Titration Questions

Question

What volume of each acid will react completely with 60.0 mL of 0.125 M Potassium Hydroxide?

a.) .150 M Nitric Acid

b.) .200 M Phosphoric Acid

Answer 1

a) KOH + HNO₃ ==> KNO₃ + H₂O

moles KOH present = 60.0 ml x 1 L/1000 ml x 0.125 mol/L = 0.0075 moles

moles HNO3 needed = 0.0075 b/c they react in a 1:1 mole ratio as seen in the balanced equation

Volume of 0.150 M HNO₃ needed: (x L)(0.150 mol/L) = 0.0075 mol and x = 0.05 L = 50 mls

NOTE: because these react in a 1:1 mole ratio, you could have used V1M1 = V2M2 as follows:

(x ml)(0.150 M) = (60 ml)(0.125 M) and x = 50 mls. But be careful using this method as it really only works if the acid and base react in a 1:1 mole ratio.

b) 3KOH + H₃PO₄ ==> K₃PO₄ + 3H₂O (note that the acid and base DO NOT react in a 1:1 mole ratio)

moles KOH present = 60.0 ml x 1 L/1000 ml x 0.125 mol/L = 0.0075 moles

moles H₃PO₄ needed = 0.0075 mol KOH x 1 mol H₃PO₄ / 3 mol KOH = 0.0025 moles

Volume of 0.200 M H₃PO₄ needed: (x L)(0.200 mol/L) = 0.0025 moles and x = 0.0125 L = 12.5 mls