`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` litre solution of `pH=6` (solution of a strong acid) is added to the `7//3` litres of water.What is the `pH` of resulting solution? Neglect the common ion effect on `H_(2)O`.

Options

A. `6.4`
B. `6.52`
C. `6.365`
D. `6.432`

Explanation

`pH=6`
`[H^(+)]=10^(-6)`
`N_(1)V_(1)=N_(1)V_(2)`
rArr `10^(-6)xx1=N_(2)[1+(7)/(3)]" "rArr" "10^(-6)=N_(2)xx(10)/(3)`
`N_(2)=(3)/(10)xx10^(-6)" "rArr " "N_(2)=3xx10^(-7)`
`[H^(+)]lt10^(-6)` so `[H^(+)]` of water is also added as common ion effect on `H_(2)O` is neglected so
`[H^(+)]=3xx10^(-7)+10^(-7)=4xx10^(-7)M`
`rArr pH=7-log4=7-0.60=6.4`