Calculate the `pH` of solution obtained by mixing `10mL` of `0.1M HCl` and `40mL` of `0.2M H_(2)SO_(4)` take `log3.4=0.53`

Explanation

Given is the case of a mixture of `2` strong acids.
Mill moles of `H^(+)` form `HCl =10xx0.1=1`
Milli moles of `H^(+)` form `H_(2)SO_(4)=40xx0.2xx2=16`
,.Total millimoles of `H^(+)` in solution =`1+16=17`
`:.[H^(+)]=(17)/(50)=3.4xx10^(-1)( :.[H^(+)]_(f)=(("Millmoles"_("Total"))/(V))`
`:.pH=-log[H^(+)]=-log0.34`
`pH=0.47`