Name:     PROFESSOR’S KEY    

 

Chemistry 3600

Second Examination

November 13, 2002

 

Directions: There are 4 questions, each worth equal points.  Some are more difficult than others.  Read over all questions before you start to answer.  Answering the harder questions last will insure that you obtain the maximum score.  You must show your work and pertinent formula's to get full credit.  Final numerical results are not as important as the methods used to find them.

 

Useful formula:

 

 

Solvated ion radii (pm) and activity coefficients for selected ions

 

Ion

Radius

Ionic Strength

0.001

0.005

0.010

0.050

0.100

H+

900

0.967

0.933

0.914

0.860

0.830

Hg22+

400

0.867

0.742

0.665

0.455

0.370

OH-

350

0.964

0.926

0.900

0.810

0.760

NO3-

300

0.964

0.925

0.899

0.805

0.755

Cl-

300

0.964

0.925

0.899

0.805

0.755

K+

300

0.964

0.925

0.899

0.805

0.755

1. Chemical Activity

(a)   Answer the following questions:

-          Why is the hydrated ion radius most often larger for ions with smaller ionic radii?

 

The higher the charge density ions hold water molecules strongly. They thus have larger solvated radii.

 

-          How do chemical activity effects affect pH?

 

pH is defined as proton chemical activity. pH=-log10 AH+

 

(b) Using activities, calculate the concentration of mercurous ion, Hg22+, in a saturated calomel (Hg2Cl2) solution that has 0.100 F KNO3. The Ksp for Hg2Cl2 is 1.2x10-18.

 

KNO3 is an inert electrolyte that, at 0.1 F, produces a solution with an ionic strength of

From the Table of activity coefficients,

We calculate the effective solubility product constant

For x mole of Hg2Cl2 dissolved we get x mole Hg22+ and 2x mole Cl-

x is the molar concentration of the Hg22+ ion.

2. Systematic Approach to Solving Chemical Equilibrium

Using the systematic approach to solving chemical equilibrium problems, find the pH of a solution where 1.0´10-8 mole of Mg(OH)2 is dissolved in water.

(a) First, set up the problem by giving the following:

1 - Pertinent chemical reactions

 

 

2 - Charge balance

 

3 - Mass balance

 

4 - Chemical equilibrium expressions

 

5 - Count unknowns and equations. Are there enough equations to solve this problem?

 

The 3 unknows are Mg2+, H3O+, and OH-. There are 3 equations so the problem is solvable.

 

(b) What is the pH of this solution?

 

3. Acid-Base Titrations/Advanced Acid-Base Chemistry

The amino acid, methionine, is diprotic with a carboxylic acid pKa1=2.20 and an ammonium pKa2=9.05. In this problem, 10 mL of a 0.1 F solution of this amino acid is titrated with 0.1 F NaOH titrant.

(a)   Below is the resulting titration curve.

Indicate the two points where the solution has the maximum buffer capacity by A and C, and indicate the first and second equivalence points as B and D.

(b)   Calculate the volume to first and second equivalence point and label the Titrant Volume axis markers with volumes.

 

(c)   There are simple expressions for the pH at points A, B, and C.  Give the pH's at these points?  Give the equations below and put numbers on the pH axis on the above plot.

 

 

(d) At what pH would you want your indicator color change to occur at in order to titrate to the first equivalence point?

 

The color change should be from pH 5 to 6

4. EDTA Titrations

(a) An EDTA titer value was determined using an magnesium sulfate standard. A 50 mL aliquot of solution containing 0.450 g of MgSO4 (FW 120.37) in 500 mL required 37.6 mL of an EDTA solution to reach the end point. How many milligrams of CaCO3 (FW 100.09) will react with 1 mL of this EDTA solution?

 

 

(b) 0.001 M Pb2+ is titrated with 0.010 F EDTA in a pH 4.00 buffered solution. What is the concentration of lead at the equivalence point?  For Pb(II)-EDTA, log Kf = 18.04 and at pH 4,  aY4- =3.8´10-9.

 

The total volume will be; 1 for the Pb2+, plus 0.1 for the EDTA solution or 1.1.

 

 

Pb2+

Y4-

Pb(II)-EDTA

Initial

0

0

0.001/(1.1) M

Change

+x

+x

-x

Final

x

x

.00091-x