What would be the pH of an aqyeous solution of sulphuric acid which is `5xx10 ^(-5) `mol litre^(-1)` in concentration.

Explanation

Sulphuric acid dissociates in water as:
`H_(2)SO_(4(aq))to2H_(aq)^(+)+SO_(4(aq))^(2-)`
Each mole of sulphuric acid gives two m oles of `H^(+)` in the solution .One litre of `H_(2)SO_(4)` Solution contains `5xx10^(-5)` moles of `H_(2)SO_(4)` which would give `2xx5xx10^(-5)=10xx10^(-5)` pr `1.0xx10^(-4)` ,moles of `H^(+)` ion in on litre of the solution.
Therefore ,`[H^(+)]=1.0xx10^(-4)` mol `litre^(-1)`
pH=`-log[H^(+)]=-log_(10)10^(-4)=-(-4xxlog_(10)10)`
`-(-4xx1)=4`