Question

The maximum amount of lead sulfate that will dissolve in a 0.241 M lead nitrate solution is

Answer 1

Consider a dissoliution of calcium oxalate as

PbSO_4(s) Pb²⁺_(aq) + SO₄^2-_(aq)

Expression for K_sp is given as   K_sp = [ Pb²⁺] [SO₄^2-]

Let x mol / dm³ be the solubility of   Pb²⁺ and  SO₄^2- but solution already have Pb²⁺ions.Thus  [ Pb²⁺]= x + 0.241

Hence    K_sp = ( x + 0.15 ) (x) = 1.6 x 10^-8 ( K_sp of PbSO₄ at 298 K = 1.6 x 10^-8)

Here    K_sp << 0.241 hence we can write x+0.241 0.241 then

K_sp = x ( 0.241 ) = 1.6 x 10^-8

x = 1.6 x 10^-8/ 0.241

x = 6.64 x 10 ^-8 M

molar solubility of PbSO₄= 6.64 x 10 ^-8 M

Here due to common ion effect solubility decreases.