100 ml of `pH=4` solution is mixed with 100 ml of `pH=6` solution. What is the pH of resulting solution?

Explanation

`N=(100xx10^(-4)+100xx10^(-6))/(200)`
`=(10^(-2)+10^(-4))/(200)`
`=(0.01+0.0001)/(200)=[H^(+)]`
`=(0.01001)/(200)`
`pH=-log[H^(+)]`
`=-"log"(0.01001)/(200)`
`=-log(0.01001)-log200`
`=4.32`