Chemistry 360
Second Examination
November 19, 1997
Directions: There are 5 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 |
OH- |
350 |
0.964 |
0.926 |
0.900 |
0.810 |
0.760 |
I- |
300 |
0.964 |
0.925 |
0.899 |
0.805 |
0.755 |
NO3- |
300 |
0.964 |
0.925 |
0.899 |
0.805 |
0.755 |
K+ |
300 |
0.964 |
0.925 |
0.899 |
0.805 |
0.755 |
Hg22+ |
400 |
0.867 |
0.740 |
0.660 |
0.445 |
0.355 |
1. Precipitation Titrations/Chemical Activity
(a) What volume (mL) of 0.100 F potassium iodide, KI, titrant is needed to reach the equivalence point for 25.0 mL of a solution of 0.050 F Hg2(NO3)2 if the reaction is Hg22+ + 2I- ® Hg2I2(s)?
(b) Using activities, calculate the molar
solubility of Hg2I2 (Ksp=4.5´ 10-29)
in 0.050 F KNO3
The
solubility for Hg2I2 is found from
We need the activity coefficients for Hg22+ and I-. The ionic strength is due to the soluble 0.050 F KNO3 solution.
At this ionic strength, the table values for the activity coefficients are
The molar solubility is found by considering the change in concentrations starting with the solid
Hg2I2 |
Hg22+ |
I- |
|
Initial |
(s) |
0 |
0 |
Change |
-x |
+x |
+2x |
Final |
(s) |
x |
2x |
From the Ksp equation
Since x=[ Hg22+], x is the molar solubility sought.
2. Sytematic Chemical Equilibrium
(a) Give the chemical equilibria, charge balance, mass balance, and all used to solve for the solubility of CaC2O4. C2O42- is a dibasic ion.
chemical equilibrium equations:
charge balance:
mass balance:
mathematical equilibrium equations:
(b) Give the number of equations and the number of unknowns. Are there enough equations to solve for all the unknowns?
6 Equations, 6 unknowns. Problem has a solution.
3. Acid-Base Equilibria
(a) Which of the following bases would be most suitable for preparing a buffer of pH 4.60? Show your work and briefly state why you chose that base.
| 1) NH3 (Kb=1.75´ 10-5) | 2) C6H4NH2 (Kb=3.99´ 10-10) |
| 3) H2NNH2 (Kb=3.0´ 10-6) | 4) C5H5N (Kb=1.69´ 10-9) |
A maximum buffer capacity occurs for equal base and conjugate acid concentrations. For the various bases, we find, 1) pKb=4.7; 2) pKb=9.4; 3) pKb=5.5; 4) pKb=8.8 Since 14-4.60 (the target pH) is 9.4, we would choose base 2.
(b) A buffer was prepared by 0.100 mole of a weak monoprotic acid (Ka=1.0´ 10-6) plus 0.050 mole of its conjugate base in 1 liter of solution. Find the pH of the resulting buffer.
(c) Calculate the pH of a solution that is 1.0´ 10-8 F H2SO4
Charge balance, mass balance, and equilibrium
Solve for proton concentration and pH
4. 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. Label the Titrant Volume axis markers.
(b) There are simple expressions for the pH at points A, B, and C. What are the pH's at these points? Give the equations below and put numbers on the pH axis on the above plot.
c) At what pH would you want your indicator color change to occur at in order to titrate to the first equivalence point?
Around the first equivalence point, the pH~6. This would be a good choice for the indicator pKa
5. EDTA Titrations
(a) A 50.00 mL sample containing Ni2+ was treated with 25.0 mL of 0.0500 M EDTA to complex all the Ni2+ and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring 5.00 mL of 0.0500 M Zn2+. What was the concentration of Ni2+ in the original solution?
(b) 0.0020 M Ca2+ is titrated with 0.010 F EDTA in a pH 7.00 buffered solution. What is the concentration of Ca2+ at the equivalence point? For Ca(II)-EDTA, log Kf = 10.69 and at pH 7, a Y4- =5.0´ 10-4.
First, 0.2 volumes of 0.010 F EDTA titrant are required for each volume of 0.0020 M Ca2+ solution at the equivalence point. EDTA reacts according to
At the equivalence point, the initial concentration of CaY is
The complex is in equilibrium
CaY |
Ca2+ |
Y- |
|
Initial |
0.00167 |
0 |
0 |
Change |
-x |
+x |
+x |
Final |
0.00167-x |
x |
x |
Assuming x<<0.00167
This page was created by Professor Stephen Bialkowski, Utah State University
Last Updated Friday, October 03, 2003