Chemistry Labs

第 1 题

破损标签的酸试剂瓶。一瓶酸的稀水溶液标签受损，仅能辨认其浓度数值。用 pH 计测得氢离子浓度恰好等于标签上的浓度值（[HX+]=c[\ce{H+}] = c）。(1.1) 若稀释 10 倍后 pH 恰好改变 1 个单位，写出该溶液中可能含有的 4 种酸的分子式。(1.2) 该稀溶液中是否可能含有硫酸（pKa2=1.99pK_{a2} = 1.99）？回答“是”或“否”；若是，计算或估算 pH。(1.3) 溶液中是否可能含有乙酸（pKa=4.76pK_a = 4.76）？回答“是”或“否”；若是，计算或估算 pH。(1.4) 溶液中是否可能含有 EDTA（乙二胺四乙酸；pKa1=1.70pK_{a1} = 1.70、pKa2=2.60pK_{a2} = 2.60、pKa3=6.30pK_{a3} = 6.30、pKa4=10.60pK_{a4} = 10.60）？回答“是”或“否”；若是，计算其浓度。
第 4/4 步：EDTA 溶液的条件
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c=[HX4A]+[HX3AX−]+[HX2AX2−]=[HX+], [HX+]=[HX3AX−]+2[HX2AX2−]⇒[HX4A]=[HX2AX2−];pH=pK1+pK22=2.15, c=0.0071 Mc = [\ce{H4A}] + [\ce{H3A-}] + [\ce{H2A^{2-}}] = [\ce{H+}],\ [\ce{H+}] = [\ce{H3A-}] + 2[\ce{H2A^{2-}}] \Rightarrow [\ce{H4A}] = [\ce{H2A^{2-}}];\quad \mathrm{pH} = \frac{pK_1+pK_2}{2} = 2.15,\ c = 0.0071\ \text{M}
分析

酸性介质中第三、四步解离可忽略。将 c=[HX4A]+[HX3AX−]+[HX2AX2−]=[HX+]c = [\ce{H4A}] + [\ce{H3A-}] + [\ce{H2A^{2-}}] = [\ce{H+}] 与 [HX+]=[HX3AX−]+2[HX2AX2−][\ce{H+}] = [\ce{H3A-}] + 2[\ce{H2A^{2-}}] 联立得 [HX4A]=[HX2AX2−][\ce{H4A}] = [\ce{H2A^{2-}}]。则 [HX+]2=Ka1Ka2[\ce{H+}]^2 = K_{a1}K_{a2}，给出 pH = (1.70+2.60)/2=2.15(1.70+2.60)/2 = 2.15（[HX+]=7.08×10−3[\ce{H+}] = 7.08\times10^{-3} M）。因此 [HX+]=c=0.0071[\ce{H+}] = c = 0.0071 mol dm−3^{-3}。回答：是。