In a titation experiment 16.7 ml of .759M HCOOH neutralizes 23.1 mL of BA(OH)2. What is...

Question

Question

In a titation experiment 16.7 ml of .759M HCOOH neutralizes 23.1 mL of BA(OH)2. What is the concentration of the BA(OH)2 soln?

Answer 1

Answer #1

2 HCOOH + Ba(OH)2 ---------------> Ba(CHOO)2 + 2H2O

Molarity of HCCOH M1 = 0.759 M

volume of HCOOH V1 = 16.7 mL

moles of HCOOH n1 = 2

molarity of Ba(OH)2 = ??

volume of Ba(OH)2 = 23.1 ml

moles = 1

from the relation

M1 V1 / n1 = M2 V2 / n2

0.759 x 16.7 / 2 = M2 x 23.1 / 1

M2 = 0.274 M

concentration of Ba(OH)2 = 0.274 M

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Answer 2

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