Two buffer solutions `A` and `B` each made with benzoic acid and sodium benzoate differ in their `pH` by two units. A has salt`:` acid`a:b. B` has salt`:` acid `=b:a.` If `agtb`, then the value of `a:b` is
`a. 3.17" "b.10.0" "c.3.0" "d. 6.0`

Explanation

`b.` Buffer `A`
`pH_(A)=pK_(a)+log.((a)/(b))" "(i)`
Buffer `B`
`pH_(B)=pK_(a)+log((b)/(a))" "(ii)`
Since `agtb`
`:. pH_(A)-Ph_(B)=log.(a)/(b)-log.(b)/(a)`
`2=log.(a)/(b)+log.(a)/(b)`
`implies2=2log.(a)/(b)`
`:. log .(a)/(b)=1`
`(a)/(b)=Antilog(1)`
`=10`
`:. (a)/(b)=10`