An aqueous solution of nitric acid is standardized by titration with a 0.167 M solution of barium hydroxide. If 11.6 mL of base are required to neutralize 11.9 mL of the acid, what is the molarity of the nitric acid solution?

Question

An aqueous solution of nitric acid is standardized by titration with a 0.167 M solution of barium hydroxide.   If 11.6 mL of base are required to neutralize 11.9 mL of the acid, what is the molarity of the nitric acid solution?

anonymous · Answer

11.6 mL = 0.0116 L
( 0.167 M barium hydroxide)(0.0116 L barium hydroxide) = 0.00194 mol barium hydroxide.

2HNO3 + Ba(OH)2 --> Ba(NO3)2(aq) + 2HOH 

(0.00194 mol barium hydroxide)(2 mol HNO3/1 mol Ba(OH)2) =  0.00388 mol nitric acid.

11.9 mL = 0.0119 L

0.00388 mol nitric acid / 0.0119 L nitric acid = 0.326 M nitric acid.

anonymous · Answer

Next time I would post that in Chemistry instead of Mathematics :D Also welcome to OS!! Check out Code of Conduct: http://openstudy.com/code-of-conduct