50 ml of `0.1 MH_(2)SO_(4)` were added to 100 ml of `0.2 MHNO_(3)`. Then the solution is diluted to 300 ml. What is the pH of the solution?

Explanation

`V_(1)=50 ml " " V_(2)=100 ml`
`N_(1)=0.1xx2=0.2N " " N_(2)=02N`
`:. N=(N_(1)V_(1)+N_(2)V_(2))/(V_(1)+V_(2))`
`=(50+0.2+100xx0.2)/(150)`
`=(30)/(150)=(1)/(5)N`
This is diluted to 300 ml then
`N_(1) V_(1)=N_(2)N_(2)`
`(1)/(5)xx150=N_(2)xx300`
`N_(2)=(30)/(300)=0.1N`
`:. pH=-log[H^(+)]`
`=-log0.1`
`=1`