40 ml of `0.2 M HNO_(3)` when reacted with 60 ml of `0.3 M NaOH` gave a mixed solution. What is the pH of the resulting solution?

Explanation

`V_(B)N_(B) gt V_(A) N_(A)`
`:. N=(V_(B)N_(B)-V_(A)N_(A))/(V_(A)+V_(B))`
`=(60xx0.3-40xx0.2)/(60+40)`
`=(18-8)//(100)=(10)/(100)=(1)/(10)=0.1N=[OH^(-)]`
`pOH=-log[OH^(-)]`
`=-log 0.1`
`=1`
`pH=14-pOH=14-1`
`=13`