KOH(aq) → K+(aq) + OH–(aq)
2 H2O(l) → ← H3O+(aq) + OH–(aq)
The strong base contributes 1.0×10–7 M hydroxide.
The water contributes a little less than 1.0×10–7 M hydroxide, so the total is
[OH–] = 1.0×10–7 + <1.0×10–7 = ?
We cannot make the simple approximation this time, so we need to do a full equilibrium calculation using the water autoionization equilibrium:
2 H2O(l) → ← H3O+(aq) + OH–(aq)