Ba(IO₃)₂(s) partially dissociates in water as Ba²⁺(aq) and IO₃⁻(aq).
Ba(IO₃)₂(s) ⇄ Ba²⁺(aq) + 2IO₃⁻(aq)
Initial
Change -X +X 2X
Equilibrium X 2X
Ksp = [Ba²⁺(aq)] x [IO₃⁻(aq)]²
6.0 x 10⁻¹⁰ = X * (2X)²
6.0 x 10⁻¹⁰ = 4X³
X = 5.313 x 10⁻⁴ mol/L
Hence, the solubility of the Ba(IO₃)₂(s) is 5.313 x 10⁻⁴ mol/L
The answer is same atomic number!
Answer:
1.2 M
Explanation:
If you use the dilution equation (M1V1=M2V2), you end up with (50)(12)=(500)(M2), and when you solve for M2 you get 1.2 M.
