Mid-Term Examination
IPS*1200
Principles of Modern Chemistry

March 2, 1999

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1. One gram of arsenic reacts with 2.366 g of Cl2. What is the empirical formula of the resulting compound?

There are 5 times as many chlorine atoms as there are arsenic atoms. Hence, the empirical formula is AsCl5. Don't let the fact that the chlorine comes as a dimer fool you. There are 2.366 g of chlorine atoms, whether they are arranged as Cl, Cl2, or anything else.

2. Diborane, B2H6, reacts violently with O2 to form B2O3. What is the maximum mass of B2O3, in grams, that can be formed from 13.8 g of B2H6 and 32.0 g of O2?

Determine the moles of boron atoms delivered by the diborane and the number of moles of oxygen atoms delivered by the O2.

To determine which reagent is limiting, consider the moles of product that could be formed from each. The 1 mole of boron atoms could produce 0.50 mol of B2O3; the 2 mol of O atoms could produce 2/3 of a mol of B2O3. Hence, the B is the limiting reagent and 0.5 mol of B2O3 will be produced. To find its mass

3. Sodium fluoride is often used as the source of fluoride in toothpaste. What is the pH of a 0.75 M solution of NaF?

The F- ion is not like the Cl- ion; it does hydrolyze; it does have basic properties. The HF molecule is its conjugate acid. From the accompanying tables we obtain its pKa as being 3.17. From this we can find the pKb of the F- ion. Then it is just a weak base concentration problem.

We can easily see that this is only slightly ionized so this approximation is O.K. It is getting close to water's autoionization contribution, but not too close to worry about yet.

4. At a certain temperature, two reactant gases, CO and H2, are introduced at the partial pressures of 0.46 atm and 1.35 atm respectively. These gases can react to form formaldehyde, H2CO, by the reaction

When the reaction reaches equilibrium, what are the partial pressures in atmospheres of all three gases?

An equilibrium table takes on the following form

CO +
H2
H2CO
0.46 atm
1.35 atm
0
-x
-x
+x
0.46 - x
1.35 - x
x

If we write out the equilibrium constant expression and then substitute with these unknown expressions, we can solve for x.

The only reasonable root is 0.11. Hence, the equilibrium pressure of H2CO is 0.11 atm.; for CO it is 0.46 - 0.11 = 0.35 atm; for H2 it is 1.35 -0.11 = 1.24 atm.

5. We mix the following three solutions together:

What is the pH of the resultant solution?

The total final volume is 15 + 40 + 50 = 105 mL. The added HCl will react with the NH3 to form more NH4+. The NH4Cl dissociates completely to give NH4+. In the end, we just have an ammonia buffer. Since the concentrations are identical, it is as if 50 + 15 = 65 mL of 1.00 M NH4Cl is mixed with 40 - 15 = 25 mL of 1.00 M NH3. The dilution of the solutions does occur but since both species are diluted equally, we can ignore dilution effects and just concentrate on amounts. Solve for these two amounts, substitute into the Henderson-Hasselbalch equation and find the resulting pH. Don't forget to find the correct pKa for the ammonium ion from the pKb of the ammonia.

6. When 0.30 mol of a certain weak acid is dissolved in 1.50 L of water, the acid is 28% dissociated. What is the pKa of the acid?

7. A 35.00 mL aliquot of 0.1500 M NaOH is titrated with 0.1000 M HNO3. What is the solution pH when 35.00 mL of titrant has been added?

Because the titrant is of a lower concentration than the analyte, the titration is not yet complete and the pH is still being controlled by the excess analyte remaining in solution.

Determine the remaining amount of base, find its diluted concentration, and determine the pOH of the solution.

8. A 25.00 mL aliquot of a 0.0750 M solution of 4-methylanilinium chloride, a monoprotic acid, is titrated with a 0.1000 M solution of KOH.4-methylaniline What is the equivalence volume of added titrant? What is the pH at the half-equivalence point? What is the pH at the equivalence point?

The titrant volume is determined when the amount of titrant equals the amount of analyte. We obtain

The pH at the half-equivalence point is just the pKa. We find this easily from the pKb of the base.
pH = pKa = 14.00 - pKb = 14.00 - 8.916 = 5.084

The pH at the equivalence point is just a calculation of the weak base concentration. Find its diluted concentration and do the square root approximation.
The Kb = 10(-8.916) = 1.21 x 10-9.
The total final volume is 0.0250 + 0.01875 = 0.04375 L.
The amount of base is just 0.0750 M x 0.0250 L = 0.001875 mol
The formal concentration of the base is, therefore, 0.001875 mol / 0.04375 L = 0.04286 M. The [OH-] concentration is just

9. A 30.00 mL aliquot of a 0.0600 M solution of morpholine is titrated with 0.1200 M HCl. What is the pH when 20.00 mL of titrant has been added?

 

The equivalence volume is 30.00 mL x (0.0600 M / 0.1200 M) = 15.00 mL. Hence, 20.00 mL of added titrant is beyond the equivalence point; 5.00 mL has been added in excess. The final volume if 20 + 30 = 50.00 mL. The 5.00 mL of 0.1200 M HCl controls the system pH. Its diluted concentration is just [H3O+] = 0.1200 M x (5.00 mL / 50.00 mL) = 0.01200 M. The pH is just 1.92.

10. Glycine is a diprotic aminoacid. A small amount is found in the blood stream which is buffered at pH = 7.40. What fraction of glycine is found in its intermediate form under these conditions?

 

Determine the Ka's.
Ka1=10(-2.350) = 4.47 x 10-3 Ka2=10(-9.778)=1.67 x 10-10

From the buffer conditions we obtain: [H3O+]=10(-7.40) = 4.0 x 10-8 M

We can rewrite the ionization constant equations

By definition, the fraction in the HA form can be written down, and the above relations substituted in. The equation can then be solved for the answer.

Last update:4 March 1999
Comments to: Dan Thomas