CHEM 213 Chemical Analysis

Final

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2 __10___ (of 10)

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5 __10__ (of 10)

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7 __20___ (of 20)

8 __10___ (of 10)

Σ_100_ (of 100)

100%

KEY Name:___________________________________________

(please print)

1. In the Bradford protein determination, the color of a dye changes from brown to blue when it binds to protein. The intensity of blue color (measured by absorbance of light at a wavelength of 595 nm) is proportional to protein concentration.

Protein (µg):

0.00 9.36 18.72 28.08 37.44 Absorbance at 595 nm:

0.466 0.676 0.883 1.086 1.280 a. Determine the equation of the least-squares straight line through these points

in the form y = [m(±sm)]x + [b(±sb)] (15 points) xi yi xiyi xi

2 yi2

0.00 0.466 0 0 0.217156 9.36 0.676 6.327 87.61 0.456976 18.72 0.883 16.530 350.44 0.779689 28.08 1.086 30.495 788.49 1.179396 37.44 1.280 47.923 1401.75 1.6384 Σ 93.60 4.391 101.275 2628.29 4.271617

x = 93.60/5 = 18.72, y = 4.391/5 = 0.8782 Sxx = Σxi

2 – (Σxi)2/N = 2628.29 – 8760.96/5 = 876.098 Syy = Σyi

2 – (Σyi)2/N = 4.271617 – 19.280881/5 = 0.4154408 Sxy = Σxiyi – (Σxi)(Σyi)/N = 101.275 – (93.60)(4.391)/5 = 19.07548 m = Sxy/Sxx = 0.0217732 b = y – mx = 0.8782 – (0.0217732)(18.72) = 0.4706052

sr =Syy - m

2Sxx= 0.4154408 - (0.0004741)(876.098)

5 - 2= 0.0059738N-2

sm = (0.0059738)2sr2

Sxx 876.098= = 0.0002

sb = 0.0059736srΣxi

2

NΣxi2 - (Σxi)

2 =2628.29

(5)(2628.29) - 8760.96= 0.0046271

Equation of best line:

y = [0.218(±0.0002)]x + 0.471(±0.005) b. An unknown protein sample gave an absorbance of 0.973. Calculate the

number of micrograms of protein in the unknown and estimate its uncertainty. (5 points)

x = (y – b)/m = (0.973 – 0.471)/0.0218 = 23.0 µg If you keep more digits for m and b, x = 23.07 µg. Uncertainty in x:

2

sc =srm

1 1 (yc - y)2

m2SxxM N+ + 0.00059738

0.02177321 1 (0.973 - 0.8782)2

(0.0217732)2876.0981 5+ +=

= 0.30 µg The final answer is 23.1 ± 0.30 µg. 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. (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. Calculate pCu2+ (to the 0.01 decimal point) at each of the following points in the titration of 50.0 mL of 0.040 M EDTA with 0.080 M Cu(NO3)2 at pH 5.00. Remember to use the appropriate α4 value from the table below. (KCuY2- = 6.31•1018)

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. 0.1 mL (4 points)

Designating the total concentration of free EDTA as [EDTA] we can write: Kf’ = [CuY2-]/([Cu2+][EDTA] = αY4-•Kf = (3.7•10-7)( 6.31•1018) = 2.3 •1012

[EDTA] = (0.040)(50.0/50.1) = 0.03998 M [CuY2-] = (0.1/50.1)(0.080) = 1.60•10-4 M [Cu2+] = [CuY2-]/(Kf’[EDTA]) = (1.60•10-4)/(2.3•1012•0.0398) = 1.74•10-15

pCu2+ = 14.76

b. 25.0 mL (4 points)

Formal concentration of CuY2- = (25.0/75.0)(0.080)= 0.0267 M Cu2+ + EDTA ↔ CuY2- Initial 0 0 0.0267 Final x x 0.0267 - x

(0.0267 – x)/x2 = 2.3•1012

→ [Cu2+] = 1.08•10-7

→ pCu2+ = 6.97

c. 26.0 mL (2 points)

[Cu2+] is equal to excess amount of titrant solution added: [Cu2+] = (1.0/76.0)(0.080) = 1.05•10-3 M → pCu2+ = 2.98

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4. What volume of 2.00 M NaOH must be added to 300 mL of 1.00 M glycolic acid (Ka = 1.47•10-4) to produce a buffer solution having a pH of 4.00?

pH = 4.00 and [H3O+] = 10-4.00 = 1•10-4

[G-] = glycolate ion, /[GH] = glycolic acid Ka(glycolic acid) = 1.47•10-4

Ka = [H3O+][G-]/[GH] or 1.47•10-4 = 1.00•10-4 [G-]/[GH]

[G-]/[GH] = 1.47

Let VNaOH = ml 2.00 M NaOH added.

amount G- formed = amount NaOH added = VNaOH mL•2.00 mmol/mL amount GH remaining = original amount GH – amount VNaOH added

= (300 mL•1.0 mmol/mL) – (VNaOH mL•2.00 mmol/mL)

Total volume of solution = (300 + VNaOH) mL

cG- = 2.00•VNaOH/(300 + VNaOH) ≈ [G-] cGH = (300 – 2.00•VNaOH)/ (300 + VNaOH) ≈ [GH]

Substituting:

[G-] 2.00•VNaOH/(300 + VNaOH) = = 1.47 [GH] (300 – 2.00•VNaOH)/ (300 + VNaOH)

2.00•VNaOH = (300 – 2.00•VNaOH)•1.47

VNaOH = 89.27 mL

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5. How would you prepare 1.00 L of a buffer with pH of 7.00 from 0.300 H3PO4 and 0.160 M NaOH? Ka1(H3PO4) = 7.11•10-3, Ka2(H3PO4) = 6.32•10-8, Ka3(H3PO4) = 4.5•10-13 (10 points)

pH = 7.00 [H3O+] = antilog (-7.00) = 1.00•10-7

[H3O+][HPO4

2-]/[H2PO4-] = 6.32•10-8

[HPO4

2-]/[H2PO4-] = 6.32•10-8/1.00•10-7 = 0.632 (1)

Let VH3PO4 and VNaOH be the volume in milliliters of the two reagents. Then: VH3PO4 + VNaOH = 1000 mL (2) From mass-balance consideration, we may write that in the 1000 mL, no. mmol NaH2PO4 + no. mmol Na2HPO4 = 0.200•VH3PO4 (3) no. mmol NaH2PO4 + 2•no. mmol Na2HPO4 = 0.160•VNaOH (4) Equation (1) can be written as:

no. mmol Na2HPO4/1000no. mmol NaH2PO4/1000

no. mmol Na2HPO4no. mmol NaH2PO4

= = 0.632 (5)

Thus we have four equations: (2), (3), (4), and (5) and four unknowns: VH3PO4, VNaOH, no. mmol NaH2PO4 and no. mmol of Na2HPO4. Subtracting equation (3) from (4) yields: no. mmol Na2HPO4 = 0.160 VNaOH – 0.200 VH3PO4 (6) Substituting equation (6) into (3) gives no. mmol NaH2PO4 + 0.160 VNaOH – 0.200 VH3PO4 = 0.200•VH3PO4no. mmol NaH2PO4 = - 0.160 VNaOH + 0.400 VH3PO4 (7) Substituting equations (6) and (7) into (5) gives (0.160 VNaOH – 0.200 VH3PO4)/(0.400 VH3PO4 - 0.160 VNaOH) = 0.632 This equation rearranges to: 0.2611 VNaOH = 0.4528 VH3PO4 Substituting equation (2) gives 0.2611(100 - VH3PO4) = 0.4528 VH3PO4 VH3PO4 = 261.1/0.7139 = 366 mL VNaOH = 100 – 366 = 634 mL

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6. Calculate the electrode potential for the titration of 50.00 mL of 0.1000 M V2+ with 0.05000 M Sn4+ at the volumes of titrant given below. The relevant half reactions are:

V3+ + e- → V2+ Eº = -0.256 Sn4+ + 2e- → Sn2+ Eº = 0.154

a. At 10.00 mL (3 points)

b. At 50 mL (4 points)

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7. 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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8. a. Generate the equilibrium constant expression for the following reaction. Calculate the numerical value for Keq. (5 points)

2V(OH)4+ + U4+ ↔ 2VO2+ + UO2

2+ + 4H2O The relevant half-cell reactions are:

V(OH)4+ + 2H+ + e- ↔ VO2+ + 3 H2O Eº = +1.00 V

UO22+ + 4H+ + 2e- ↔ U4+ + 2H2O Eº = +0.334 V

b. Calculate the electrode potential of the above system at the equivalence points. (5 points)

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