Question

To 1 L of water, 3.0x10^-6 mol of Pb(No3)2, 4.0x10^-5 mol of k2CrO4 and 1.0 mol NaCl are added. What will happen? ignore the increase in volume due to added substances

Answer 1

Ans:

[Pb(NO₃)₂ ] = [Pb²⁺] = 3.0 x 10^-6 mole / L = 3.0 x 10^-6 M

[K₂CrO4 ] =  [CrO₄^2-] =4.0 x 10^-5 mole / L = 4.0 x 10^-5 M

[NaCl] = 1.0mol / L = 1.0 M

BY MIXING THESE THREE SALTS WE MAY GET TWO SPARINGLY SOLUBLE SALTS PbCl₂ or PbCrO₄ or BOTH.

Ksp = solubility product constant or Solubility product = Constant at constant temperature.

= It is ionic product of saturated solution of sparingly soluble salt.

Ksp (PbCl₂) = 1.7 x 10^-5

Ksp (PbCrO₄) = 2.3 10^-13

When,

ionic product < Ksp (Solution is unsaturated and no ppt formed )

ionic product = Ksp ( Solution is saturated and no ppt formed )

Ionic product > Ksp ( Solution is holding more amount of ions than solubility of the salt and hence solution is supersaturated and unstable which immediately gives presipitate of the salt (solid) )

Ionic product = [Pb²⁺] [CrO₄^2-] =   3.0 x 10^-6 M x 4.0 x 10^-5 M = 1.2 x 10^-10 M²

Ksp (PbCrO₄) = 2.3 10^-13 M²

So, Ionic product > Ksp and hence Ppt of Lead chromate will be formed.

Ionic product = [Pb²⁺] [Cl ^- ] ²  = 3.0 x 10^-6 M x (1.0 M)² = 3.0 x 10^-6 M³

Ksp (PbCl2) = 1.7 x 10 ^-5 M³

So, Ionic product   <   Ksp and hence Ppt of Lead chloride will not form.

Therefore by mixing all three salts in 1 L water only Lead Chromate will precipitate out

and solution contains dissolved ions Pb²⁺, K+, Na⁺, Cl^-, CrO₄^2-

Note that although lead chromate precipitate out it is always in equilibrium with aq. Pb²⁺ and aq. CrO₄^2-