The ionization of `H_(2)SO_(4)` is completed in 2 steps .
(i)`H_(2)SO_(4)toH^(+)+HSO_(4)^(-1)`
`k_(a)(1)=1.4xx10^(-2) `mol `L^(-1) s^(-1)`
(ii)`HSO_(4)^(-)=3.5xx10^(-2)` mol `L^(-1)S^(-1)`
Then which of the following equation given below is true for rate?

Options

A. Rate=`K_(a)(1)[H_(2)SO_(4)]`
B. Rate =`K_(a)(2)[HSO_(4)^(-)]`
C. Rate =`K_(a)(1)[H_(2)SO_(4)]`
D. Rate=`K_(a)(2)[HSO_(4)^(-)]`

Answer

B

Explanation

The rate of the second reaction is very les.So the rate can be taken same as other less reaction.
(ii)`HSO_(4)^(-1)toH^(+)+SO_(4)^(2-)`Ka (2) `3.5xx10^(-2)`
`therefore` Rate will be =`K_(a)(2)HSO_(4)^(-1)`
So,option (B) is correct.