`100 ml` of `0.15 M` HCl is mixed with `100 ml` of `0.05 M` HCl, is the pH of the resulting solution?

Explanation

`[H^(+)]=(V_(1)N_(1)+V_(2)N_(2))/((V_(1)+V_(2)))`
`((100)(0.15)+(100)(0.05))/(100+100)=10^(-1)N`
`P^(H)=-log[H^(+)]=-log[10^(-1)]=1`