#3 and 4 Pbso4(s) 2e--, Pb(s) so42- E0- -0.350 diluted Zn2+ 2e- E0 0.763V condit

Question

#3 and 4

Pbso4(s) 2e--, Pb(s) so42- E0- -0.350 diluted Zn2+ 2e- E0 0.763V conditions and form Zn(s) certain reductor to 3. A 2.16 g sample containing both Fe and v was dissolved through a Walden Ce reach to 500.00 mL. A first 50.00 mL aliquot was taken and passed mL of form Fe and V reach M Fe and VO2 ions. The titration of this solution 16.84 reductor to to end point. A second 50.00 mL aliquot was passed through a Jones solution ions. The titration of the second solution required 44.26 of 0.1000 M Ce an end point. Calculate the percentage of Fe and Vin the sample. 5) solution 4. th 0.1000 M EDTA in a Derive a titration curve for 30.00 mL of 0.0500 M Co of 10.00, 15.00 and buffered to pH 10. Determine the pco values after the of EDTA. 2.82 x 1016. pts)

Explanation / Answer

3. percent Fe

moles Ce4+ used = moles of Fe present

                             = 0.1 x 16.84 ml = 1.684 mmol

In 500 ml = 1.684 x 500/50 = 16.84 mmol

mass Fe present = 16.84 x 55.845/1000 = 0.940 g

%Fe = 0.940 x 100/2.16 = 43.52%

percent V

moles Ce4+ used = (1/3)moles of V present

                             = 0.1 x 44.26 ml/3 = 1.475 mmol

In 500 ml = 1.475 x 500/50 = 14.75 mmol

mass V present = 14.75 x 50.94/1000 = 0.751 g

%V = 0.751 x 100/2.16 = 34.78%

4. Kf' = Kf x alpha[Y4-]

         = 2.82 x 10^16 x 0.36 = 1.01 x 10^16

a) at EDTA = 10 ml

moles Co2+ = 0.05 M x 30 ml = 1.5 mmol

moles EDTA = 0.1 M x 10 ml = 1 mmol

excess [Co2+] = 0.5 mmol/40 ml = 0.0125 M

pCo2+ = -log[Co2+] = -log(0.0125) = 1.90

b) at EDTA = 15 ml

moles Co2+ = 0.05 M x 30 ml = 1.5 mmol

moles EDTA = 0.1 M x 15 ml = 1.5 mmol

Equivalence point

[CoY2-] formed = 1.5 mmol/45 ml = 0.033 M

       Co2+ + EDTA <==> CoY2-

I        -            -                0.033

C     +x         +x                -x

E       x          x              0.033-x

So,

considering x being a small number,

Kf' = 1.01 x 10^16 = 0.033/x^2

x = [Co2+] = 1.81 x 10^-9 M

pCo2+ = -log[Co2+] = 8.74

c) at EDTA = 20 ml

moles Co2+ = 0.05 M x 30 ml = 1.5 mmol

moles EDTA = 0.1 M x 20 ml = 2 mmol

[CoY2-] formed = 1.5 mmol/50 ml = 0.03 M

excess [EDTA] = 0.5 mmol/50 ml = 0.01 M

Kf' = [CoY2-]/[Co2+][EDTA]

1.01 x 10^16 = 0.03/0.01[Co2+]

[Co2+] = 2.97 x 10^-16 M

pCo2+] = 15.53

pCo2+ = -log[Co2+] = -log(0.0125) = 1.90

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