We have taken a saturated solution of AgBr , K_ sp is 12 10⁻¹⁴ . If 10⁻⁷ mole of AgNO ₃ are added to 1 litre of this solution, find conductivity (specific conductance) of this solution in terms of 10⁻⁷ Sm ⁻¹ units. Given, aligned _ ( Ag ⁺ ) ^ & =6 10⁻³ Sm ^2 ~mol ⁻¹, _ ( Br ⁻ ) ^ =8 10⁻³ Sm ^2 ~mol ⁻¹, _ ( NO ₃⁻ ) & =7 10⁻³ Sm ^2 ~mol ⁻¹, aligned

Step-by-step solution

Suppose the solubility of AgBr in 10⁻⁷ MAgNO ₃ is mol / L aligned & s ~mol / L AgBr Ag ⁺+ Br ⁻ & AgNO ₃ Ag ⁺+ NO ₃⁻ & 10⁻⁷ M 10⁻⁷ M 10⁻⁷ M & Total [ Ag ⁺ ]= (s+10⁻⁷ ) M & K_ sp of AgBr = [ Ag ⁺ ] [ Br ⁻ ] & 12 10⁻¹⁴= (s+10⁻⁷ )(s)=s^2+10⁻⁷ s & s^2+10⁻⁷ s-12 10⁻¹⁴=0 & aligned On solving s=3 10⁻⁷ M Hence, [ Br ⁻ ]=3 10⁻⁷ M =3 10⁻⁷ 10^3 ~m ^3=3 10⁻⁴ ~m ^3 [ Ag ⁺ ]= (3 10⁻⁷+10⁻⁷ ) M =4 10⁻⁷ 10^3 ~m ^3=4 10⁻⁴ ~m ^3 [ NO ₃⁻ ]=10⁻⁷ M =10⁻⁷ 10⁺³=1 10⁻⁴ ~m ^3 = k C or k= C aligned k_ Br ⁻ & =3 10⁻⁴ 8 10⁻³ Sm ⁻¹=24 10⁻⁷ Sm ⁻¹