Analytical Chemistry Mid Term Discussion

Name : RedID #

Show your work, answers alone will not receive partial credit. Where indicated, please enter your answer the box beneath the question. There are blank pages near the end of the exam to use as scratch paper. Answers without the units indicated (when necessary) will not be graded.

The last page of the exam contains a selection of data tables and equations for your use – feel free to remove the page from the exam.

Question

Value

Score

1

10

 

2

15

 

3

12

 

4

6

 

5

7

 

Total

50

 

By signing below I acknowledge that I will abide by the SDSU Academic Code of Honor as it pertains to academic integrity. For this exam I will not make use of any unauthorized materials, nor will I engage in any practices that otherwise fall within the broader definition of “cheating”.

_____________________________________________________________________________________

Signature Date

If this pledge is not signed the instructor reserves the right to not grade the exam and award a score of 0 for this exam in the calculation of the final grade.


  1. (10 points) A mixture is made of two weak bases. A 15.00 mL aliquot of the base mixture is analyzed by titration with a 0.140 M strong acid. The resulting titration data is plotted in the graph below. Use this graph to determine the answers to the questions in the table below.

word image 819

Numerical answers do not need to be perfectly accurate, but they should be reasonable.

 

Question

Answer

a

Is there more of the weaker base, or the stronger base, in the mixture?

 

b

Approximately how much titrant (in mL) was needed to titrate only the the weaker base?

 

c

What is the concentration of the stronger of the bases in the mixture?

 

d

What is the approximate pKa of the stronger base?

 

e

What is the approximate pKa of the weaker base?

 

 

Titrant volume added (mL)

Solution pH

(a)

0.00

 

(b)

1.98

 

(c)

5.67

 

(d)

10.00

 

(e)

13.77

 

(f)

18.00

 
  1. (15 points) A 25.00 mL aliquot of a solution containing 40.00 mM phosphoric acid is titrated with a 0.200 M LiOH titrant. Determine the pH of the solution during the course of the titration at the volumes indicated in the table to the right.


P

O

pKa

2.148

HOOH 7.198

12.375

OH

 

Titrant volume added (mL)

Cell Potential (V)

(a)

34.92

 

(b)

54.00

 

(c)

70.23

 

3) (12 points) A 30.00 mL solution containing 45.00 mM iodine and buffered to pH 8.94 is titrated with a solution of Pb4+. The concentration of lead(IV) in the titrant is 75.00 mM. The cell potential is monitored with a platinum wire cathode, and a S.C.E. (0.241 V) anode as the reference electrode. Determine what potential will be measured for each of the titration volumes listed in the table to the right.


Use the half reactions below to determine the cell potentials.


Pb4+ + 2e ⇌ Pb2+ E° = 1.69 V


IO3– + 3H2O + 6e– ⇌ I– + 6OH– E° = 0.269 V

  1. (6 points) A 50.00 mL solution containing 5.00 mM each of Ag+ and Co3+ buffered at pH 9.00 is mixed with 50.00 mL of 12.50 mM EDTA. What will be the concentration of the metals, and EDTA that are not part of a complex?


See page 7 for EDTA constant tables


 

Species

Concentration

(a)

Ag+

 

(b)

Co3+

 

(c)

EDTA

 
  1. (7 points) A 20.00 mL solution containing 50.00 mM AgNO3 and 100.00 mM Pb(NO3)2 is titrated by precipitation with 250.0 mM KCl titrant. Determine the concentrations of the metals (Ag+ and Pb2+) free in solution at the given titration volumes.

 

Titration

volume (mL)

[Ag+]

[Pb2+]

(a)

4.00

  

(b)

22.00

  

TlL PbL2

3.74

4.78

1.8 # 10“4

1.7 # 10“5

Solubility Table pKsp Ksp

Chlorides: L ! Cl

CuL 6.73 1.9 # 10“7

AgL 9.74 1.8 # 10“10

Hg2L2 17.91 1.2 # 10“18

word image 820 word image 821

Order a unique copy of this paper
(550 words)

Approximate price: $22