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

`pH` calculation upon dilute of strong acid solution is generally done

1.	`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?A. `4.52`B. `4.365`C. `4.4`D. `4.432`
Answer» 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`

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