What volume (ml) of a 0.2450 m koh (aq) solution is required to completely neutralize 55.25 ml of a 0.5440 m h3po4 (aq) solution?

Answer: It depends on what came after "0.5440 M H ... ". If it was a monoprotic acid, like HCl, the calculation would go like this: (55.25 mL) x (0.5440 M acid) x (1 mol KOH / 1 mol acid) / (0.2450 M KOH) = 122.7 mL KOH If it was a diprotic acid, like H2SO4, like this: (55.25 mL) x (0.5440 M acid) x (2 mol KOH / 1 mol acid) / (0.2450 M KOH) = 245.4 mL KOH If it was a triprotic acid, like H3PO4, like this: (55.25 mL) x (0.5440 M acid) x (3 mol KOH / 1 mol acid) / (0.2450 M KOH) = 368.0 mL KOH