0.25 L Barium nitrate solution contains 0.02 mol nitrate ions
Further explanation
Given
0.04 M Barium nitrate, Ba(NO₃)₂(aq)
Required
The volume of Barium nitrate
Solution
Ionization of Barium nitrate in 1 L solution :
Ba(NO₃)₂ ⇒ Ba²⁺ + 2NO₃²⁻
mol :
0.04 0.04 0.08
There is 0.08 mol of nitrate ions in 1 L solution
So for 0.02 mol nitrate ions :
= 0.02/0.08 x 1 L
= 0.25 L