The solubility of lead chromate : s= 5.29 x 10⁻⁷ mol/L
Further explanation
Given
Ksp for PbCl,=2.8x10*-13 mole/dmº
PbCrO4
Required
The solubility
Solution
P b C r O ₄ ( s ) ⇌ P b²⁺( a q) +C r O ₄ ⁻²( a q )
s s s
s= solubility
Ksp P b C r O ₄ = s²
K s p = [ P b ²⁺] [C r O ₄ ⁻² ]= 2.8x10⁻¹³
s = √2.8x10⁻¹³
s= 5.29 x 10⁻⁷ mol/L