Eastern Mediterranean University
Department of Chemistry
Department of Chemistry
CHEM247 Analytical Chemistry I
Fall 2012-2013 Midterm Exam
10-11-2012
10.30-12.00
ANSWER KEY
INSTRUCTIONS:
1.
Write your student number, name and surname, and sign the question booklet.
2.
The exam consists of 5 questions. To get full marks, answer all questions. Show all your steps.
3.
The Periodic Table provided may be necessary to answer some of the questions.
4.
Use of mobile phones, exchange of calculators or erasers is not allowed.
5.
You can see your papers in the first10 days after the announcement of the results.
Periodic Table of Elements
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
1
2
H
1.008
3
Li
6.94
11
Na
22.99
19
K
39.10
37
Rb
85.47
55
Cs
132.91
87
Fr
223.02
He
4.003
10
Ne
20.18
18
Ar
39.95
36
Kr
83.80
54
Xe
131.29
86
Rn
222.02
4
5
6
7
8
9
Be
9.01
12
Mg
24.30
20
Ca
40.08
38
Sr
87.62
56
Ba
137.33
88
Ra
226.03
B
10.81
13
Al
26.98
31
Ga
69.72
49
In
114.82
81
Tl
204.38
C
12.01
14
Si
28.09
32
Ge
72.61
50
Sn
118.71
82
Pb
207.2
N
14.01
15
P
30.97
33
As
74.92
51
Sb
121.75
83
Bi
208.98
O
16.00
16
S
32.07
34
Se
78.96
52
Te
127.6
84
Po
208.98
F
19.00
17
Cl
35.45
35
Br
79.90
53
I
126.90
85
At
209.99
21
Sc
44.96
39
Y
88.91
71
Lu
174.97
103
Lr
260.11
22
Ti
47.88
40
Zr
91.22
72
Hf
178.49
57
Lanthanides
La
138.91
89
Actinides
Ac
227.03
23
V
50.94
41
Nb
92.91
73
Ta
180.95
58
Ce
140.12
90
Th
232.04
24
Cr
52.00
42
Mo
95.94
74
W
183.85
59
Pr
140.91
91
Pa
231.04
25
Mn
54.94
43
Tc
98.91
75
Re
186.2
60
Nd
144.24
92
U
238.03
26
Fe
55.85
44
Ru
101.07
76
Os
190.2
61
Pm
146.92
93
Np
237.05
27
Co
58.93
45
Rh
102.91
77
Ir
192.22
62
Sm
150.36
94
Pu
244.06
28
Ni
58.69
46
Pd
106.42
78
Pt
195.08
63
Eu
151.97
95
Am
243.06
29
Cu
63.54
47
Ag
107.87
79
Au
196.97
64
Gd
157.25
96
Cm
247.07
30
Zn
65.39
48
Cd
112.41
80
Hg
200.59
65
Tb
158.93
97
Bk
247.07
66
Dy
162.50
98
Cf
251.08
67
Ho
164.93
99
Es
252.08
68
Er
167.26
100
Fm
257.10
69
Tm
168.93
101
Md
258.10
70
Yb
173.04
102
No
259.10
FALL 2012-2013
CHEM247 Midterm Exam
Page 2
Question 1
a) A concentrated solution of aqueous ammonia is 28.0% w/w NH3 and has a density of 0.899 g/mL. What is the
molar concentration (in mol/L) of NH3 in this solution?
(2 points)
b) Calcium carbonate is relatively insoluble in water. At 25°C, only 5.80 mg will dissolve in 1.00 liter of water.
What volume of water is needed to dissolve 5.00 g of calcium carbonate?
(2 points)
If 5.80 mg per 1 L is the maximum solubility, 5.00 g per V L must have the same mass to volume ratio.
Therefore
c) What is pNa for a solution of 3.45×10−4 M Na2SiO3?
(1 point)
pNa = −log10[Na+] = −log10[ 2×3.45×10−4] = −(−3.16) = 3.16
Question 2
a) Define accuracy and precision, and how we measure them. Explain the difference between them.
(2.5 points)
Accuracy is the closenes of measured values to the true or accepted value. It is measured in terms of error,
E=xtrue−xexp It s affected by systematic errors. It can usually be detected by correct calibration or by using
standard materials and can be eliminated.
Precision describes the reproducibility of measurements. It is the closeness to each other of the results
obtained in replicate analysis that have all been analyzed in exactly the same way. It is caused by random
errors and it is measured by standard deviation. It may be minimised but it can never be completely
eliminated. It is unrelated to accuracy of a method or analysis. Graphical comparison of accuracy and
precision:
b) What are the possible systematic and random errors in measuring mass on an electronic analytical balance. How
can we minimise or correct for these errors? Name at least two for each type of error.
(2.5 points)
 Buoyancy error is a systematic error. Causes measured mass to be smaller than true mass. The relative
error for samples with a density of less than 2 g/cm3 becomes significant. Can be corrected
mathematically by considering upthrust due to displaced air by the sample
 Temperature effect causing measured mass to be less than true value. Caused by convection currents
from hot bodies and reduced density of heated air inside the weighing compartment.Cool to room
temperature in a dessicator before weighing.
 Static electricity causes erratic measurements. Wipe glass container with moist chamois.
 Contamination from hands if vessels held by hand directly.
 Uncalibrated or non-level balance giving biased reading.
 Vessel not centered on the balance, but placed randomly on the pan each time, causing small bt random
errors.
 Balance in or on a non-vibration proof bench such that reading may not stabilise due to floorvibrations.
____________________________________________________________________________________________
FALL 2012-2013
CHEM247 Midterm Exam
Page 3
Question 3
a) Explain the differences in the use of burette, pipette and measuring cylinder in analytical chemistry? (2 points)
All three are used for dispensing liquids byut they differ in accuracy, precision or
purpose.
A burette is used to deliver continuously varying amounts of titrant
accurately during a titration. The quantity of solution delivered is found by
subtracting final reading from the initial reading level on the burette.
A pipette is usually used to deliver accuarately a fixed volume (aliquat) of
liquid.
Both pipettes and burettes have high accuracy and high precision.
A measuring cylinderis used to measure out approximate volumes of general
purpose liquid reagents. It is neither accurate nor precise.
measuring
cylinder
burette pipette
b) The results for the determination of morphine in five separate tablets of MS-Contin are: 24.0 mg, 24.6 mg, 23.9
mg, 25.3 mg and 26.2 mg. Calculate the mean, median, range, standard deviation, and variance for these data
and write your answers in the table below.
(3 points)
mean
̅
median
∑
24.8 mg
middle value
24.6 mg
range
xlargest−xsmallest
standard deviation
√
2.3 mg
∑
(
(
0.96 mg
̅)
)
variance
(
∑
(
̅)
)
0.93 mg
Question 4
Two analysts, A and B, determine the molar concentration of an unknown hydrochloric acid solution by titrating
five 25.0 mL portions with 0.100 M NaOH solution to phenolphthalein endpoint. The average volume of NaOH
used to reach endpoint by each analyst ( ̅ and ̅ ) and the standard deviations (sA and sB) are given below:
Analyst
Average Std dev
20.32
0.22
VA / mL
21.44
0.62
VB / mL
a) Calculate the 95% confidence interval for the mean volume of NaOH ( ̅ and ̅ ) for each analyst, using the
critical t value for 5 degrees of freedom,
.
(2 points)
̅̅̅̅
√
̅̅̅̅
√
̅
̅
b) Do the two confidence intervals for ̅ and ̅ overlap? Is there a significant difference between the two mean
values or can they be considered to be equal? Support your answer with a brief explanation.
(1 point)
The two intervals do not overlap. Therefore we can reject the null hypothesis that the two means are equal,
and say that there is a significant difference at the 95% level between the two means ̅ and ̅ .
FALL 2012-2013
CHEM247 Midterm Exam
Page 4
c) If the “true” titration volume of NaOH, VTrue, is 20.0 mL, calculate the “true” concentration of the HCl
solution.
(2 points)
At equivalence point, the amount of acid is equal to the amount of base. In other word nHCl = nNaOH.
Therefore [HCl]×VHCl = [NaOH]×VNaOH
And
[HCl] = [NaOH]×VNaOH/VHCl = 0.100×20.0/25.0 = 0.0800 M
Question 5
a) Calculate the pH and pOH of a solution that is 0.0250 M in NaOH. Kw = 1×10−14 M2
(2 points)
Since NaOH is a strong electrolyte, all of it is ionised in solution. Therefore [OH−] = 0.0250 M
From self-ionization of water; [H+] = Kw/[OH−] == 10−14/0.025 = 4×10−13 M
Thus
And
pOH = −log10[OH−] = −log10[ 0.025] = −(−1.60) = 1.60
pH = −log10[H+] = −log10[4×10−13] = −(12.40) = 12.40
or pH = 14 – pOH = 14−1.60 = 12.40
b) Acetic acid (HA) is a weak acid. Its acid dissociation constant is Ka = 1.75×10−5 M.
HA(aq)  A−(aq) + H+(aq)
Write three expressions for Ka, mass balance and charge balance. Using these equations calculate the pH of a
0.250 M acetic acid solution. List any assumptions you make about concentrations in your calculations.(3 points)
Eqn 1
Acid dissociation constant
[ ] [ ]
[ ]
With the assumptions on
the right Eqn 1 becomes
[ ]
Eqn 2
Mass Balance
Eqn 3
Charge balance
CHA = [HA] + [A−]
[H+] = [OH−] + [A−]
Assumption that
[HA] [A−] gives
CHA
[H+]
[HA]
For eqn 3: Since solution is acidic, we know that [H+]
Assumption that
[H+] [OH−] gives
[A−]
[OH−] thus we can assume that [H+]
For eqn 2: Since HA is a weak acid, only a small fraction will ionize, in other words [HA]
therefore we can assume that CHA [HA]
[A−].
[A−],
With these assumptions we can substitute for [HA] and [A−] in Eqn 1 and obtain a simplified eqn. After
rearranging the substituted eqn 1 for [H+] we get
[H+] = √
√
Thus, pH =−log10[H+] = −log10[2.09×10−3] = −(−2.68) = 2.68