`pH` calculation upon dilute of strong acid solution is generally done by equating `n_(H)` in original solution & diluted solution.However . If strong acid solution is very dilute then `H^(+)` from water are also to be considered take `log3.7=0.568` and answer the following questions.
A `1` litres solution of `pH=4`(solution of a strong acid ) is added to the `7//3` litres of water.What is the `pH `of resulting solution?

Options

A. `4.52`
B. `4.365`
C. `4.4`
D. `4.432`

Explanation

Initial `pH=4`
`[H^(+)]=10^(-4)`
`N_(1)V_(1)=N_(2)V_(2)rArr 10^(-4)=N_(2)xx[1+(7)/(3)]`
`10^(-4)=N_(2)xx(10)/(3)rArr N_(2)=3xx10^(-5)" " gt10^(-6)`
so `[H^(+)]` of water is not consider
`[H^(+)]=3xx10^(-5)`" " so" "`pH=5-log(3)`" "`=5-0.48=4.52`