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)
Kw = [H3O+]e[OH–]e = 1.0×10–14
Initial0 1.0×10–7
Change+x +x
Equilibriumx 1.0×10–7 + x
1.0×10–14 = x(1.0×10–7 + x)
0 = x2 + 1.0×10–7x – 1.0×10–14
This requires the quadratic equation to solve:
x = 6.2×10–8 or –1.6×10–7
Only the positive value makes sense chemically, so:
[H3O+]e = x = 6.2×10–8 M
[OH–]e = 1.0×10–7 + x = 1.0×10–7 + 6.2×10–8
= 1.6×10–7 M