a) What volume of a 0.180 M nitric acid solution is required to neutralize 22.4

Question

a) What volume of a 0.180 M nitric acid solution is required to neutralize 22.4 mL of a 0.172 M barium hydroxide solution? (Answer should be in mL)

b)What volume of a 0.154 M calcium hydroxide solution is required to neutralize 12.7 mL of a 0.181 M hydroiodic acid solution? (Answer in mL)

c)An aqueous solution of hydroiodic acid is standardized by titration with a 0.182 M solution of barium hydroxide. If 17.7 mL of base are required to neutralize 21.9 mL of the acid, what is the molarity of the hydroiodic acid solution?

d) An aqueous solution of barium hydroxide is standardized by titration with a 0.184 M solution of nitric acid.
If 12.3 mL of base are required to neutralize 17.9 mL of the acid, what is the molarity of the barium hydroxide solution?

Explanation / Answer

Since you have 2 that say 0.172M and 2 that say 0.18M, I thought I'd break the tie. Quite simply it is calculated as 0.0224 L x 0.172 moles/L = 0.00385 moles of HNO3 according to balanced equation you need 1/2 of that amount of Ba(OH)2 or 0.00192 moles 0.00192 moles/x L = 0.18 M no of liters =0.01067 hence volume =10.67 mL

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