Answer:
0.031M of Ba(OH)₂
Step-by-step explanation:
The reaction of nitric acid (HNO₃) with barium hydroxide (Ba(OH)₂) is:
2 HNO₃ + Ba(OH)₂ → 2H₂O + Ba(NO₃)₂
Where 2 moles of nitric acid reacts per mole of barium hydroxide
Moles of nitric acid used to neutralize Ba(OH)₂ are:
0.0250L ₓ (0.10mol / L) = 0.00250 moles of HNO₃
As 2 moles of acid reacts per mole of base:
0.00250 moles of HNO₃ ₓ (1 mole Ba(OH)₂ / 2 moles HNO₃) = 0.00125moles of Ba(OH)₂
In 0.0400L:
0.00125 moles Ba(OH)₂ / 0.0400L = 0.031M of Ba(OH)₂