[tex]pH_1 = 8 \\
pH_2 = 9[/tex]
[tex]pOH_1 = 14 - 8 = 6 \\
pOH_2 = 14 - 9 = 5[/tex]
[tex]OH^-_1 = 10^{-6} M[/tex]
[tex]OH^-_2 = 10^{-5} M[/tex]
Ratio = [tex]\frac{10^{-5}}{10^{-6}} = 10[/tex]
so, OH- concentration will become 10 times the initial one