Chem 55 Exam 2 Solutions

CHEM 55 Fall 2007 Exam 1 Copy # 1
Name: KEY
Signature_________________________________ Date __________
1. A trainee in a medical lab measured the six samples and compared her results with
those of an experienced worker.
Trainee:
avg = 14.5 7 mM s = 0.5 3 mM n = 6 samples
Experienced: avg = 13.9 5 mM s = 0.4 2 mM n = 5 samples
Is there evidence at the 95% confidence level that the trainee’s measurements are
different from those of the experienced worker Show all work for credit.
5
x av1 := 14.57 s 1 := 0.53 n 1 := 6 x av2 := 13.95 s 2 := 0.42 n 2 := 5
n 1 ⋅n 2
x av1 − x av2 ⋅
n 1 + n 2
t calc := degrees_of_freedom := n 1 + n 2 − 2
2
s 1 ⋅ n1
+
n 1 + n 2
2
s 2 ⋅ n2
t calc = 2.119 t tab = 1.833
since tcalc > ttab, the difference is judged significant
at the 95% confidence level
2. An unknown sample of Cu +2 gave an absorbance of 0.262 in an atomic absorption
analysis. Then 1.00 mL of solution containing 100.0 ppm of Cu 2+ was mixed with 95.0
mL of the Cu +2 unknown and diluted to the mark in a 100.0 mL volumetric flask. The
absorbance of the new solution was 0.500.
a. Denoting the unknown concentration as [Cu +2 ] i , write and expression for the
concentration of [Cu +2 ] after dilution to 100 mL, denote this concentration [Cu +2 ] f
b. Do the same for the spike concentration denoting it [S] i and [S] f .
c. Calculate the concentration of [Cu] i in the original unknown solution.
10
I x := 0.262
I xs := 0.500 S i := 100
95
1.00
a. X f X i ⋅ b. S
100
f S i ⋅
100
I x
X i
I xs
X f + S f
0.262 0.500
X i X i ⋅0.95
+ 1.00
0.5⋅ X i 0.262⋅
0.95⋅ X i + 0.262 0.262⋅ 0.95 = 0.249
0.5⋅
X i
0.249⋅ X i + 0.262 ( 0.5 − 0.249) ⋅ X i 0.262
0.262
X i := X
( 0.5 − 0.249)
i = 1.04 ppm

CHEM 55 Fall 2007 Exam 1 Copy # 1
Name: Au,Hai Nguyen
3. Calculate the molarity of La(IO 3 ) 3 at saturation in water (K SP (La(IO 3 ) 3 = 1.0x10 -11 ).
Ksp
⎛
⎜
⎝
x⋅ ( 3x) 3 27x 4 x
1.0⋅
10 − 11
27
⎞
⎟
⎠
1
4
⎛
⎜
⎝
Ksp
27
⎞ ⎟⎠
1
4
= 7.801×
10 − 4 = [La+3]
5
4. Calculate the molarity of La(IO 3 ) 3 at saturation in water containing 0.050 M LiIO 3 given
K SP (LaIO 3 ) = 1.0x10 -11 .
Ksp x⋅
0.050 3
1.0⋅
10 − 11
= 8×
10 − 8 = [La+3]
0.050 3
5
5. Repeat the above calculation but including the effects of ion activity and given that the
ionic radius of La +3 is 900 pm and that of IO 3 - is 450 pm.
5
− 0.51⋅z 2 ⋅ μ
1+
α⋅ μ
305
z:= 3 α := 900
γLa := 10
γLa = 0.24
− 0.51⋅z 2 ⋅ μ
1+
α⋅ μ
305
z:= −1
α := 405
γIO3 := 10
γIO3 = 0.82
Ksp
x⋅ γLa⋅0.050 3 ⋅γIO3 3
1.0⋅
10 − 11
0.050 3 ⋅γIO3 3 ⋅γLa
= 6.1 × 10 − 7 = [La+3]

CHEM 55 Fall 2007 Exam 1 Copy # 1
Name: Au,Hai Nguyen
6. A solution of NaOH was standardized by titration of a known quantity of the primary
standard potassium hydrogen phthalate (KHP, FM 204.221 g/mol).
a. Titration of 0.824 g of potassium hydrogen phthalate required 38.314 mL of NaOH to
reach the endpoint detected by phenolphthalein. Find the concentration of NaOH in
molarity (M).
10
mol⋅
KHP
0.824⋅
g⋅
⋅
204.221g ⋅ ⋅KHP
1⋅
NaOH
1⋅KHP
⋅
1
=
38.31410 ⋅
− 3 ⋅L⋅
NaOH
mol NaOH
0.105 ⋅ L
b. A 100.00 mL aliquot of H 2 SO 4 solution required 57.911 mL of the NaOH solution to
reach the endpoint. Calculate the concentration of the H 2 SO 4 .
5
57.91110 ⋅
− 3 0.1053mol ⋅ ⋅ NaOH 1⋅H2SO4
1
⋅L⋅
NaOH⋅
⋅ ⋅
=
L⋅
NaOH 2⋅
NaOH
100.0⋅
10 − 3 ⋅L⋅
H2SO4
mol H2SO4
0.0305 ⋅ L

CHEM 55 Fall 2007 Exam 1 Copy # 1
Name: Au,Hai Nguyen
7. How many milliliters of 0.100 M KI are needed to react with 40.0 mL of 0.0400 M
Hg 2 (NO 3 ) 2 if the reaction is Hg 2 2+ + 2I - Hg 2 I 2 (s)
5
40.0⋅
10 − 3
mol⋅
Hg2NO32 2KI ⋅ 1L ⋅ ⋅KI
⋅L⋅ Hg2NO32⋅
0.0400⋅
⋅ ⋅
=
L⋅
Hg2NO32 Hg2NO32 0.100⋅
mol⋅
KI
0.0320L⋅
KI
8.
a. Find the activity coefficient of H + in a solution containing 0.010 M HCl plus 0.040 M
KClO 4 . (H + = 900 pm)
5
0.010⋅ 1 2 + 0.010⋅( −1) 2 + 0.040⋅
1 2 + 0.040⋅( −1) 2
μ := μ = 0.050
2
− 0.51⋅z 2 ⋅
1+
α⋅
μ
305
z:= 1 α := 900
γH:= 10
γH = 0.85
μ
b. What is the pH of this solution
5
( )
pH := −log 0.010⋅
γH pH = 2.069

CHEM 55 Fall 2007 Exam 1 Copy # 1
Name: Au,Hai Nguyen
Values of Student's t
% confidence interval
°freedom 50 80 95 99
1 1.00 3.08 12.71 63.66
2 0.82 1.89 4.30 9.92
3 0.76 1.64 3.18 5.84
4 0.74 1.53 2.78 4.60
5 0.73 1.48 2.57 4.03
6 0.72 1.44 2.45 3.71
7 0.71 1.41 2.36 3.50
8 0.71 1.40 2.31 3.36
9 0.70 1.38 2.26 3.25
10 0.70 1.37 2.23 3.17
20 0.69 1.33 2.09 2.85
50 0.68 1.30 2.01 2.68
100 0.68 1.29 1.98 2.63
∞ 0.67 1.28 1.96 2.58
Values of Q for rejection of data:
Q
90% confidence
Number of
Observations
0.76 4
0.64 5
0.56 6
0.51 7
0.47 8
0.44 9
0.41 10
Average, Sample standard deviation
and confidence interval formulae:
x =
n
∑
i=
1
n
n
x
∑
( xi
− x)
i=
1
s =
n −1
CI = x ±
ts
n
( x − μ)
t =
s
i
n
2
Error propagation formulae:
for : y = x1
+ x2
for : y = x1⋅
x2
for : y = log( x)
for : y = 10
for : y = ln( x)
for : y = e
for : y = x
x
a
x
e
e
y
y
% e
y
e
y
% e
y
=
=
ex
ey
=
x
= y ⋅e
x
e
2
x1
= a ⋅%
e
% e
x
+ e
2
x1
ex
ey
= 0.434
x
= 2.303⋅
y ⋅e
x
2
x2
+ % e
2
x2

CHEM 55 Fall 2007 Exam 1 Copy # 1
Name: Au,Hai Nguyen
For standard additions:
I
X
I
S + X
=
[ X ] [ X ] + [ S]
where
[ X ]
[ S]
f
i
f
f
Vx
= [ X ]
i
and
V
VS
= [ S]
i
V
f
For Internal Standads:
AX
AS
= F
[ X ] [ S]
where
A
A
X
S
= Sample Signal
= Internal S tan dard
Signal
Detection Limit:
3s
Cm
=
m
and
y
DL
= y
BLANK
+ 3s
Spike Recovery
C
% recovery
=
spiked sample
C
− C
added
unspiked sample
Ion activity in general:
for
aA + bB ↔ cC + dD
K
EQ
c c d d
[ C]
γ
C[
D]
γ
D
=
a a b b
[ A]
γ [ B]
γ
A
B
pH
= −log[
H
+
] γ
H +
Ion activity coefficient:
2
− 0.51z
μ
log( γ ) =
α μ
1 +
305
Ionic strength:
1
μ =
2
∑
i
Ci z 2
i