According to the fact that [H+]*[OH-] is always equal Kw\(= 1x10^(-14)\), you can easily find concentration of [H+] by changing from \(1*10^(-9)\)to \(1*10^(-11)\)
H increases, Kw is constant, therefore \(10^-14 / 10^-11 = 10^-3.\)
And you will have \(Hence [H+] = 10^-3\) where pH is 3
