Question

Concentration of SO₄^-2 in Ba⁺² solution
M₁V₁ = M₂V₂
1 $\times$50 = M₂ $\times$ 500
M₂ = $\frac{1}{10}$
for just precipitation
I.P = Ksp
[Ba⁺²] [SO₄^-2] = K_sp (BaSO₄)
[Ba⁺²] $\times$$\frac{1}{10}={10}^{-10}$
[Ba⁺²] = 10^-9 Min 500ml solution
For calculation of [Ba⁺²] in original solution (450 ml)
M₁ $\times$ 450 = 10^-9 $\times$ 500
M₁ = $\frac{500}{450}\times {10}^{-9}$ = 1.11 $\times$ 10^-9 M
[M₁= molarity of Ba⁺² in original solution  (450ml)]