How many milliliters of 0.0150 M KOH are required to react exactly with 75.0 mL of 0.0150 M H₃PO₄? H₃PO₄(aq) + 3KOH(aq) → K₃PO₄(aq) + 3H₂O(l)

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First, we need to determine the moles of H₃PO₄ and KOH: moles of H₃PO₄ = M x V = 0.0150 M x 75.0 mL = 0.001125 mol moles of KOH = (3 mol H₂O/mol KOH) x 0.001125 mol = 0.003375 mol Now we can use the molarity and moles of KOH to determine the volume needed: volume of KOH = moles of KOH / M = 0.003375 mol / 0.0150 M = 0.225 L = 225 mL Therefore, the answer is C) 225 mL.

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