25.0 mL of a 0.10 M solution of KOH is added to 15.0 mL of 0.15 M solution of HCl. What is the pOH of the resulting solution?

excess of KOH, 6.25 med mer

pOH = 2.2
pH = 11.8
C.

Initial number of moles of KOH = (0.10 mol/L) × (25.0/1000 L) = 0.0025 mol
Initial number of moles of HCl = (0.15 mol/L) × (15.0/1000 L) = 0.00225 mol

KOH + HCl → KCl + H₂O
Mole ratio KOH : HCl = 1 : 1

Initial number of moles of KOH > Initial number of moles of HCl
Hence, KOH is in excess.
Number of moles of excess KOH = (0.0025 - 0.00225) mol = 0.00025 mol
Volume of final solution = (25.0 + 15.0) mL = 40.0 mL = 0.0400 L
Molarity of KOH in the final solution = (0.00025 mol) / (0.040 L) = 0.00625 M

KOH completely dissociates in the solution to give OH⁻ ions.
Hence, [OH⁻] in the final solution = 0.00625 M
pOH = -log[OH⁻] = -log(0.00625) = 2.2
pH = pKw - [OH = 14.0 - 2.2 = 11.8

The answer: None of the 4 options is the answer.
(If asked for pH instead of pOH, the answer is: C: 11.8)