課程名稱︰分析化學丁
課程性質︰選修
課程教師︰林金全
開課學院：農學院
開課系所︰園藝系
考試日期（年月日）︰100/12/1
考試時限（分鐘）： 120分鐘
是否需發放獎勵金：是
試題 :
一、(9pt)
(a) (3.14±0.05)^1/3
(b) log( (0.104±0.006)^1/2 / 0.0511±0.0009 )
(c) sin(80.2±0.5)＋cos(62.5±0.8)
二、(10pt)
the CdSe content (g/L) of nanocrystal was measured by two methods for six
different samples. do the two methods differ significantly at the 95%
confidence level ?
－－－－－－－－－－－－－－－－－－－－－－－－－－－－－－
method 1 method 2
sample anodic stripping atomic absorption
A 0.88 0.83
B 1.15 1.04
C 1.22 1.39
D 0.93 0.91
E 1.17 1.08
F 1.51 1.31
三、(10pt)
in a preliminary experiment, a solution containing 0.0837M X and 0.0666M S
gave peak areas of Ax=423 As=247.
to analyze the unknown, 10.0mL of 0.146M S were added to 10.0mL of unknown,
and the mixture was diluted to 25.0mL in a volumetric flask. this mixture
gave the chromatogram for which Ax=553 As=582. find the concentration of X
in the unknown.
四、(12pt)
for the following reactions
PbI2 → Pb2+ ＋ 2I- Ksp=7.9x10^-9
Pb2+ ＋ I- → PbI+ K1=1.0x10^2
Pb2+ ＋ 2I- → PbI2 K2=1.4x10^3
Pb2+ ＋ 3I- → PbI3- K3=8.3x10^3
Pb2+ ＋ 4I- → PbI4 2- K4=3.0x10^4
(a)given a concentration of 0.010M I-, find the concentration of PbI+ and PbI4 2-
(b)the total [Pb] is a function of [I-], find the minimum [Pb]
五、(15pt)
if river water is satured with calcite [Ca2+] is governed by the following equilibria
CaCO3 → Ca2+ ＋ CO3 2- Ksp=4.5x10^-9
CO2(g) → CO2(aq) K=0.032
CO2(aq) ＋ H2O → HCO3- ＋ H+ K1=4.46x10^-7
HCO3- → CO3 2- ＋ H+ K2=4.69x10^-11
(a)find the equilibrium constent for the reaction
CaCO3 ＋ CO2(aq) ＋ H2O → 2HCO3- ＋ Ca2+
(b)the mass balance for reaction A is [HCO3-]=2[Ca2+]. find [Ca2+] in equilibrium
with atomspheric CO2 if Pco2 = 3.8x10^-4 bar.
(c)the concentration of Ca2+ in the Don River is 80 mg/L. what effective Pco2
is in equilibrium with this much Ca2+ ?
六、(10pt)
a dibasic compound B, has pKb1=4.00 pKb2=6.00. find the fraction in the form
BH2 2+ at pH 7.00.
七、(10pt)
the kjeldahl procedure was used to analyze 0.256mL of a solution containing
37.9mg protein/mL. the liberated NH3 was collected in 5mL of 0.0336M HCl, and
the remaining acid required 6.34mL of 0.01M NaOH for complete titration.
what is the weight percent of nitrogen in the protein ?
八、(12pt)
sulfide ion was determined by indirect titration with EDTA. to a solution
containing 25mL of 0.04332M Cu(ClO4)2 plus 15mL 1M acetate buffer were added
25mL of unknown sulfide solution with vigorous stirring. the CuS precipitate
was filtered and washed with hot water. then ammonia was added to the filtrate
until the blue color of Cu(NH3)4 2+ was observed.
titration of the filtrate with 0.03927M EDTA required 12.11mL to reach the
murexide end point, calculate the molarity of sulfide in the unknown.
九、(12pt)
the voltage of the cell shown here is -0.246V.the right half-cell contains the
metal ion, M2+, whose standard reduction potential is -0.266V.
M2+ ＋ 2e- → M E=-0.266
calculate Kf for the metal-EDTA complex.
(a)pyrophosphoric acid : pKa1=0.9,pKa2=2.28,pKa3=6.70,pKa4=9.40
(b)EDTA : ａy4- = 4.2x10^-3 at pH 8.00
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