Respuesta :

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

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