a) Calculate pH of `10^(-8) M H_2 SO_4`

Explanation

`underset(10^-8 M)(H_2 SO_4) overset(H_2O)(hArr)underset(2xx10^(-8))(2H_3O^+)+underset(10^(-8)M)(SO_(4)^(2-))`
In this case the concentration of `H_2SO_4` is very low and hence `[H_3O^+]` from water cannot be neglected
`therefore [H_3O^]=2xx10^(-8)("from" H_2 SO_4)+10^(-7)` (from water)
` = 10^(-8)(2+10)`
`=12xx10^(-8)=1.2xx10^(-7)`
`pH=-log[H_3 O^+]`
`=-log_(10)(1.2xx10^(-7))`
`=7 -log_(10)1.2`
`=7-0.0791=6.9209`.