常温下,向20毫升0.1摩尔每升NACN溶液中逐滴加入0.1摩尔每升盐酸

2025-04-03 00:47:06
推荐回答(2个)
回答1:

HCN水溶液极小部分电离,电离出H+,HCN=可逆=H++CN-
CN盐水溶液,首先完全电离,完全电离出CN-离子,然后极小部分CN-水解,水解出OH-,CN-+H2O=可逆=OH-+HCN
根据水解方程式cOH-=cHCN,由于CN-用去水解的部分极小近似等于完全电离时的浓度,
平衡常数,kh=cOH- x cHCN / cCN-
而HCN溶液电离平衡常数的计算式。Ka=cH+ x cCN- /cHCN
可见水解平衡常数可以反映出水解程度—水解出OH-的量,
电离平衡常数反映出电离程度—电离出H+的量
把电离水解平衡常数计算式合并即 ka x kh=Kw
如图没有加盐酸时PH就大于7,说明OH->H+ — Kh>ka
B
电荷守恒 Na+ +H+ =CN-+ Cl-+OH-
点2中性OH-=H+
Na+=CN-+Cl-,Na+和Cl-是完全电离的
2点时盐酸大于10ml,即溶液中有大于0.001mol的Cl-
Na+依旧为0.002mol,这样cCN-小于0.001mol
cCl->cCN-
C
3点等于是完全反应,生成0.002molNaCl和0.002mol的HCN的混合
NaCl依旧完全电离Na+=Cl-=0.002,
HCN依旧主要以分子形式存在cHCN约定于0.002,而极小部分电离
HCN=可逆=H++CN-.根据这个方程式,按说cH+=cCN-
虽然最渺小但最后还有H2O=可逆=H++OH-
这样Na+>HCN>H+>CN-

回答2:

A错误 加10ml溶液还是碱性,这时候NaCN和HCN是1;1.说明NaCN水解大于HCN电离
B正确 中性
C正确 1:1混合
D正确 电荷守恒

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