What will be the pOH of a solution that contains 0.2 moles of potassium hydroxide and 0.05 moles of sulfuric acid in 100 cm3 of solution? (Complete dissociation).
The reaction between acid and base is as follows:
2KOH + H2SO4 → K2SO4 + 2H2O
From here, two moles of potassium hydroxide interact with one mole of sulfuric acid.
From here, the number of moles of potassium hydroxide left in the solution after the neutralization equals:
n(KOH) = 0.2 mol - (2 × 0.05 mol) = 0.1 mol
From here, the molar concentration of hydroxide ions equals:
[OH-] = 0.1 mol 100 cm3 = 0.1 mol / 0.1 dm3 = 1 M
pOH = -log[OH-] = -log = 0