Mid-Term Examination
IPS*1200
Principles of Modern Chemistry

March 1, 2000

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1. Aluminum reacts wioth bromine to form Al2Br6. If you have 54.0 g of Al, how many moles of Br2 are needed for complete reaction?

The relevant molar masses are

M(Al) = 26.982 g/mol

M(Br) = 79.904 g/mol

M(Al2Br6) = 533.39 g/mol

The moles of aluminum that are available are .

Since 2 moles of Al make up 1 mole of Al2Br6, we will require 6 moles of Br atoms for complete reaction. However, since the Br comes as a dimer molecule, we will need only 3 moles of Br2.

2. The isotopic abundance of 37Cl in naturally occuring chlorine is 24.6 mol% (meaning 24.6% of the number of atoms are of this isotope. This is in contrast to a weight percent, for instance.) How many 37Cl atoms are in a 100.0 g sample of CFCl3?

The molar mass of fluorotrichloromethane is . The moles of CFCl3 present is therefore . Since there are three atoms of chlorine per molecule, we must have . We know that 24.6% of all of these chlorine atoms are of the Z=37 isotopic variety, meaning . With Avogadro's number, we can convert this to the number of atoms:.

3. The solid fuel in the booster stage of the space shuttle is a mixutre of ammonium perchlorate and aluminum powder. Upon ignition, the unbalanced reaction that occurs is

Balance this reaction.

There are, of course, many ways to approach this problem. The first thing I saw was that with 1 mole of nitrogen gas, we would required two moles of ammonium perchlorate. This would then force there to be 2 moles of HCl to balance the chlorine.

I next looked to balance the hydrogens. There are 8 on the LHS. We need 3 H2O (along with the 2 in HCl) to balance.

Next I balance the oxygen. There are also 8 on the LHS. The are 3 oxygens in the water. So we need to get 5 from the only other source, Al2O3. We can only do that if we have 5/3 for the stoichiometric coefficient. We the only need to balance the aluminum. With a stiochiometry of 5/3, and 2 Al atoms per molecule, we need 10/3 moles of Al. We have

We make these integers by multiplying through my 3 and obtain the final answer for the balanced equation:

4. A reaction chamber with a volume of 1.00 contains a mixture consisting of 0.500 mol of N2(g) and 0.800 mol H2 (g). When equilibrium is achieved, the concentration of NH3 (g) is 0.150 M. Calculate the concentration equilibrium constant Kc for the reaction .

Set-up a reaction equilibrium table. Because the reaction vessel's size is 1 litre, the moles is also the concentration in molarity (moles per litre).
N2
3 H2
2 NH3

initial

0.500

0.800

0.0

change

-x

-3 x

+2 x

final

0.500-x

0.800-3x

0.150

From the ammonia, we can clearly see that x = 0.150/2 = 0.075 M. We can readily solve for the final concentration of the other two components [N2] = 0.500 - 0.075 = 0.425 M and [H2] = 0.800 - 3 x 0.075 = 0.575 M.

The concentration equilibrium constant is defined as . This is the value of Kc.

5. If morphine C17H19O3N is present in the blood at micromolar concentrations, it responds to the buffered blood at pH = 7.40 by partitioning itself between its basic (M) and acidic (MH+) forms. What fraction of the morphine is in the acidic form in the blood?

We look up the pKb of morphine to be 5.79. Its pKa is therefore 14 -5.79 = 8.21. The Henderson-Hasselbalch equations is applied here and we solve for the ratio of the two components.

This is the ration of base form to acid form. Since the solution pH (=7.40) is to the acid side of the pKa (=8.21), it makes sense that there is more acid form than base form present. To find the fraction in the acid form, we procced as follows:

The morphine is 86.6% in its acidic form.

6. What is the pH of a solution which is 1.00 M in chloroacetic acid?

Look up the pKa and find it to be 2.87. The Ka is therefore . We first try the square-root approximation for the pH.

We need to check that this is a valid approximation and do so by checking to see that the fractino dissociated is less than 5%.

Since it is less than 5% dissociated the approximation is valid and the pH is found to be 1.43.

7. What is the pH of a solution which is 0.000100 M in acetic acid?

The pKa is 4.74 so the Ka is .We try the square root approximation again.

Check for validity.

This is way over the 5% guideline, so we must go back and use the quadratic approximation. The equation that leads to this is

This concentration is still a factor of 3000 greater than 10-7 so ignoring the autodissociation of water is a valid approximation and this is the correct answer.

8. A volume of 37.7 mL of a 0.148 M solution of Ba(OH)2 was titrated with a dilute solution of HCl of an unknown concenttation. A volume of 30.2 mL of the HCl solution was required to reach the endpoint of the titration. What was the concentration of the HCl in M?

The moles of Ba(OH)2 present is n(Ba(OH)2) = 0.0377 L x 0.148 M = 5.58 x 10-3 mol. But since the hydroxide concentration is twice the formal concentration of Ba(OH)2, the amount of hydroxide present is n(OH- ) = 2 x 5.58 x 10-3 mol = 1.12 x 10-2 mol. What volume of the HCl is necessary to deliver the correct number of moles of H3O+ to neutralize this solution?

9. An aqueous solution is formed with 0.500 mol of TRIS-HCl (the hydrochloride salt of TRIS -tris(hydroxymehtyl)aminomethane). What volume of 0.250 M NaOH must be added to form a buffer solution fixed at a pH of 7.50?

We apply the Henderson-Hasselbalch equation again. We find from the table that the pKb is 5.92 so we can calculate the pKa to be 14 - 5.92 = 8.08. We have

Each molecule of OH- added will turn one BH+ into a B. The amount of added base will equal the amount of the basic form. The unreacted portion of BH+ remains in the acidic form. We can write

416 mL of NaOH is needed to form the solution at the appropriate buffer pH.

10. Nicotine is a diprotic base. A 25.00 mL sample of the fully basic form (concentration is 0.125 M) is titrated with 0.150 M HCl. Calculate the expected solution pH at the two equivalence points.

The first equivalence point is determined by recognizing that it is just like the dissolution of the intermediate species. The two pKa's for nicotine are found from tables to be 14 - 5.98 = 8.02 and 14 - 10.88 = 3.12. Since these two are well-separated, the most approximate equation will likely be valid. The first equivalence point pH is therefore

The pH at the second equivalence point is determined by recognizing that it is just like a weak acid solvation problem. Don't forget to account for the dilution due to added titrant. The square root approximation will probably be valid since the solution concentration will not be unusually small. To determine the dilution, we find the number of moles of nicotine titrated, recall that twice that amount of acid was needed to reach this second endpoint, and find the needed volume, given the concentration.

The formal concentration of NH+ would have been . The square root approximation can be used to find the resulting pH.

This is the pH at the second equivalence point.

Last update:1 March 2000
Comments to: Dan Thomas