What volume of a 0.240 m solution of barium nitrate is needed to prepare 0.500 l a solution that is 0.0800 m in nitrate ion?

Formula for Barium Nitrate = Ba (NO3) 2

Thus based on stoichiometry:

1 mole of Ba (NO3) 2 contains 2 moles of NO3-

Therefore, concentration of nitrate ion NO3 - would be = 2*0.240 = 0.480 M

Use the relation:

V1M1 = V2M2

V1 = V2M2/M1 = 0.500 L * 0.0800/0.480 = 0.0833 L

Thus, 0.0833 L or 83.3 ml solution of Ba (NO3) 2 would be required.