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盐的水解在电解质溶液一章中既是重点、又是难点知识。它是弱电解质的电离平衡、水的电离和溶液的 pH 值知识的综合运用。所以,教好这一节的知识既能使学生巩固和加深所学的基础知识又能加深对化学反应的认识(如双盐水解等)还能解释有关的实际问题。因此,教好盐的水解确实很重要。如何教好呢?1.做好演示实验,进行初步分析可分别在 CH_3COONa 晶体、NH_4Cl 晶体、NaCl 晶体中加入蒸馏水形成溶液后,再进一步测定各溶液的 pH 值,分别为:pH=8,pH=6,pH=7。结论:某些盐的水溶液呈现微弱的酸碱性。由上述实验得出盐与水作用生成酸和碱的反应叫盐的水解。如果将 CH_3COONa 的水溶液蒸发仍可得到 CH_3COONa 的晶体。由此说明盐的水解是可逆的。以 CH_3COONa 水解为例写出化学方
The hydrolysis of salt is a key and difficult point in the electrolyte solution chapter. It is a comprehensive application of knowledge of ionization balance of weak electrolytes, ionization of water, and pH of solutions. Therefore, teaching this section of knowledge can not only enable students to consolidate and deepen the basic knowledge they learned, but also deepen their understanding of chemical reactions (such as double-saline solution), and can also explain the actual problems involved. Therefore, it is very important to teach the hydrolysis of salt. How to teach it? 1. Do a demonstration experiment and conduct preliminary analysis. After adding distilled water to CH_3COONa crystals, NH_4Cl crystals, and NaCl crystals to form a solution, further measure the pH of each solution: pH=8, pH = 6, pH = 7. Conclusion: The aqueous solution of some salts shows a slight acidity. From the above experiments, the reaction of salt and water to generate acid and alkali is called hydrolysis of salt. If crystals of CH_3COONa are evaporated, crystals of CH_3COONa can still be obtained. This shows that the hydrolysis of the salt is reversible. Write a chemical recipe with CH_3COONa hydrolysis as an example