1. (8 marks) Exactly 100 mL of 0.25 M NaOH was mixed with 160 mL of 0.20 M HCl.

(a) (4 marks) What is the molarity of the NaCl formed?

Answer. This is a limiting reagent problem.

moles NaOH = 0.025 moles

moles HCl = 0.032 moles

therefore NaOH is limiting and 0.025 moles of NaCl can be formed. 0.007 moles of HCl will be in excess.

Total volume is approximately the sum of the volumes of the reactants. Therefore,

molarity of NaCl = 0.025 mol NaCl /(0.100 L + 0.160 L) = 0.096 M NaCl
 
 

Answer. Since 0.007 moles of HCl is in excess, the [H⁺] concentration is

0.007 moles /(0.260 L)

= 2.69 x 10^-2 M (1 s.f. justified)

pH = - log [H⁺] = 1.6 (1 s.f.)

pOH = 14 - pH = 12.4 (1 s.f.)

2. (16 marks) Assume that the following reaction is at equilibrium:

How do the final concentrations compare with those of the original equilibrium, and in which direction does the reaction shift to reach the new equilibrium after the following changes are made?

(a) (4 marks) What happens if you remove some NO? Correct answer shown in bold below
 

(i) The NOCl concentration: increases decreases does not change

(ii) The NO concentration: increases decreases does not change

(iii) The Cl₂ concentration: increases decreases does not change

(iv) The equilibrium: shifts to the right does not change shifts to the left
 

(i) The NOCl concentration: increases decreases does not change

(ii) The NO concentration: increases decreases does not change

(iii) The Cl2 concentration: increases decreases does not change

(iv) The equilibrium: shifts to the right does not change shifts to the left

(c) (4 marks) What happens if the temperature is raised? Circle the correct answer below

 
(i) The NOCl concentration: increases decreases does not change

(ii) The NO concentration: increases decreases does not change

(iii) The Cl2 concentration: increases decreases does not change

(iv) The equilibrium: shifts to the right does not change shifts to the left

(d) (4 marks) What happens if you increase the pressure? Circle the correct answer below
 

(i) The NOCl concentration: increases decreases does not change

(ii) The NO concentration: increases decreases does not change

(iii) The Cl2 concentration: increases decreases does not change

(iv) The equilibrium: shifts to the right does not change shifts to the left
 

(a) (3 marks) Define the terms "Bronsted acid," "Bronsted base," and "conjugate acid-base pair."

Answer A Bronsted acid is a proton donor; a Bronsted base is a proton acceptor. A conjugate acid-base pair is two species that differ by a single proton.

In the reaction

H₂CO₃ is a Bronsted acid becauses it donates a proton to the Bronsted base H₂O to form the products. H₂O accepts the proton so it is a Bronsted base. Examples of conjugate acid-base pairs are H₂CO₃/HCO₃^- and H₃O⁺/H₂O.

(i) Based on the K_a values shown, arrange the four Bronsted acids in order of their acid strengths:

Answer

 
(ii) Arrange the four Bronsted bases in order of their base strengths

Answer

Answer

 

HCO2-	+ NH4+	--->	HCO2H	+ NH3
Base 1	Acid 1		Acid 2	Base 2
weaker base	weaker acid		stronger acid	stronger base

From parts (a) and (b): NH₃ is a stronger base than HCO₂^-; HCO₂H is a stronger acid than NH₄⁺. Therefore this reaction does not proceed to the right since the direction is stronger acid + stronger base ---> weaker acid + weaker base.

4. (14 marks)

(a) (3 marks) The following reaction has K_c = 13.2 at 700°C.

What is the equilibrium concentration of HI when [H₂] = 0.65 M and I₂ = 0.42 M?

Answer.

=1.9 to correct s.f.

(c) (6 marks) What is the pH of a solution that is 0.15 M in trimethylamine, (CH₃)₃N, and 0.20 M in trimethylammonium chloride, (CH₃)₃NHCl? The K_b for trimethylamine is 7.4 x 10^-5.

5. (12 marks) At the right is a plot of boiling point vs. period of the periodic table for the hydrides of Groups VA (nitrogen to antimony) and VIA (oxygen to tellurium) of the periodic table.     (a) (2 marks) On the diagram, sketch where you would expect the boiling points of the corresponding noble gases Ne, Ar, Kr, and Xe, to occur. What kind of intermolecular forces would be important between noble gas atoms?  Answer: Approximate boiling points of rare gases are shown in red. Only intermolecular forces are induced dipole -induced dipole (dispersion forces)

(b) (6 marks) Explain, in terms of intermolecular forces, the trend of boiling points from NH₃ to SbH₃.

Answer Ammonia has the highest boiling point in the series because the N-H covalent bond is quite polar and so the molecules undergo hydrogen-bonding to each other.

When ammonia boils, the H-bonds are broken and NH₃ molecules go into the gas phase.

PH₃, AsH₃, and SbH₃ boil lower than NH₃ because the electronegativities of P, As, and Sb are less than the electronegativity of N and the molecules are less polar than NH₃. Thus the dipolar interactions are not as strong as the H-bonds in liquid ammonia.

On going from NH₃ to SbH₃, the polarizability increases because the number of electrons in the molecules increases. These factors increase the dispersion interactions (induced dipole-induced dipole interactions) so that the boiling point increases from PH₃ to SbH₃. [Note the b.p. trend for Ne through Xe, where the increase is entirely due to dispersion interactions.]

Answer. There are two reasons. We know from periodic trends that oxygen is more electronegative than nitrogen, so the O-H bond should be more polar than the N-H bond. This will make the H-bonding stronger and more difficult to break in H₂O than in NH₃.

The second reason is that we know from the structure of ice that each water molecule is involved in four hydrogen bonds, as shown below. To boil water all four of these must be broken to give H2O molecules. An ammonia molecule is involved in only two water molecules, as shown in part (b).

(a) (4 marks) Which of the following solutes dissolved in 1.0 kg of water would cause the greatest lowering of the freezing point? Explain your answer!

 
1.0 mol Fe(NO₃)₃

1.5 mol CaCl₂

1.5 mol CH₃OH

2.5 mol CH₃CO₂H

1.0 mol NaOH

1.0 m x 4 ions/mol = 4.0 m total particle molality for Fe(NO₃)₃

1.5 m x 3 ions/mol = 4.5 m total particle molality for CaCl₂

1.5 m for CH₃OH

2.5 m for CH₃CO₂H

1.0 m x 2 particles per mol = 2.0 m total particle molality for NaOH

(b) (6 marks) When 10.2 g of an unknown compound X is dissolved in exactly 200 g of water, the freezing point was depressed to -2.82 ^oC. What is the molar mass of X? The freezing point of water is 0 ^oC and its K_f is -1.86 ^o kg/mol

Answer. Freezing point depression is related to the molality of the solution,

Thus the molality of the solution = (-2.82 ^oC / -1.86 ^oC) = 1.516 mol solute / kg solvent

7. (16 marks).

(a) (12 marks) Draw a Lewis structure for each of the following, including non-zero formal charges if any. Give the electron count first (valence shell electrons only). The structures will not be marked if the correct electron count is not given.

(i) CH₃CO₂H

(ii) Ca(NO₃)₂

(iii) NH₄HSO₃

(iv) IF₅

 
(i) Which compound(s), if any, produce 3 ions when dissolved in water? Answer: Ca(NO₃)₂

(ii) Which compound(s), if any, ionize only slightly in water? Answer: CH₃CO₂H

8. (10 marks)

(a) (2 marks) Define the following terms:

(i) Triple point The pressure and temperature at which gas, liquid, and solid phase coexist in equilibrium.

(ii) Critical temperature The temperature beyond which it is impossible to liquify a gas no matter what pressure is applied. (The pressure at this point is the critical pressure.)

Label the areas in which gas, liquid, and solid will be found. Also label the vaporization line, sublimation line, melting line, critical point, triple point

This page is http://chemiris.labs.brocku.ca/~chemweb/courses/chem181/Midterm_Exam_2_1998.html
Created March 13, 2001 by M. F. Richardson
© Brock University, 2001