C.2.1 x 10-4
2.7 10-2
Since Lead (II) Chloride has the formula PbCl2, the equilibrium equation for its dissolution is: PbCl2 <=> Pb+2+2Cl- so the equilibrium-constant expression is Ksp= [Pb+2][Cl-]
AlCl3 is soluble in water so to find the solubility of Alcl3 ( not KSp) ,the among of this compound dissolving in definite volume of water should be given.
Ksp= [Ag]^2 [CrO4] / [Ag2CrO4]
ksp= [Ca2+][Cl-]^2 = (x)((2x)^2) Ksp =4x^3 where x= the amount soluble of one mole of product
due to the solubility product constant(ksp)
Since Lead (II) Chloride has the formula PbCl2, the equilibrium equation for its dissolution is: PbCl2 <=> Pb+2+2Cl- so the equilibrium-constant expression is Ksp= [Pb+2][Cl-]
7.9 10-5
9.9 x 10-11
1.2x10-2
Ksp
the higher the Ksp value the more soluble a compound is.
It gives us an indication of its solubility in water. A large solubility constant (Ksp) means it is easily water-soluble. A small Ksp means it is generally insoluble in water.
1.0 x 10-12
1.2x10-2
You need the Ksp of copper sulphide. From that you can use the equation for solubility product - Ksp = [Cu2+].[S-] where the Cu2+ term becomes 25M.
4.5 x 10<sup> -7 </sup>
Solubility Product Constant, Ksp is the equilibrium constant for a solid substance dissolving in an aqueous solution. Molar solubility is the number of moles of a substance (the solute) that can be dissolved per liter.MnAm⇔nMm++mAn-Ksp = [Mm+]n[An-]m