私立中原大學八十八學年度博士班招生考試命題紙

所組別： 化學系博士班 科目： 分析化學 考試時間： 06月09日第2節
  1. Calculate Kal values for a 1.0 M solution of H3PO4(=2.148).Please use successive approximations until the final pH does not vary by more than 0.0001 unit.
    Note：(1)Use the equation [H+]=
    (2)Use Davies, equation to calculate the activity where I is the ionic strength of the solution.
    (3)　　　　　　　　　　　　　　　　　　　　　　　　　(55 points)

  2. In a series of experiments on the determination of tin in foodstuffs,samples where boiled with hydrochloric acid under reflux for different times. Some of the results are shown below:　　　　　　　　　　(15 points)
    Refuxing time(min)　　　Tin found (mg/kg)
    30　　　55,　57,　59,　56,　56,　59
    75　　　57,　55,　58,　59,　59,　59
    Does the mean amount of tin found differ significantly at the 0.05 probability level for the two boiling times?
    Note：(1)a pooled standard deviation:
    (2)with a degrees of freedom n1+n2-2.

    Table A.1. The t-distribution
    Value of t for a confidence interval of
    Critical value of for P values of
    Number of degrees of freedom    		90%
    0.10    		95%
    0.05    		98%
    0.02    		99%
    0.01
    16.3112.7131.8263.66
    22.924.306.969.92
    32.353.184.545.84
    42.132.783.754.60
    52.022.573.364.03
    61.942.453.143.71
    71.892.363.003.50
    81.862.312.903.36
    91.832.262.823.25
    101.812.232.763.17
    121.782.182.683.05
    141.762.142.622.98
    161.752.122.582.92
    181.732.102.552.88
    201.722.092.532.85
    301.702.042.462.75
    501.682.012.402.68
    1.641.962.332.58
    The critical values of are appropriate for a two-tailed test. For a one-tailed test the value is taken form the column for twice the desired P-value, e.g.for a one-tailed test,P=0.05,5 degrees of freedom, the critical value is read from the P=0.10 column and is equal to 2.02.

  3. Derive an expression for the slope for the least-squares fitting of data to the line y=x+B of given, fixed intercept B, assuming that only Gaussian errors in y needed be considered.　　　　　　　　　　　(15 points)

  4. A tablet weighing 1000 mg was analyzed for aspirin, caffeine and paracetamol content.The tablet was crushed and extracted by a solvent mixture(methanol 50%,water50%)and 100 mg of phenaceten was added as the internal standard.The solution was then made up to 100 ml and subsequently analysed by HPLC.A solution of a mixture of standards was also analysed. This solution was obtained by taking 250 mg of each componentand dissolving,making up to 100 ml using methanol/water(50:50). Using the results shown in Table, calculate the composition of the tablet. Note:the tablet contains the three active compounds and a binder.　　　(15 points)
    Table
    Run1　　standard mixture 25%(w/w)of each component
    Compound nameRetention time　　Peak area
    Paracetamol1.51　　38801
    Caffeine2.10　　14236
    Aspirin2.65　　26301
    Phenacetin3.70　　39594
    Run2　　tablet solution
    Paracetamol1.54　　68994
    Caffeine2.11　　19050
    Aspirin2.70　　53480
    Phenacetin(internal standard)3.80　　84922

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