If we have \([H+][OH-]= Kw = 1.0 x 10^-14\)
Then \([H+]= Kw/ [OH-]= 1.0x 10^-14\)\(/ 1 x 10^-11 =1 x 10^-3 mol/L\)
And here is the solution - as you can see it is an acidic one :
\(pH = - log [H+]= - log 1 x 10^-3 = 3 \ \textless \ 7\)
I am pretty sure this answer will help you! Regards.
