂𝐻 − ]
[C2H3O2− ]
pOH = - log[OH-]
14 – pOH = pH
Kb =
Kb = 5.56 x 10-10
HC2H3O2 (aq) + OH0M
0M
+X
+X
X
X
𝑋2
0.060𝑀
= 5.56 x 10-10
pOH = - log(5.78 x 10-6 M)
14 – 5.24 = 8.76
X = 5.78 x 10-6 M= [OH-]
pOH = 5.24