Fall 2015 Exam II * practice exam complete by 5-5-16 by 9:00pm please 1. A 32.54

Question

Fall 2015

Exam II * practice exam complete by 5-5-16 by 9:00pm please

1. A 32.54 mL aliquot of 0.7698 M ethylenediamine dichloride (ClH3NCH2 CH2NH3NCl) is titrated with 0.9999 M LiOH.

a. What are the titration reactions to form the fully deprotonated sample? What are the volumes needed for each reaction? (20 pts.)

b. What is the pH when 35.07 mL of titrant has been added? (20 pts.) Name:_____________________ TA:_________________________

c. What is the pH when 50.10 mL of titrant has been added? (20 pts.)

2. A mass of sodium pentanoate (CH3CH2 CH2 CH2CO2Na · H2O, FW 124.1135) was dissolved in 50.00 mL of water to make the solution pH = 8.8525. How much was dissolved? (20 pts.) Name:_____________________ TA:_________________________

3. A 30.00 mL portion of 0.4435 M Sn(NO3)2 was titrated with 0.5432 M EDTA at pH = 10.

a. What is the titration reaction and equivalence volume? (10 pts.)

b. If Vadded = 0.5Ve what is the pSn2+? (20 pts.) Name:_____________________ TA:_________________________

c. What is the p[Titrant] at the equivalence volume? (20 pts.)

d. What is the pSn2+ when 26.00 mL of titrant has been added? (15 pts.) Name:_____________________ TA:_________________________

4. The following electrochemical cell was set up in the lab.

Li(s)Li+Tb3+Tb(s)

a. Write the Nernst equations for these ½ reactions. (4 pts.)

b. Write the full Nernst equation to find the overall cell potential and the net cell reaction. (11 pts.)

c. If the solution in the right hand side of the equation has 0.4500 M Tb(NO3)3 and there is 0.2250 M LiNO3 in the left, what is the cell potential? (20 pts.)

d. What is the equilibrium constant (K) for the galvanic cell reaction? (20 pts.)

Explanation / Answer

1. Titration

a. Reaction : NH3+CH2CH2NH3+ + 2OH- ----> NH2CH2CH2NH2 + 2H2O

Volume required to reach half equivalence point = 0.7698 M x 32.54 ml/0.9999 M = 25.05 ml of LiOH

Volume required to reach equivalence point = 2 x 25.05 = 50.10 ml

b. pH after 35.07 ml LiOH added

moles of ethylenediamineHCl = 0.7698 M x 32.54 ml = 25.02 mmol

moles of LiOH = 0.9999 M x 35.07 ml = 35.06 mmol

[B] formed = (35.06 - 25.02)mmol/67.61 ml = 0.15 M

[BH+] formed = 25.02 mmol/67.61 ml = 0.37 M

pH = pKa + log(base/acid)

     = 9.93 + log(0.15/0.37)

     = 9.54

c. pH after 50 ml LiOH added

moles of ethylenediamineHCl = 0.7698 M x 32.54 ml = 25.02 mmol

moles of LiOH = 0.9999 M x 50 ml = 50 mmol

[B] formed = (50 - 25.02)mmol/82.54 ml = 0.302 M

[BH+] formed = 25.02 mmol/82.54 ml = 0.303 M

pH = pKa + log(base/acid)

     = 9.93 + log(0.302/0.303)

     = 9.93

2. pH = 8.8525

pOH = 14 - pH = 5.1475

pOH = -log[OH-]

[OH-] = 7.12 x 10^-6 M

RCO2- + H2O <==> RCO2H + OH-

Kb = [RCO2H][OH-]/[RCO2-]

1.44 x 10^-5 = (7.12 x 10^-6)^2/[RCO2-]

[RCO2-] = 3.52 x 10^-6 M

moles of pentanoic acid = 3.52 x 10^-6 M x 50 ml = 1.76 x 10^-4 mmol

mass of pentanoic acid dissolved = 1.76 x 10^-4 x 124.1135 g/mol = 0.022 mg

3. at pH 10 alpha[Y4-] = 0.36

logKf for Sn2 = 18.3

Kf = 2 x 10^18

Kf' = Kf x alpha[Y4-] = 7.18 x 10^17

a. Reaction : Sn2+ + EDTA <==> SnY2-

Equivalence volume of titrant = 0.4435 M x 30 ml/0.5432 M = 24.49 ml

b. When EDTA = 0.5Ve

Volume of EDTA added = 12.245 ml

[Sn2+] = (0.4435 M x 30 ml)/2 x 42.245 ml = 0.16 M

pSn2+ = -log(0.16) = 0.80

c. p[Titrant] at equivalence point

[Titrant] = 0.5432 M x 24.49 ml/54.49 ml = 0.244 M

p[Titrant] = -log[titrant] = 0.61

d. after 26 ml of EDTA added

excess [EDTA] = (26 - 24.49) ml x 0.5432 M/56 ml = 0.0146 M

[SnY2-] formed = 0.4435 M x 30 ml/56 ml = 0.24 M

Kf' = [SnY2-]/[Sn2+][EDTA]

7.18 x 10^17 = (0.24)/(0.0146)[Sn2+]

[Sn2+] = 2.29 x 10^-17 M

pSn2+] = 16.64

4. electrochemical cell

a. Oxidation : Li ---> Li+ + e-

Reduction : Tb3+ + 3e- ---> Tb

b. E = Eo - 0.0592/n log((Li+)^3/Tb3+)

3Li + Tb3+ ---> 3Li+ + Tb

c. Cell potential

E = (-2.28 - (-3.04)) - 0.0592/3 log(0.2250)^3/0.45)

   = 0.791 V

d. nFEo = RTlnK

3 x 96485 x 0.76 = 8.314 x 298 lnK

K = 3.64 x 10^38

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