Answer :
[tex]\text{ Kw = }\lbrack\text{H}_3\text{O}^+\rbrack\lbrack\text{OH}^-\rbrack[/tex][tex]\lbrack\text{OH}^-\rbrack\text{ = }\frac{\text{ Kw}}{\lbrack\text{ H}_3\text{O}^+\rbrack}=\frac{1\times10^{-14}}{1\times10^{-10}}=1\times10^{-4}[/tex]
The answer is [OH-] = 1.0E-4