一种醋酸钠溶液的PH为8.52,求1.0L此溶液中含多少无水醋酸钠。

2025-03-16 17:23:29
推荐回答(3个)
回答1:

一种醋酸钠溶液的PH为8.52,求1.0L此溶液中含5.07molL无水醋酸钠

CH₃COOH,Ka=1.8x10^-5,

醋酸盐中CH₃COO-的水解常数Kh=Kw/Ka=5.56x10^-10

PH=8.52,POH=5.48,cOH-=10^-5.48

根据水解常数计算式

CH₃COO- + H₂O = CH₃COOH + OH-

x-10^-5.48............10^-5.48...10^-5.48

x-10^-5.48≈x

Kh=cCH₃COOH *c OH-/cCH₃COO-=

(10^-5.48)^2=5.56x10^-10*x

x=5.07molL。

扩展资料:

水解过程用离子方程式表示为: Ac-+ H₂O ⇌ HAc + OH-。 

当水解达到平衡时:

Kh为水解反应的平衡常数,即水解常数。对于一元弱酸强碱盐的水解反应,  (Kw是水的离子积常数,Ka是弱酸的解离常数)。在常温下,水的离子积常数Kw为一常数,故弱酸强碱盐的水解常数值Kh取决于弱酸的电离常数Ka的大小。

参考资料来源:百度百科-水解常数

回答2:

CH3COOH,Ka=1.8x10^-5,
醋酸盐中CH3COO-的水解常数Kh=Kw/Ka=5.56x10^-10
PH=8.52,POH=5.48,cOH-=10^-5.48

根据水解常数计算式
CH3COO- + H2O = CH3COOH + OH-
x-10^-5.48............10^-5.48...10^-5.48
x-10^-5.48≈x
Kh=cCH3COOH *c OH-/cCH3COO-=
(10^-5.48)^2=5.56x10^-10*x
x=5.07molL-

回答3:

c(NaAc)=[HO-]^2/Kb=(10^-5.48)^2/(10^-14/(1.78*10^-5))=0.02mol/L
m(NaAc)=0.02*82=1.64克。

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