Chem201, Winter 2006
Midterm N1
01/26/06
Name Answer key______________
SID___________________________
1. A solution is prepared by dissolving 25.8 grams on magnesium chloride (MgCl2) in
water to produce 250.0 mL of solution. Molecular weight of the MgCl2 is 95.3 g/mol.
Molecular weights of Mg and Cl are 24.3 g/mol and 35.5 g/mol, respectively.
a. Calculate the molarity of the chloride ion in the solution. (3points)
n Cl- = 2 n
MgCl2
= 2 (m MgCl2/ MW MgCl2) = 0.541 moles
MCl- = n Cl- / Volume = 0.541 moles / 0.25 L = 2.17 M
b. What is the concentration of the Cl- in ppm? (3points)
Cl- ppm = mass Cl- (mg) / volume = n Cl- MWCl- / volume = 0.541 x 35.5 x 1000 /
0.25L =
76840 ppm = 7.68 x 104 ppm
c. Calculate the pCl- value for this solution. (3points)
pCl- = -log [2.17] = -0.34
2. A bottle of a concentrated aqueous sulfuric acid is, labeled 98.0 wt % H2SO4
(Molecular weight is 98.09 g/mol) has a concentration of 18.0 M.
a. How many milliliters of reagent should be diluted to 1.000 L to give 1.00 M
H2SO4 ? (5 points)
Vcon = Vdil x (Mdil / Mcon ) = 1000 mL x ( 1.00 M / 18.0 M) = 55.6 M
b. Calculate the density of 98.0 wt % H2SO4 (5 points)
Mass of the 1 liter of H2SO4: (18 moles) (98.09) = 1.77 x 10 3 gr.
Mass of the 1 mL of H2SO4: 1.77 g
d = mass / weight % = 1.77 g / (0.98 g H2SO4 /g solution) = 1.8 g/mL
3. How many milliliters of 3.00 M sulfuric acid are required to react with 4.35 g of solid
containing 23.2 g wt % Ba(NO3 )2 if the reaction is:
Ba2+ + SO42-
BaSO4 ?
(5 points)
Molecular weights of BaSO4 is 233.0 g/mole and Ba(NO3 )2 is 261.3 g/mol.
Mass Ba(NO3 )2 is 0.232 x 4.35 = 1.01 g
moles Ba 2+ =
mass Ba(NO3 ) 2
1.01g
=
= 3.86 !10"3 moles
MW Ba(NO3 ) 2 261.34g /mol
moles H 2 SO4 = moles Ba 2+ = 3.86 !10"3 moles
moles H 2 SO4 3.86 !10"3 moles
volume H 2 SO4 =
=
= 1.29 !10"3 L = 1.29 mL
M H 2 SO4
3 moles /L
4. A sample is certified to contain 94.6 ppm of a contaminant. Your analysis gives
values of 98.6, 98.4, 97.2, 94.6 and 96.2 ppm. Do you results differ from the
expected result at following confidence levels: i) 95%, ii) 99% and iii) 99.9%.
(9points)
x = 97.0
"(x ! x)
2
i
s = 1.65 =
t calc =
n !1
|µ ! x |
| 94.6 ! 97.0 |
n=
5 = 3.25
s
1.65
t table
95%
= 2.776 < 3.25 Significant difference
t table
99%
= 4.604 > 3.25
t table
99.9%
= 8.610 > 3.25
No significant difference
No, significant difference
5. Using the appropriate statistical test, decide whether the value 216 should be rejected
from the set of result: 192, 216, 202, 195 and 204? (3 points)
Gap = 12
Range = 24
Qcalc =
Gap
12
=
= 0.5 < Qtable = 0.64
Range 24
Value to be retained.
6. The following data was collected when performing a spectrophotometric analysis for
cobalt.
x
y
Analysis No
mg Co / liter
Absorbance
4
5.23
0.095
5
10.52
0.198
6
15.41
0.295
a. Using the least squares method of linear regression, generate the equation to define
the line for the absorbance vs. concentration. (10 points).
!x
i
= 5.23 + 10.52 + 15.41 = 31.16
!y
i
!x
i
= 0.095 + 0.198 + 0.295 = 0.588
2
= (5.23) 2 + (10.52) 2 + (15.41) 2 = 375.49
!x y
i
i
= (5.23" 0.095) + (10.52 " 0.198) + (15.41" 0.295) = 7.13
n=3
x
D= i
xi
m=
2
x i 375.49
=
31.16
n
31.16
= 155.52
3
xi yi
yi
xi
7.13
÷D=
n
0.588
2
375.49
xi yi
÷D=
31.16
yi
x
b= i
xi
31.16
÷155.52 = 0.0197
3
7.13
0.588
÷155.52 = !0.0089
Thus, linear regression line is:
y = 0.0197x ! 0.0089
b. Based on the equation you have generated, calculate the concentration of the Co in
the sample if the absorbance is:
i) 0.155 (2 points)
x=
0.155 ! (!0.0089)
= 8.32 mg /L
0.0197
ii) 0.265 (2 points)
x=
0.265 ! (!0.0089)
= 13.90 mg /L
0.0197
7. Chloroform is an internal standard in the determination of the pesticide DDT in a
polarographic analysis. A mixture containing 0.500 mM chloroform and 0.800 mM
DDT gave signals of 15.3 A for chloroform and 10.1 A for DDT. An unknown
solution (10.0 mL) containing DDT was placed in a 100.0 mL volumetric flask and
10.2 L of chloroform (FW 119.39 g/mol, density = 1.484 g/mL) were added. After
diluting to the mark with solvent, polarographic signals of 29.4 and 8.7 A were
observed for the chloroform and DDT, respectively. Find the concentration of DDT
in unknown. (10 points)
Chloroform is S, and DDT is X:
Ax
A
10.1 µA
15.3 µA
=F s !
=F
! F = 0.412
X
S
0.800 mM
0.500 mM
Concentration of the chloroform in unknown:
(10.2 !10"6 L) (1484 g /L) /119.39 g /mol
= 0.00126 M = 1.26 mM
0.100 L
For the unknown mixture:
Ax
A
8.7 µA
29.4 µA
=F s !
= 0.412
! X = 0.909 mM
X
S
X
1.268 mM
DDT in unknown:
0.909 mM !
100 mL
= 9.09 mM
10 mL
8. A beaker contains 250.0 mL of 0.150 molar silver ion (Ag + ). To this beaker is added
250.0 mL of 0.300 molar bromide ion (Br-). What is the concentration of Ag + in the
final solution? K sp for the AgBr is 5.0 × 10 -13? (5 points)
AgBr ! Ag + + Br "
Final concentration of the Ag + and Br-:
[Ag + ] = 0.150 M
250.0
= 0.075 M
250.0 + 250.0
[Br ! ] = 0.300 M
250.0
= 0.150 M
250.0 + 250.0
Br- ion is in excess: 0.150 – 0.075 = 0.075 M. [Ag + ] = x and [Br-] = (x+0.075)
K sp = (x)(x + 0.075) = 5.0 !10"13
Assuming x << 0.075, we have
K sp = (x) (0.075) = 5.0 !10"13
x = [Ag + ] = 6.67 × 10 -12 .
9. Iron in the +2 oxidation state reacts with potassium dichromate to produce Fe3+ and
Cr3+ according to the equation:
6 Fe2+ + Cr2 O72- + 14 H +
6 Fe3+ + 2 Cr3+ + 7 H2 O
How many milliliters of 0.1658 molar K2Cr2 O7 are required to titrate 200.0 mL of
0.2500 molar Fe2+ solution? (5 points)
n Fe 2+ = 6 ! nCr O 2" # M Fe 2+ VFe 2+ = 6MCr O 2" VCr O 2"
2
Therefore,
7
2
7
2
7
VCr O 2! =
2
7
M Fe 2+ " VFe 2+
6 " MCr O 2!
2
=
7
0.2500 M " 200.0 mL
= 50.26 mL
6 " 0.1658 M
10. A mixture having a volume of 10.0 mL and containing 0.100 M Ag + and 0.100 M
Hg 22+ was titrated with 0.100 M KCN to precipitate Hg 2 (CN)2 (K sp = 5×10 -40 ) and
AgCN (K sp = 2.2×10 -16 ). Calculate the concentration of the CN- at each of the
following volumes of added KCN:
2+
Hg2 + 2CN ! " Hg2 (CN) 2
Ag + + CN ! " AgCN
Hg 22+ will precipitate first and the equivalence point is at 20.00 mL. And the second
equivalence point is at 30 mL. At 5 and 15 mL there is an excess of unreacted Hg22+.
a. 5.00 mL (5points)
" 20 ! 5 %
" 10 %
2+
[Hg 2 ] = $
' (0.100) $
' = 0.050 M
# 20 &
# 10 + 5 &
[CN ! ] =
K sp
[Hg2 2+ ]
=
5 "10!40
= 1.0 "10!19 M
0.050
b. 15.00 mL (5points)
" 20 !15 %
" 10 %
2+
[Hg 2 ] = $
' (0.100) $
' = 0.010 M
# 20 &
# 10 + 15 &
[CN ! ] =
K sp
[Hg2 2+ ]
=
5 "10!40
= 2.23"10!19 M
0.010
c. 35.00 mL (5points)
At 35.00 mL, there are 5 mL excess of the [CN-]:
"
%
5.00
[CN ! ] = $
' (0.100) = 0.011 M
#10.00 + 35.00 &
11. Calculate the concentration of Ag+ in saturated solutions of Ag2CO3
(Ksp = 8.1×10-12) in:
2
2!
2
2
K sp = [Ag+ ]2" Ag+ [CO3 ]" CO 2! = x2" Ag+ x" CO 2! = x3" Ag+ " CO 2!
3
3
3
x
x
x = [Ag + ] = 3
K sp
! Ag + 2! CO 2"
3
Corresponding activity coefficients are taken from table (see supplemental information).
(a) 0.001 M KNO3 (5points)
µ=
1
1
2
c i z = ((0.001"12 ) + (0.001"12 )) = 0.001
!
i
2
2
[Ag + ] = 3
K sp
! Ag + 2! CO 2"
=3
8.1#10 -12
= 0.216 mM
(0.964) 2 (0.867)
3
(b) 0.01 M KNO3 (5points)
µ=
1
1
2
c i z = ((0.01"12 ) + (0.01"12 )) = 0.01
!
i
2
2
+
[Ag ] = 3
K sp
! Ag + 2! CO 2"
8.1#10 -12
=
= 0.247 mM
(0.898) 2 (0.665)
3
3
(c) 0.1 M KNO3 (5points)
µ=
1
2 1
c i z = ((0.1"12 ) + (0.1"12 )) = 0.1
!
i
2
2
[Ag + ] = 3
K sp
! Ag + 2! CO 2"
3
=3
8.1#10 -12
= 0.339 mM
(0.75) 2 (0.37)
Supplemental information