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 3/4 步：超稀乙酸
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c=[HA]+[AX−]=[HX+], [HX+]=[AX−]+[OHX−]⇒[HA]=[OHX−];[HX+]3=KaKw+Ka[HX+]2⇒[HX+]=5.64×10−7 M (pH=6.25)c = [\ce{HA}] + [\ce{A-}] = [\ce{H+}],\ [\ce{H+}] = [\ce{A-}] + [\ce{OH-}] \Rightarrow [\ce{HA}] = [\ce{OH-}];\quad [\ce{H+}]^3 = K_a K_w + K_a[\ce{H+}]^2 \Rightarrow [\ce{H+}] = 5.64\times10^{-7}\ \text{M}\ (\mathrm{pH} = 6.25)
分析

由物料衡算 c=[HA]+[AX−]=[HX+]c = [\ce{HA}] + [\ce{A-}] = [\ce{H+}] 与电荷守恒 [HX+]=[AX−]+[OHX−][\ce{H+}] = [\ce{A-}] + [\ce{OH-}] 联立得 [HA]=[OHX−][\ce{HA}] = [\ce{OH-}]。代入 KaK_a 表达式得 [HX+]3=KaKw+Ka[HX+]2[\ce{H+}]^3 = K_a K_w + K_a[\ce{H+}]^2。经两次迭代得 [HX+]=c=5.64×10−7[\ce{H+}] = c = 5.64\times10^{-7} M（pH = 6.25）。回答：是。