課程名稱︰普通化學乙下
課程性質︰系定必修
課程教師︰林金全
開課學院：
開課系所︰材料系
考試日期（年月日）︰2008年3月18日
考試時限（分鐘）：160分鐘
是否需發放獎勵金：是
（如未明確表示，則不予發放）
試題 :
1. The equilibrium constant Ka for the reaction
Fe(H2O)6^3+(aq) + H2O(l) <-> Fe(H2O)5(OH)^2+(aq) + H3O+(aq)
is 6.0 x 10^-3
a. Calculate the pH of a 0.10 M solution of Fe(H2O)6^3+. (4%)
b. Calculate the pH necessary for 99.90% of the iron(III) to be in the form
Fe(H2O)6^3+. (4%)
c. Will a 1.0 M solution of iron(II) nitrate have a higher or lower pH than
a 1.0 M solution of iron(III) nitrate? Explain. (4%)
2. Arsenic acid (H3AsO4) is a triprotic acid with Ka1 = 5 x 10^-3, Ka2 =
8 x 10^-8, and Ka3 = 6 x 10^-10. Calculate [H+], [OH-], [H3AsO4], [H2AsO4-]
, and [AsO4^3-] in a 0.20 M arsenic acid solution. (12%)
3. One method for determining the purity of aspirin (empirical formula,
C9H8O4) is to hydrolyze it with NaOH solution and then to tritrate the
remaining NaOH. The reaction of aspirin with NaOH is as follows:
Boil
C9H8O4(s) + 2OH-(aq) -------> C7H5O3-(aq) + C2H3O2-(aq) + H20(l)
Aspirin 10 min Salicylate ion Acetate ion
A sample of aspirin with a mass of 1.427 g was boiled in 50.00 mL of 0.500
M NaOH. After the solution was cooled , it took 31.92 mL of 0.289 M HCl to
titrate the excess NaOH. Calculate the purity of the aspirin. (12%)
4. For solutions containing salts of the form NH4X, the pH is determined by
using the equation
pH = (pKa(NH4+) + pKa(HX))/2
a. Derive this equation. (5%)
b. Use this equation to calculate the pH of the following solutions:
ammonium formate, ammonium actetate, and ammonium bicarbonate.
( Ka(NH4+) = 5.6 x 10^-10, Ka(HCO2H) = 1.8 x 10^-4, Ka(HC2H3O2) =
1.8 x 10^-5, Ka(H2CO3) = 4.3 x 10^-1) (5%)
c. Solutions of ammonium acetate are commonly used as pH = 7.0 buffers.
Write equations to show how an ammonium acetate solution neutralizes
added H+ and OH-. (5%)
5. The dilute solution of HCl on the laboratory shelf happens to be 16.8% by
weight. Its density is 1.085 g/mL. Calculate the molarity of the solution.
(H =1.0, Cl = 35.5) (10%)
6. Calculate solubility of AgCl and concentration of Ag(NH3)+ and Ag(NH3)2+ in
5.0 M NH3, based on the following equations: (15%)
AgCl(s) <-> Ag+(aq) + Cl-(aq) Ksp = 1.6 x 10^-10
Ag+(aq) + NH3(aq) <-> Ag(NH3)+(aq) K1 = 2.1 x 10^3
Ag(NH3)+(aq) + NH3(aq) <-> Ag(NH3)2+(aq) K2 = 8.2 x 10^3
7. It takes 208.4 kJ of energy to remove one mol of electrons from the atoms
on the surface of rubidium metal. If rebidium metal is irradiated with
254-nm light, what is the maximum kinetic energy the released electrons
can have? (h = 6.62608 x 10^-34 Js) (12%)
8. Compare the wavelength for an electron (mass = 9.11 x 10^-31 kg) traveling
at a speed of 1.0 x 10^-7 m/s with that for a ball (mass = 0.10 kg)
traveling at 35 m/s. (12%)
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