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在高中化学教学中,溶液中离子浓度大小的比较是一个重要的知识点。有关 NaHS 溶液中各种离子浓度大小比较的问题,经常在各种教辅资料的习题以及考题中出现。一般都认为 c(OH~-)>c(S~(2-))。关于0.1 mol/L NaHS 溶液中 c(OH~-)和 c(S~(2-))的大小判断常见的分析思路是:NaHS 溶液中存在如下平衡:HS~-的水解:HS~-+H_2OH_2S+OH~-K_(h_2)=1.1×10~(-7)HS~-的进一步电离:HS~-H~++S~(2-)K_(a_2)=1.1×10~(-12)H_2O的电离:H_2OH~++OH~-由于水解常数大于电离常数,所以可推知 HS~-的
In high school chemistry teaching, the comparison of ion concentration in solution is an important point of knowledge. NaHS solution for various ion concentration in the size of the comparison, often in a variety of supplementary materials, exercises and test questions appear. Generally considered c (OH ~ -)> c (S ~ (2-)). The common analytical approach to judge the size of c (OH ~ -) and c (S ~ (2-)) in 0.1 mol / L NaHS solution is that there is the following balance in NaHS solution: HS ~ - Hydrolysis: Further ionization of H 2 OH 2 S + OH ~ -K h 2 = 1.1 × 10 -7 HS ~ -: HS ~ -H ~ ++ S ~ (2-) K a -1 = 1.1 × 10 ~ (-12 ) H_2O ionization: H_2OH ~ + + OH ~ - As the hydrolysis constant is greater than the ionization constant, it can be deduced that HS ~ -