Chem 213 Chemical Analysis

Final December 1, 2004

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100% KEY

Name:___________________________________________ (please print)

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1. A reliable assay shows that the ATP (adenosine triphosphate) content of a certain cell type is 111 µmol/100 mL. You develped a new assay, which gave the following values for replicate analyses: 117, 119, 111, 115, 120 µmol/100 mol. Can you be 95% confident that your result differs form the “known” value? (10 points) (A simple “yes” or “no” without any calculations is not a sufficient answer and will receive no credit!)

s = 3.58, x = 116.4

tcalculated =known value - x

sN

=111 - 116.4

3.585 = 3.37

For 4 degrees of freedom and 95% confidence, ttable = 2.227. Because tcalculated (3.37) > ttable (2.227), the difference is significant. Thus, the answer is yes.

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2. A solution with an ionic strength of 0.10 M containing 0.010 M phenylhydrazine has a pH of 8.13. Using activity coefficients correctly, find the pKa for the phenylhydrazonium ion found in phenylhydrazine hydrochloride. Assume that γBH+ = 0.80 (appropriate value for OH- can be found in Kielland’s table, page 11). (10 points)

NHNH2

PhenylhydrazineB

NHNH3+Cl-

Phenylhydrazine hydrochlorideBH+Cl-

The reaction of phenylhydrazine with water is

B + H2O ↔ BH+ + OH- Kb We know that pH = 8.13, so we can find [OH-].

[OH-] = aOH-/γOH- = (Kw/10-pH)/γOH- = 1.78•10-6 M (using γOH- = 0.76 for µ = 0.10 M) Kb = ([BH+]•γBH+•[OH-]•γOH-)/([B]•γB) = [(1.78•10-6)(0.80)(1.78•10-6)(0.76)]/[(0.010 - 1.78•10-6)(1.00)] = 1.93 •10-10

Ka = Kw/Kb = 5.19•10-5

pKa = 4.28

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3. Suppose that 0.010 M Mn2+ is titrated with 0.005 M EDTA at pH 7.00. Kf(MnY2-) = 1013.87.

pH α4 pH α4 pH α40 1.3•10-23 5 3.7•10-7 10 0.36 1 1.9•10-18 6 2.3•10-5 11 0.85 2 3.3•10-14 7 5.0•10-4 12 0.98 3 2.6•10-11 8 5.6•10-3 13 1.00 4 3.8•10-9 9 5.4•10-2 14 1.00

a. What is the concentration of free Mn2+ (pMn2+) at the equivalence point. (5

points) One volume of Mn2+ will require two volumes of EDTA to reach the equivalence point. The formal concentration of MnY2- at the equivalence point is (1/3)(0.010) = 0.00333 M

Mn2+ + EDTA ↔ MnY2-

x x 0.00333 - x (0.00333 – x)/x2 = αY4-Kf = (5.0•10-4)1013.87 = 3.7•1010

x = [Mn2+] = 3.0•10-7

b. What is the quotient [H3Y-]/[H2Y-] in the solution when the titration is just

63.7% of the way to the equivalence point? Use the equilibrium given below for your calculation: (5 points)

= 10-2.66

Because the pH is 7.00, the ratio [H3Y-]/[H2Y-] is constant throughout the entire titration.

[H2Y2-][H+]/[H3Y-] = K4

[H3Y-]/[H2Y-] = [H+]/K4 = 10-7.00/10-2.66 = 4.6•10-5

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4. How would you prepare exactly 1.00 L of a buffer with a pH of 9.60 from 0.300 M Na2CO3 and 0.200 M HCl? The acid dissociation constants of H2CO3 are: Ka1 = 4.45•10-7; Ka2 = 4.69•10-11. (10 points)

pH = 9.60 [H3O+] = antilog (-9.60) = 2.512•10-10

[H3O+][CO3

2-]/[HCO3-] = 4.69•10-11

[CO3

2-]/[HCO3-] = 4.69•10-11/2.512•10-10 = 0.1867

Let VHCl and VNa2CO3 be the volume in milliliters of the two reagents. Then: VHCl + VNa2CO3 = 1000 mL [CO3

2-] ≈ c(Na2CO3) = (VNa2CO3 • 0.300 – VHCl • 0.200)/1000 [HCO3

-] ≈ c(HCO3-) = VHCl • 0.200/1000

Substituting these relationships into the ratio [CO3

2-]/[HCO3-] gives

0.300V - 0.200 VHCl

0.200 VHCl= 0.1867

Na CO2 3

0.300 VNa2CO3 – 0.200 VHCl = 0.03734 VHCl 0.300(1.000 – VHCl) = 0.23734 VHCl VHCl = 300/0.5373 = 558 mL VNa2CO3 = 1000 – 558 = 442 mL Thus mix 442 mL of 0.300 M Na2CO3 with 558 ml of 0.200 M HCl.

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5. An 8.13-g sample of an ant-control preparation was decomposed by wet-ashing with H2SO4 and HNO3. The As in the residue was reduced to the trivalent state with hydrazine. After removal of the excess reducing agent, the As(III) required a 23.77-mL titration with 0.02425 M I2 in a faintly alkaline medium.

H3AsO3 + I2 + H2O ↔ H3AsO4 + 2I- + 2H+

Express the result of this analysis in terms of percentage of As2O3 in the original sample. (10 points) 2 mol I2 ≡ 2 mol H2AsO3 ≡ 1 mol As2O3 0.02377 L I2 x 0.02425 mol I2/L = 0.000576422 mol I2 0.000576422 mol I2 x 1 mol As2O3/2 mol I2 = 0.000288211 mol As2O3 0.000288211 mol As2O3 x 197.841 g As2O3/mol As2O3 = 0.05702 g As2O3 0.05702 g As2O3/8.13 g sample x 100% = 0.701% As2O3

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6. Consider the titration of 50.0 mL of 0.050 M malonic acid (Ka1 = 1.42•10-3, Ka2 = 2.01•10-6) with 0.100 M NaOH. The titration reaction occurring is:

HOCCH2CO2H + OH- → -O2CCH2CO2H + H2O -OCCH2CO2H + OH- → -O2CCH2CO2

- + H2O Designate malonic acid as H2M and use the following values:.

a. At 0.0 mL (3 points) H2M ↔ H+ + HM-

0.050 – x x x x2/(0.050 – x) = K1 → x = 7.75•10-3

pH = 2.11

b. At 8.0 mL (3 points) H2M + OH- → HM- + H2O

Initial: 25 8 - - Final: 17 o 8 -

pH = pKa1 + log([HM-]/[H2M]) = 2.487 + log(8/17) = 2.52 c. At 12.5 mL (4 points)

Vb = ½ Ve → pH = pKa1 = 2.85 d. At 25.0 mL (4 points)

At the first equivalence point, H2M has been converted to HM-.

[H3O+] =Ka1Ka2F + Ka1Kw

Ka1 + F

where F = (50/75)(0.050) = 0.0333 M [H3O+] = 5.23•10-5 M → pH = 4.28

e. At 50.0 mL (3 points) At the second equivalence point, H2M has been converted to M2-:

M2- + H2O ↔ HM- + OH-

(50/100)(0.050) –x x x

x2/(0.025 – x) = Kb1 = Kw/Ka2 → x = 1.12•10-5 M pH = -log(Kw/x) = 9.05

f. At 56.3 mL (3 points)

There are 6.3 mL of excess NaOH [OH-] = (6.3/106.3)(0.100) = 5.93•10-3 M →pH = 11.77

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7. Given the formation constants Fe3+ + Y4- ↔ FeY- Kf = 1.3•1025

Fe2+ + Y4- ↔ FeY2- Kf = 2.1•1014

Calculate E0 for the process FeY- + e- ↔ FeY2-

The relevant redox equilibrium is Fe3+ + e- ↔ Fe2+ E0 = +0.771 V

(10 points)

[Fe3+] = [FeY-]/1.3•1025•[Y4-] and [Fe2+] = [FeY2-]/2.1•1014•[Y4-] E = 0.771 – 0.0592 log[Fe2+]/[Fe3+] = 0.771 – 0.0592 log([FeY2-]1.3•1025[Y4-])/2.1•1014[Y4-][FeY-] E= E0

FeY- when [FeY2-], [FeY-], and [Y4-] = 1.000 E0 = 0.771 – 0.0592 log(1.3•1025/2.1•1014) = 0.771 – 0.639 = 0.13 V

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8. Calculate the potentials after the following volume of 0.02000 M MnO4-

solution was added to 50 mL of a 0.05000 M U4+ solution. The relevant redox equilibria are:

UO2

2+ + 4H+ + 2e- ↔ U4+ + 2 H2O E0 = +0.334 V MnO4

- + 8H+ + 5e- ↔ Mn2+ + 2H2O E0 = +1.51 V a. 25.00 mL (3 points) Pre-equivalence point region

UO22+

U4+

[U4+]

c =25 mL x 0.02 M KMnO4 x 5 mmol UO2

2+/2 mmol KMnO4

75 mL= [UO2

2+] = 0.01667 M

c =50 mL x 0.0500 M UO2

2+ - 25 mL x 0.02 M KMnO4 x 5/275 mL

= [U4+] = 0.01667 M

E = E0 - log = 0.334 - 0.0296 logUO22+/U4+

0.05922 [UO2

4+][H+]40.01667

0.01667(1.00)4

E = 0.334 V b. 50.00 mL (4 points)

2E0 + 5E0 0.0592 1UO22+/U4+ MnO4

-/Mn2+

Eeq = - log7 7 [H+]12

= - 0.00846 log 1 = 1.170.668 + 7.55

7 c. 60.00 mL (3 points)

[Mn2+]

60 mL x 0.02 M KMnO4 - (50 mL x 0.05 M UO22+ x 2/5)

110 mL[MnO4

-] =

50 mL x 0.05 M UO22+ x 2/5

110 mL[Mn2+] =

= 0.0018 M

= 0.0091 M

E = E0 - log = 1.51 - 0.01184 logMnO4-/Mn2+

0.05925 [MnO4

-][H+]80.0091

0.0018(1.00)8

E = 1.50 V

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9. The complex formed between Cu(I) and 1,10-phenathroline has a molar absorptivity of 7000 L•mol-1•cm-1 at 435 nm, the wavelength of maximum absorption. Calculate a. the absorbance of an 8.50•10-5 M solution of the complex when measured in a 1.00-cm cell at 435 nm. (3 points) A = ε•b•C = 7.00•103•1.00•8.50•10-5 = 0.595 b. the percent transmittance of the solution in part a. (2 points) T = (antilog -0.595)•100% = 25.4% c. the concentration of a solution that in a 5.00 cm cell has the same absorbance as the solution in part a. (3 points) c = A/(ε•b) = 0.595/(7.00•103•5.00) = 1.70•10-5 M d. the pathlength through a 3.40•10-5 M solution of the complex that is needed for an absorbance that is the same as the solution in part a. b = A/(ε•c) = 0.595/(7.00•103•3.40•10-5) = 2.50 cm

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