A chemist needs to determine the concentration of a solution of nitric acid, HNO

Question

A chemist needs to determine the concentration of a solution of nitric acid, HNO3. She puts 835 mL of the acid in a flask along with a few drops of indicator. She then slowly adds 0.600 M Ba(OH)2 to the flask until the solution turns pink, indicatig the equilvalence point of the titration. She notes that 245 mL of Ba(OH)2 was needed to reach the equilvalence point.

In this titration, the concentration of base is known and can be used to calculate the unknown acid concentration:

concentration of base--> moles of base---> moles of acid----->concentration of acid

How many moles of Ba(OH)2 are present in 245 mL of 0.600 M Ba(OH)2?

Explanation / Answer

Number of moles of Ba(OH)2 , n = Molarity x volume in L

= 0.600 M x 245 mL x 10-3 L/mL

= 0.147 mol

Ba(OH)2 + 2 HNO3 ----> Ba(NO3)2 + 2 H2O

From the balanced reaction,

1 mole of Ba(OH)2 reacts with 2 moles of HNO3

0.147 moles of Ba(OH)2 reacts with 2x0.147 =0.294 moles of HNO3

So concentration of HNO3 = number of moles / volme in L

= 0.294 mol / 0.835 L

= 0.352 M

The number of moles of Ba(OH)2 are present in 245 mL of 0.600 M Ba(OH)2 is , n = Molarity x volme in L

n = 0.600 M x 0.245 L

= 0.147 moles

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