-- dus --> [OH-] = 10 -2,60 = 0,00251 mol/L
volume (voor verdunnen) = 100 mL
volume (na verdunnen) = 100 mL + 900 mL = 1000 mL
-- dus --> de oplossing wordt 1000 / 100 = 10x verdund
[OH-] = 0,00251 mol/L /10 = 0,000251 mol/L
-- dus --> pOH = -log (0,000251) = 3,60 -- dus --> pH = 14,00 - 3,60 = 10.40