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關於儀器分析的問題～感謝
關於儀器分析的問題～感謝參考書上寫很省略求較詳細的計算式～＿～麻煩了感謝
1.首先 [H2In] = 4 * 10^(-3) * 1/100 = 4 * 10^(-5) M因為 1mL H2In solution 99mL buffer solution2. 因為 Ka1 = [H ][HIn-]/[H2In] = 10^(-4) 由pK1=4而來Ka2 = [H ][In(2-)]/[HIn-] = 10^(-8)把 [H ] = 10^(-5) 帶入上述二式: 因為buffer solution pH=5得 [HIn-] = 10[H2In]
[In(2-)] = 10^(-3)[HIn-]3.因為在 solution 中 [H2In] [HIn-] [In(2-)] = 4 * 10^(-5) M所以 (100 1000 1)[In(2-)] = 4 * 10^(-5)由此可得:[In(2-)] = 3.63 * 10^(-8)[H2In] = 3.63 * 10^(-6)[HIn-] = 3.63 * 10^(-5)4.帶入 A(absorbance) = EbC Beer
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