Chemistry Exam

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5.111 Principles of Chemical Science

EXAM # 3
============================================================================ Write legibly your name and your TA's name below. Do not open the exam until the start of the exam is announced. The exam is closed notes and closed book. You have 50 minutes (1 academic hour) to complete it.

● Read each part of each problem carefully.
● Write your answers legibly in the corresponding spaces of the attached sheets. ● For problems requiring calculations, you must show these calculations clearly and indicate all values, including physical constants used to obtain your quantitative result. Significant figure and unit usage must be correct.

● If you do not understand what the problem is asking, raise your hand and a proctor will come to your desk.
● Some relevant equations and the periodic table are given on the last two pages. You may detach these pages once the exam has started.
============================================================================ 1.

(22 points) _____

2.

(12 points) _____

3.

(27 points) ____

4.

(15 points) _____

5.

(24 points) _____

Total (100 points)

_________________
Your Name(print):___ANSWER

KEY

Your TA’s Name (print):___________________

Fall 2009, 5.111 Exam #3 solution key
Problem 1 (22 points total; 2 points each question).

Page 2 of 10

Give one example of each of the following:
(i) a pure liquid with an extremely low surface tension
Answer: any molecule with low interatomic/intermolecular interactions Examples include – liquid He, Li, H2,
(ii) a strong Lewis base
Answer: Any Group I or Group II hydroxide e.g., LiOH, NaOH, KOH, Ca(OH)2 (iii) a pure substance which can be melted by reducing pressure at a given temperature Answer – Almost any compound but water; examples include sulfur, N2, phosphorus, CO2 (iv) an important colligative property of solutions

Answer: vapor lowering pressure, boiling point elevation, freezing point depression, osmosis (v) a pure substance with no intermolecular (interatomic) London forces Answer: No such thing – all atoms and molecules have London forces (but H+ and ideal gases are exceptions)

(vi) a molecular solid with no intermolecular hydrogen bonding Answer: Any molecule that does not contain a H bonded to N, F, O and is a molecular solid Examples include: I2, naphthalene, dry ice (CO2), wax

(vii) a pure substance whose solubility in water drops as the temperature is raised Answer: Almost any gas, e.g., O2, H2, N2, Cl2, Ne, Ar
(viii) a good solvent for ethanolamine, H2NCH2CH2OH
Answer: Almost any polar, hydrogen-bonding solvent; examples include water, ethanol, methanol (ix) a weak polyprotic acid
Answer: examples include H3PO4, H2S, H2SO3, H2CO3
(x) a group of chemical compounds that increase the solubility of hydrophobic substances in water Answer: detergents (soaps) organic solvents
(xi) a compound besides water capable of intermolecular hydrogen bonding Answer: Any compound that contains a H bonded to N, F or O. examples include: NH3, ethanol, , glycerol, HF

Fall 2009, 5.111 Exam #3 solution key
Problem 2 (12 points).

Page 3 of 10

A buffer solution can be prepared by starting with a weak acid and converting some of it to its salt by titrating with a strong base. The fraction of the original acid that is converted to the salt is designated f. (a) (6 points) Derive an equation similar to the Henderson-Hesselbalch equation but expressed in terms of f rather than concentrations.

Answer: Think of what happens in the reaction table: Assume you start with [HA] Reaction
HA(aq)
+
OH-1 (aq)
A-1 (aq)
+

Initial
[HA]0
f
After Rx’t
[HA] 0 -f
f

H2O ()

Henderson-Hesselbalch equation is pH = pKa + log ([base]/[acid]). In this particular instance, the base forms from the reaction of a weak acid with a strong base. Dividing both the numerator and the denominator under the logarithm by the initial concentration of the weak...
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