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Chem201, Winter 2006 Name Answer key______________ Midterm N1 01/26/06 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) nCl- = 2 n MgCl2 = 2 (m MgCl2/ MW MgCl2) = 0.541 moles MCl- = nCl- / 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 = nCl- 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+ + SO4
2- 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 Ba2+
=mass Ba(NO3)2
MW Ba(NO3)2=
1.01g
261.34g /mol= 3.86 !10
"3moles
moles H2SO
4= moles Ba
2+= 3.86 !10
"3moles
volume H2SO
4=moles H
2SO
4
M H2SO
4
=3.86 !10
"3moles
3moles /L=1.29 !10
"3L =1.29 mL
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
s = 1.65 =
(xi! x)2"
n !1
tcalc
=| ! x |
sn =
| 94.6 ! 97.0 |
1.655 = 3.25
ttable
95%= 2.776 < 3.25 Significant difference
ttable
99%= 4.604 > 3.25 No significant difference
ttable
99.9%= 8.610 > 3.25 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
Range=12
24= 0.5
yi =! 0.095 + 0.198 + 0.295 = 0.588
xi
2=! (5.23)2 + (10.52)2 + (15.41)2 = 375.49
xiyi =! (5.23 " 0.095) + (10.52 " 0.198) + (15.41" 0.295) = 7.13
n = 3
D =xi
2xi
xi
n
=375.49 31.16
31.16 3=155.52
m =xiyi xi
yi nD =
7.13 31.16
0.588 3155.52 = 0.0197
b =xi2
xiyi
xi yiD =
375.49 7.13
31.16 0.588155.52 = !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)
0.0197= 8.32 mg /L
ii) 0.265 (2 points)
x =0.265 ! (!0.0089)
0.0197=13.90 mg /L
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
X= F
As
S!
10.1A
0.800 mM= F
15.3A
0.500 mM! F = 0.412
Concentration of the chloroform in unknown:
(10.2 !10"6L) (1484 g /L) /119.39 g /mol
0.100 L= 0.00126 M =1.26 mM
For the unknown mixture:
Ax
X= F
As
S!8.7 A
X= 0.412
29.4 A
1.268 mM! X = 0.909 mM
DDT in unknown:
0.909 mM !100 mL
10 mL= 9.09 mM
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? Ksp 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
250.0 + 250.0= 0.075 M
[Br!] = 0.300 M
250.0
250.0 + 250.0= 0.150 M
Br- ion is in excess: 0.150 0.075 = 0.075 M. [Ag+] = x and [Br-] = (x+0.075)
Ksp = (x)(x + 0.075) = 5.0 !10"13
Assuming x
VCr2O7
2! =M
Fe2+ "V
Fe2+
6 " MCr2O7
2!
=0.2500 M " 200.0 mL
6 " 0.1658 M= 50.26 mL
10. A mixture having a volume of 10.0 mL and containing 0.100 M Ag+ and 0.100 M Hg2
2+ was titrated with 0.100 M KCN to precipitate Hg2(CN)2 (Ksp = 510-40) and
AgCN (Ksp = 2.210-16). Calculate the concentration of the CN- at each of the
following volumes of added KCN:
Hg2
2++ 2CN
!" Hg
2(CN)
2
Ag+
+ CN!" AgCN
Hg22+ 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)
[Hg22+] =
20 ! 5
20
"
# $
%
& ' 0.100( )
10
10 + 5
"
# $
%
& ' = 0.050 M
[CN!] =
Ksp
[Hg2
2+]
=5 "10
!40
0.050=1.0 "10
!19M
b. 15.00 mL (5points)
[Hg22+] =
20 !15
20
"
# $
%
& ' 0.100( )
10
10 +15
"
# $
%
& ' = 0.010 M
[CN!] =
Ksp
[Hg2
2+]
=5 "10
!40
0.010= 2.23 "10
!19M
c. 35.00 mL (5points)
At 35.00 mL, there are 5 mL excess of the [CN-]:
[CN!] =
5.00
10.00 + 35.00
"
# $
%
& ' 0.100( ) = 0.011M
11. Calculate the concentration of Ag+ in saturated solutions of Ag2CO3 (Ksp= 8.110
-12) in:
!+!+!+ ===!+
23
23
23
23222
3
22 ][][COAgCOAgCOAgsp
xxxCOAgK """"""
x x
x = [Ag+] =
Ksp
!Ag
+
2!CO3
2"
3
Corresponding activity coefficients are taken from table (see supplemental information). (a) 0.001 M KNO3 (5points)
=1
2ciz!i
2=1
2(0.001"12) + (0.001"12)( ) = 0.001
[Ag+] =
Ksp
!Ag
+
2!CO3
2"
3 =8.1#10-12
(0.964)2(0.867)
3 = 0.216 mM
(b) 0.01 M KNO3 (5points)
=1
2ciz!i
2=1
2(0.01"12) + (0.01"12)( ) = 0.01
[Ag+] =
Ksp
!Ag
+
2!CO3
2"
3 =8.1#10-12
(0.898)2(0.665)
3 = 0.247 mM
(c) 0.1 M KNO3 (5points)
=1
2ciz!i
2=1
2(0.1"12) + (0.1"12)( ) = 0.1
[Ag+] =
Ksp
!Ag
+
2!CO3
2"
3 =8.1#10-12
(0.75)2(0.37)
3 = 0.339 mM
Supplemental information