Exam 2 - 90 points total

CHEM104

Spring 2002

$ Instructions

Complete 6 questions of your choice.

At least one problem must be chosen part A (discussion questions) and from Part B (quantitative problems).

Only 6 questions will be graded. All questions are worth 15 points. 

Part A: Discussion QUESTIONS

1. None of your group members must have gotten any sleep last night! They can't seem to figure out why salts and complexes of the metal ions sodium(+), aluminum(3+), Ti(4+), copper(+) and zinc(2+) are all colorless, i.e., white solids. And they can't come up with some reasons why Cr(3+) compounds can be found in almost every color of the rainbow. They need your help: can you explain it makes sense for these metals compounds to have these colors?

 First, check out teh electron configuration: sodium(+), aluminum(3+), Ti(4+) all have an inert gas configuration and so have 0 d electrons. Copper(+) and zinc(2+) both have a d10 configuration.

What is needed to make a metal compound colored are partially filled d-orbitals. In these cases an electron can absorb a small amount of energy (which corresponds to the visible, i.e., colored region of the electromagnetic spectrum) and "promote" or jump and electron from a lower level d- orbital to a higher level d-orbital. Therefore the d0 and the d10 metals ions above are white because their electron configuration do not allow for such an electron promotion event.

Cr(3+) makes colored compounds because it has a partially filled set of d- orbitals: d3. The different colors result when this ion bonds to different types of ligands that have different lignad field stengths, delta o. for an octahedral geometry.

2. Your grandmother loves to drink tea! But after 60 years, her teapot is completely coated on the inside with a white scum. You the excellent chemistry student can help! Explain to her that the scum is called scale, what the scale is and how by adding vinegar the scale will be removed from her teapot.

(With gratitude, she will make you a nice chocolate cake to eat with some tea.)

 The scum--or scale-- is calcuim carbonate, CaCO3. Removing the scale requires dissolving the the solid CaCO3: CaCO3(s) <==> Ca(2+) + CO3 2-

Since the carbonate formed is the conjugate base of an acid, addition of an acid will protonate the carbonate and remove it from the dissolution equilibrium:

CO3 2- + H+ --> HCO3-

This pushes the equilibrium for CaCO3 dissolution forward and helps the scale dissolve for removal.

 

3. You're helping a classmate finish her lab report and all she has to do is label the titration plot.

But she cannot remember if the titration was sodium phosphate to which was added hydrochloric acid or phosphoric acid to which was added sodium hydroxide.

a) Point out all the aspects of the titration plot (found at end of exam) that identify which titration reaction was done.

The titration was NaOh added to phosphoric acid. You can determine this because the pH begins acidic and because there are two equivalence points, meaning the acid si at least dibasic.

 

The species in the first buffer region are H3PO4/ H2PO4-.

The species in the second buffer region are H2PO4 - / HPO4 2-.

The species in the third buffer region are HPO4 2-/ PO4 3-.

b) Not done yet: she must still identify the species present in the buffer regions of the titration plot and choose a good indicator to be used from the list below.

indicator pKa

thymol blue 1.7

alizarin red 11.7

phenolphthalein 9.4

methyl red 5.0

 

Good thing she has you for a friend! Help her out then you can both go sit in the sun.

 Methyl red would be optimal for th efirst endpoint, phenolphthalein for the 2nd endpoint. Although alizarin red falls approximately at teh third endpoint, becaus eit is barely visible in the plot, the color change will be exceedingly gradual and impossible to see.

4. If you can imagine it, you might be able to make it.

Can you imagine how to invent a device that changes color when it become magnetic?

In other words, can you imagine under what conditions coordination complexes could change color as well as also change their magnetic properties from diamagnetic to paramagnetic? Use any metal, any geometry, and any ligands you wish in your invention. To convince me–the fund-raiser for your project–also give specific examples of colors which may pair with the diamagnetic and paramagnetic behaviors.

It is the ligand field strength that determines both the color and the magnetic properties. Therefore it should be possible to change both simultaneously.

 

For example, Fe(2+) has 6 d electrons.

If the ligand is the very strong field strength CN, making Fe(CN)6 4-would be a diamagnetic complex. Since the delta o is expected to be large, it would absorp in the short wavelength violet-blue and reflect yellow light.

 

If the ligand were H2O which has a weak ligand field, the 6 d electrons would not be all paired and a paramagnetic complex with 2 paired and 4 unparied electrons. The small delta o menas that long wavelength, or red light, woudl be absorbed and greenish blue light reflected.

 

5. Your mother doesn't believe it: "How could hyper-ventilating* change the pH of my blood?"

You explain it to her; she's your mother.

**hyper-ventilation: excessively rapid and deep breathing, resulting in the decrease of carbon dioxide in the blood.

 The importance of CO2 is that it is the basis of the buffer system in the blood, and this buffer system is required to keep a steady pH. Since hyperventilation decreases the amount of the CO2 products in the blood, H2CO3 and HCO3 -, addition of acid or base into the bloodstream may not be neutralized and teh pH can change. Note also , the H2CO3 formed from CO2 + H2O <==> H2CO3 contributes acidity to the pH and this will be altered with less CO2 dissolved in the blood.

 

 Part B

PROBLEMS:

1. The pale green precipitate you may find around the faucets in your bathroom is Cr(OH)3.

Can 1.0 mg (0.001 g) of Cr(OH)3 be completely dissolved in 100 mL water?

Ksp for Cr(OH)3 is 6 x 10-31

 Click here for Problem 1 answer

2. 4.00 g sodium phosphate, Na3(PO4), is dissolved into 200 mL of water. To this solution is added 50 mL of a 0.300 M HCl solution. What it the pH of the final solution?

[For phosphoric acid, pKa1 = 2.12; pKa2 = 7.21; pKa3 = 12.32]

 Click here for Problem 2 answer

3. Certain foods, like spinach and rhubarb, contain a high concentration of oxalic acid. When the concentration of oxalate (C2O4 2- ) in your body is too high, Ca(oxalate) can precipitate. This precipitate has another name: kidney stones!

a) Draw the structure showing how oxalate binds to Ca(2+).

b) What concentration of Ca2+ will cause the formation of kidney stones at a buffered physiological pH (7.4) when 0.001 g H2C2O4 is dissolved in 10 mL of water?

For oxalic acid, pKa1 = 1.19; pKa2 = 4.21.

For Ca(oxalate), Ksp = 2.6 x 10-9.

Click here for Problem 3 answerpage 1 page2, page3

4. a) Write a balanced equation that describes the reaction of the compound cobalt trichloride with excess ammonia to make a coordination complex having the following properties:

- the coordination number is six in a common geometry

- it diamagnetic

- it makes four ions when dissolved in water.

b) Draw the crystal field diagram for this complex and label all the orbitals.

c) Draw the orbitals used in the CFT diagram.

d) Suggest a reaction that would form a paramagnetic complex starting from cobalt trichloride.

 Click here for Problem 4 answer

5. Mother Nature is so clever! A buffer was needed the maintain constant pH for her mammalian life-forms and she made use of one of the most abundant gases that blankets the earth: CO2.

a. Write the equilibria reaction equations that are involved in converting the gas to a useful buffer for the plasma of mammals.

b. Identify the species that make up the buffer system in mammalian blood.

c. If normal blood pH is 7.43, what is the ratio of buffer species, A- / HA, ?

d. If the blood pH falls to the dangerous (possibly fatal) level of pH 7.0, what is the ratio of buffer species, A- / HA, ?

e. If there are 5 L of blood in a mammal and at pH 7.43 the concentration of buffer component HA = 3.72 x 10-6 M, how much acid or base was added to the blood to make the pH fall from 7.43 to 7.0?

For HA: Ka1 = 4.2 x 10-7. Note that Ka2 is not needed in this problem.

Click here for Problem 5 answer