DEPARTMENT OF CHEMISTRY, CFS, IIUM
SEMESTER III, 2013/2014
CHE 0325/SHE 1325
QUIZ 5
NAME:………………………………………………………………………………………..
MATRIC NO……………… GROUP:…………
Answer all questions.
[30 mins]
PART I [5 marks]
1
Which of the following are classified as strong acids?
I
III
H2SO4
CH3COOH
II
IV
HF
HBr
A
D
I and II
III and IV
B
D
II and III
I and IV
D/L1/4a
1
1
Which of the following are classified as strong bases?
I
III
NH3
Ca(OH)2
II
IV
NaOH
CH3NH2
A
D
I and II
III and IV
B
D
II and III
I and IV
B/L1/4a
Which of the following are classified as weak acids?
I
III
HF
H2SO4
II
IV
HBr
CH3COOH
A
C
I and II
III and IV
B
D
II and III
I and IV
D/L1/4a
2
Which of the following is a conjugate acid-base pair?
A
C
2
HSO4-/SO42H3PO4/HPO4-
B
D
H3O+/H2OCH3CH2COOH/CH3COOA/L2/4b
Which of the following is not a conjugate acid-base pair?
A
C
HCN/CNH2SO4/HSO4-
B
D
H3O+/H2O
H3PO4/HPO4-
D/L2/4b
2
Which of the following is a conjugate acid-base pair?
A
C
3
H2F/FHSO4-/SO4-
B
D
H3O+/H2O
H3PO4/HPO4-
B/L2/4b
What is the acid dissociation constant, Ka expression for the following
equilibrium?
H3PO4(aq) + H2O(l) ⇌ H2PO4-(aq) + H3O+(aq)
A
C
3
[H2PO4-]/[H3PO4]
[H2PO4-][H3O+]/[H3PO4]
B
D
[H2PO4-][H3O+]
[H2PO4-][H3O+]/[H3PO4][H2O]
C/L2/4d
What is the acid dissociation constant, Ka expression for the following
equilibrium?
HF(aq) + H2O(l) ⇌ F-(aq) + H3O+(aq)
A
C
3
[F-]/[HF]
[F-][H3O+]/[H2O]
B
D
[F-][H3O+]/[HF]
[F-][H3O+]/[HF][H2O]
B/L2/4d
What is the acid dissociation constant, Ka expression for the following
equilibrium?
HCN(aq) + H2O(l) ⇌ CN-(aq) + H3O+(aq)
A
C
4
B
D
[CN-][H3O+]/[HCN]
[CN-][H3O+]/[HCN][H2O]
B/L2/4d
What is the concentration of hydroxide ion, OH- of an acid with pH value of
2.95?
A
C
4
[CN-]/[HCN]
[CN-]/[HCN][H3O+]
1.12 × 10-3 M
8.91 × 10-12 M
B
D
4.70 × 10-1 M
8.93 × 10-5 M
C/L2/4f
What is the concentration of hydronium ion, H3O+ of a solution with pOH
value of 5.65?
A
C
2.24 × 10-6 M
7.52 × 10-1 M
B
D
4.47 × 10-9 M
9.22 × 10-1 M
B/L2/4f
4
What is the concentration of hydroxide ion, OH- of an acid with pH value of
9.81.
A
C
5
1.55 × 10-10 M
8.10 × 10-10 M
B
D
6.46 × 10-5 M
9.20 × 10-5 M
B/L2/4f
Which of the following graphs describes the relationship between pH and
pOH?
A
B
C
D
B/L2/4f
5
Which of the following graphs describes the relationship between pH and
pOH?
A
B
C
D
D/L2/4f
5
Which of the following graphs describes the relationship between pH and
pOH?
A
B
C
D
C/L2/4f
PART II [10 marks]
SET A
1
(a)
Define buffer
[2]
A solution of weak acid and its conjugate base or weak base
and its conjugate acid [1] that will slightly change the pH when
a small amount of strong acid or base is added [1].
L1/4k
(b)
A 1.00 L buffer solution is made up of 0.600 M HOCl and 0.200 M
KOCl. Calculate the initial pH of the buffer solution. [Ka = 3.50 × 10-5]
[2]
pH
= pKa + log ([A-]/[HA])
= pKa + log ([OCl-]/[HOCl])
= -log (3.50 × 10-5) + log [(0.200)/(0.600)]
[1]
= 3.98
[1]
L2/4m
2
40.0 mL of 0.150 M benzoic acid, C6H5COOH is titrated with 0.100 M NaOH.
(Kb of C6H5COO- = 1.59 × 10-10)
(a)
Predict whether the salt produced is acidic, basic or neutral.
[1]
Basic
L2/4i
(b)
Calculate the pH after 60.0 mL of NaOH is added.
[4]
n C6H5COOH initial = (0.0400 L × 0.150 M) = 6.00 × 10-3 mol
n OH- added
= (0.0600 L × 0.100 M) = 6.00 × 10-3 mol
This is at the equivalence point [1]
whereby 6.00 × 10-3 mol of OH- reacts with 6.00 × 10-3 mol of
C6H5COOH to form 6.00 × 10-3 mol of C6H5COO-. C6H5COO- will
hydrolyze in water.
[C6H5COO-]
= 6.00 × 10-3 mol / (0.040 + 0.060) L
= 0.0600 M
Conc, M
C6H5COO-(aq)
I
0.06
-
0
0
C
-x
-
+x
+x
F
0.06 – x
-
x
x
+
H2O(l)
⇌
C6H5COOH
+
OH-
For correct hydrolysis (equation/rxn with water) of C6H5COO- [1]
[Initial]/Kb >>> 400. We can neglect x and assume
[C6H5COO-]initial ≈ [C6H5COO-]eq
x2/0.06
x
pOH = -log OH-
(c)
= 1.59 × 10-10
= 3.09 × 10-6 M
= 5.51
= Kb
= [OH-]
pH = pKw – pOH = 8.49
[1]
[1]
Identify a suitable indicator that can be used in this titration.
Phenolphtalein
L3/4n
[1]
L1/4o
SET B
1
(a) Define buffer
[2]
A solution of weak acid and its conjugate base or weak base and
its conjugate acid [1] that will slightly change the pH when a
small amount of strong acid or base is added [1].
L1/4k
(b) A 1.00 L buffer solution is made up of 0.300 M HOCN and 0.150 M
KOCN. Calculate the initial pH of the buffer solution. [Ka = 3.5 × 10-4]
[2]
pH
= pKa + log ([A-]/[HA])
= pKa + log ([OCN-]/[HOCN])
= -log (3.50 × 10-4) + log [(0.150)/(0.300)]
[1]
= 3.15
[1]
L2/4m
2
45.0 mL of 0.200 M nitrous acid, HNO2 is titrated with 0.150 M KOH. (Kb of
HNO2 = 2.50 × 10-11)
(a)
Predict whether the salt produced is acidic, basic or neutral.
Basic
[1]
L2/4i
(b)
Calculate the pH after 60 mL of KOH is added.
[4]
n HNO2 initial
n OH- added
= (0.0450 L × 0.20 M) = 9.00 × 10-3 mol
= (0.0600 L × 0.15 M) = 9.00 × 10-3 mol
This is at the equivalence point [1] whereby 9.00 × 10-3 mol of OHreacts with 9.00 × 10-3 mol of HNO2 to form 9.00 × 10-3 mol of NO2-.
NO2- will hydrolyze in water.
[NO2-]
Conc, M
= 9.00 × 10-3 mol / (0.045 + 0.060) L
= 0.0857 M
NO2-(aq) +
H2O(l)
⇌
HNO2(aq)
+
OH-(aq)
I
0.0857
-
0
0
C
-x
-
+x
+x
F
0.0857– x
-
x
X
For correct hydrolysis (equation/rxn with water) of NO2- [1]
[Initial]/Kb >>> 400. We can neglect x and assume
[NO2-] initial ≈ [NO2-]eq
x2/0.0857 = 2.50 × 10-11
x
= 1.46 × 10-6 M
pOH = -log OH- = 5.83
= Kb
= [OH-]
pH = pKw – pOH = 8.17
[1]
[1]
L3/4n
(c) Identify a suitable indicator that can be used in this titration. [1]
Phenolphtalein
L1/4o
SET C
1
(a) Define buffer
[2]
A solution of weak acid and its conjugate base or weak base and
its conjugate acid [1] that will slightly change the pH when a
small amount of strong acid or base is added [1].
L1/4k
(b) A 1.00 L buffer solution is made up of 0.450 M HCOOH and 0.150 M
HCOOK. Calculate the initial pH of the buffer solution.
[Ka = 1.80 × 10-4]
pH
[2]
= pKa + log ([A-]/[HA])
= pKa + log ([HCOO-]/[HCOOH])
= -log (1.80 × 10-4) + log [(0.150)/(0.450)]
[1]
= 3.27
[1]
L2/4m
2
60.0 mL of 0.150 M NH3 is titrated with 0.200 M HCl.
[Ka(NH4+) = 5.56 × 10-10]
(a)
Predict whether the salt produced is acidic, basic or neutral.
Acidic
[1]
L2/4i
(b)
Calculate the pH after 45.0 mL of HCl is added.
[3]
This is at the equivalence point [1] whereby 9.00 × 10-3 mol of
HCl reacts with 9.00 × 10-3 mol of NH3 to form 9.00 × 10-3 mol
of NH4Cl. NH4Cl will hydrolyze in water.
[NH4+]
= 9.00 × 10-3 mol / (0.045 + 0.060) L
= 0.0857 M
Conc, M
I
NH4+
0.0857
H2O
NH3
H3O+
-
0
0
C
-x
-
+x
+x
F
0.0857 - x
-
x
X
For correct hydrolysis (equation/rxn with water) of NH4+[1]
x2/0.0857 = 5.56 × 10-10
= Ka
x
= 6.90 × 10-6 M = [H3O+ ]
pH = -log [ H3O+] = 5.16
(c)
[1]
[1]
Identify a suitable indicator that can be used in this titration.
Methyl red
L3/4n
[1]
L1/4o