Question

The solubility of pbso4 water is 0.038g/l.Calculate the solubility product constant of pbso4

Answer 1

When the solubility of PbSO₄ water is 0.038g/l. To Calculate the solubility product constant of  lead (II) sulphate, we need to find the number of moles of  of lead (II) sulphate then write the equation for the dissociation of it as well.

Number of moles= mass/ relative formula mass

relative formula mass of  PbSO₄ =208 + 32 + 64 =304

therefore moles = 0.038/304

                           = 0.000125 moles

dissociation of  PbSO₄ ⇒Pb²⁺ + SO₄²⁻

        ksp=(0.000125)²

             = 1.5 x 10^-8