For a `0.072 M NaSO_(4)` solution,select the incorrect option(s):`(K_(a_(1))` and `K_(a_(2))` of `H_(2)SO_(4)=oo` & `1.2xx10^(-2))`
A. `pH=1.62`
B. `pH=7.39`
C. Degree of hydrolysis `h=(1)/(sqrt(6))`
D. `h=3.4xx10^(-6)`

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1 Answers

`1.2xx10^(-7)=(7.2xx10^(-2)alpha^(2))/(1-alpha)rArr 6alpha^(2)+alpha-1=0rArr alpha=(1)/(3)`
So `pH=-log ((0.072)/(3))=1.62`

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