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-]
The solubility of BaCO3 can be calculated by taking the square root of the Ksp value, which is 7.94 x 10^-5 mol/L. This represents the maximum amount of BaCO3 that can dissolve in water at equilibrium.
The solubility of AlPO4 can be calculated by taking the square root of its Ksp value. In this case, the solubility of AlPO4 is equal to approximately 3.13 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
To calculate the molar solubility of copper(II) sulfide, you need to consider the solubility product constant (Ksp) of CuS. Once you have the Ksp value, set up an equilibrium expression for the dissociation of CuS into Cu^2+ and S^2- ions. Use the initial concentration of CuCl2 to determine the concentration of Cu^2+ ions and then solve for the molar solubility of CuS.
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