Chemistry 3600
Second Examination
November 4, 1998

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 Mn2+ 600 0.870 0.749 0.675 0.485 0.405 OH- 350 0.964 0.926 0.900 0.810 0.760 ClO4- 350 0.867 0.740 0.660 0.445 0.355 K+ 300 0.964 0.925 0.899 0.805 0.755

1. Chemical Activity

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

Smaller ions have a higher charge density which attracts them more strongly to the water molecules.

• How do chemical activity effects affect the molar solubility of ionic substances?

In general, increasing the ionic strength of solution decreases the activity coefficients, increasing the molar solubility.

• How does the presence of salt affect the pH of a solution? Give the equations and describe the results.

The short answer is that the pH=-log10AH+ will increase because the salt ions lower chemical activity. The long (and correct) answer is that the pH may increase or decrease. This is because the activity coefficient of OH- is always less than that of H+. Since KW=AOH-AH+ ; pKW=pH-log10[OH-]g OH

b. Using activities, calculate the pH of a 0.01 F KClO4 solution saturated with Mn(OH)2. Mn(OH)2 has Ksp=1.6´ 10-13

The ionic strength of solution is dominated by the 0.01 F KClO4

The Mn(OH)2 equilibrium is*

 Mn(OH)2 (s) Û Mn2+ (aq) 2 OH- (aq) Initial solid 0 0 Change -x + x + 2x Final solid x 2x

Using the table or calculating with the Debye-Huckle equation,

Solving for x

To obtain the pH, we first get [OH-], then obtain the pH from Kw

2. Acid-Base Equilibria

a. Which of the following would be most suitable for preparing a buffer of pH 2.5? Show your work and briefly state why you chose that species.

1) H2C2O4 (Ka1=5.60´ 10-2 Ka2=5.42´ 10-5) 2) H3PO4 (Ka1=7.11´ 10-3 Ka2=6.32´ 10-8)

3) B(OH)3 (Ka1=5.81´ 10-10 Ka2=1.82´ 10-13) 4) H2CO3 (Ka1=4.45´ 10-7 Ka2=1.69´ 10-9)

The best acid to use is the one with a pKa» pH. Of the choices, H3PO4, with pKa1=2.15, is the closest to the target value of pH=2.5

b. A buffer was prepared by addition of 0.050 mole NaOH to 500 mL of a 0.10 F acetic acid (Ka=1.75´ 10-5) solution and then diluted to 1 L. Find the pH of the resulting buffer.

It turns out that this is not a buffer solution (due to a typographic error in the examination) Nonetheless, we can calculate the resulting pH. First, 50 mmole of the strong base (OH-) reacts with the 500 mL´ 0.1 F HOAc=50 mmole HOAc producing 50 mmole OAc-. The molar concentration of the acetate is 0.05 M. Acetate is a base that extracts a proton from water via

We can use the table method to determine the governing equation

 OAc- + H2O Û HOAc OH- Initial .05 0 0 Change -x + x + x Final 0.05 - x x x

Now we solve the chemical equilibrium equation

Assuming that x<<0.05

And the pH is

A diprotic acid has pKa1=4.00 and pKa2=7.00. In this problem, 10 mL of a 0.1 F solution of this acid is titrated with 0.1 F NaOH titrant. The resulting titration curve is shown below.

a. 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. Label the Titrant Volume axis markers.

c. 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.

d. At what indicator pKa would you want your color-change indicator to have in order to titrate to the second equivalence point?

Indicator changes at pH~pKIn The second equivalence point pH is ~10, so a pKIn~10 would work.

4. Precipitation and EDTA Titrations

a. 50 mL of a solution containing Cl- and Br- are titrated with 0.01 F AgNO3 solution. The two equivalence points, found using an electrochemical apparatus, occur at 12.5 mL and 18.0 mL total volume of titrant added. What are the molar concentrations of Cl- and Br- in the unknown solution? The silver-halide solubility products are 1.8´ 10-10 for AgCl and 5.0´ 10-13 for AgBr.

Based on the Ksp, AgBr is the first to precipitate (first equivalence point) and AgCl will precipitate second. The silver-halide reaction is

The 1:1 ratio allows us to calculate halide concentrations based on the amount of silver required to reach the equivalence points

b. 0.010 M Fe3+ is titrated with 0.010 F EDTA in a pH 7.00 buffered solution. What is the concentration of iron at the equivalence point? For Fe(III)-EDTA, log Kf = 14.32 and at pH 7, a Y4- =5.0´ 10-4.

The initial concentration of the Fe(III)-EDTA complex will be 0.01 M / 2 = 0.005 M due to the dilution that occurs with titration. The Fe(III) concentration is found from

 FeY- Û Fe3+ Y4- Initial .005 0 0 Change -x + x + x Final 0.005 - x x x

The EDTA grabs protons and is distributed as the many acid forms. To account for this, we use the fraction, a , distributed as Y4- in the equilibrium equation.

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