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Question
The acid ionization constant of Zn2+ is 1.0×10−9. Calculate the pH of a 0.001 M solution of ZnCl2.
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Solution
The correct option is A 6
ZnCl2 is a salt of strong acid and weak base.
Given, Kh=10−9
Hence,
Kh=h2C
h2=KhC=10−90.001=10−6
h=10−3
[H+]=nhC=1×10−3×0.001=10−6mol/dm3 where, n=1
Therefore, pH=−log10[H+]=−log10(10−6)=6
ZnCl2 is a salt of strong acid and weak base.
Given, Kh=10−9
Hence,
Kh=h2C
h2=KhC=10−90.001=10−6
h=10−3
[H+]=nhC=1×10−3×0.001=10−6mol/dm3 where, n=1
Therefore, pH=−log10[H+]=−log10(10−6)=6
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