`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`