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