1.

A weak acid HX has the dissociation constant 1xx10^(-5)M. It forms a salt NaX on reaction with alkali. The percentage hydrolysis of 0.1 M solution of NaX is :

Answer»

`0.0001 %`
`0.01%`
`0.1%`
`0.15%`

Solution :`NaX+H_(2)O rarr NaOH+HX`
HX is weak acid so that NaX is salt of weak acid and strong BASE
`K_(H)=(K_(w))/(K_(a))=(1.0xx10^(-14))/(10^(-5))=1.0xx10^(-9)`
Now `K_(h)=CH^(2)`
`10^(-9)=1.0 h^(2)`
`h^(2)=(1.0xx10^(-9))/(0.1)=10^(-8)` or `h=10^(-4)`
Percentage of HYDROLYSIS `=10^(-4)xx100`
`=10^(-2)%`or`=0.01 %`


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