Question

1. At 25 deg Celsius, the solubility product constant of PBI2 is 1.4 x 10^-8. Calculate the molar solubility in water.

2. The solubility of AgI at 25 deg Celsius is 9.1 x 10^-4 M. Calculate the Ksp of AgI at that temperature.

Answer 1

Ans (1) The Solubility product of PbIg is 1.4 X10 Molar Solubility in water = ? PbIg is sparingly soluble Salt at 25°C temperature, The equillibouium equation for Pb Ig can be written as - P61909 = Pb2+ (aq) +21" S 25 Here 'S'be molası solubility and Ksp is solubility product and is equal to the paroduct of [Pb + cars] and [I” Cag: )) Ksp = [ Pb2+ ] [1")* 1-4X10% = .(25)2 453 14x108 so = .4x16% 0.35 X108 s= /3.5 8109 s = 1151 X10M] The molar solubility in water of Pb Is is 1.51 M

Ans. The solubility of AgI = 9.1810 M 1 ksp of Agi = ? The equillibrium equation can be written as - AgI Ag + +1 S S Molay Solubility 's' is 9-1x109 m Ksp = [Ag+] [I] Ksp - sos Ş S= 9.1x10 4M Ksp = (9.1810-4) ² = s² Ksp = 82-81X108 Ksp = 8.281 X10** So, the solubility product of Ho I is 8.38! X16