課程名稱︰普通化學甲
課程性質︰第一階段暑修課
課程教師︰林英智、何東英、陳振中
開課系所︰不限
考試時間︰07/15, 2005
試題 :
1. If a 50 mL sample of 0.1M HCN (Ka = 6.2 X 10^-10) is titrated with 0.1M
NaOH, calculate the pH of the solution at (a) 8.00 mL of 0.1M NaOH has been
added, (b) at the halfway point in the titration, and (c) at the equivalent
point. (10%)
2. A chemist has synthesized a monoprotic weal acid and wants to determine its
Ka value. To do so, the chemist dissovles 2.00 mmol of the solid acid in
100.0 mL of water and titrates the resulting solution with 0.0500M NaOH.
After 20.0 mL of NaOH has been added, the pH is 6.00. What is the Ka value
for the acid? (10%)
3. Calculate the change in pH that occurs when 0.010 mol of gaseous HCl is
added to 1.0 L of each of the following solutions.
Solution A ; 5.00M HC2H3O2 and 5.00M NaC2H3O2
Solution B ; 0.050M HC2H3O2 and 0.050M NaC2H3O2
For acetic acid, Ka = 1.8 X 10^-5 (10%)
4. Calculate the pH of a 0.10M NH4Cl silution. The Kb value for NH3 is
1.8 X 10^-5 (10%)
5. Calculate the pH of a 5.0M H3PO4 solution and determine equilibrium
concentrations of the species H3PO4, H2PO4^-, HPO4^2-, PO4^3-.
(Ka1 = 7.5 X 10^-3, Ka2 = 6.2 X 10^-8, Ka3 = 4.8 X 10^-13) (10%)
6. Lactic acid (HC3H5O3) is a waste product that accumulate in muscle tissue
during exertion, leading to pain and a feeling of fatigue. In a 0.100M
aqueous solution, lactic acid is 3.7% dissociated. Calculate the value of
Ka for this acid. (10%)
7. Calculate the pH of a 1.00 X 10^-2M H2SO4 solutoin (Ka2 = 1.2 X 10^-2)
(10%)
8. The partial pressure of an equilibrium mixture of N2O4(g) and NO2(g) are
Pn2o4 = 0.34 atm and pno2 = 1.20 atm at certain temperature. The volume
of the container is doubled. Find the partial pressures of the two gases
when a new equilibrium is established. (10%)
9. Calculate the pH of an aqueous solution containing 1.0 X 10^-2M HCl,
1.0 X 10^-2M H2SO4, and 1.0 X 10^-2M HCN (H2SO4 Ka2 = 1.2 X 10^-2) (10%)
10.For the synthesis of ammonia at 500C, the equilibrium constant is
6.0 X 10^-2 L^2/mol^2. Predict the direction in which the system will
shift to reach equilibrium in each of the following cases.
a. [NH3] = 1.0 X 10^-3M ; [N2] = 1.0 X 10^-5M ; [H2] = 2.0 X 10^-3M
b. [NH3] = 2.0 X 10^-4M ; [N2] = 1.5 X 10^-5M ; [H2] = 3.54 X 10^-1M
c. [NH3] = 1.0 X 10^-4M ; [N2] = 5.0M ; [H2] = 1.0 X 10^-2M
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