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Answered 2012-02-22 15:02:44

it's soluble in water, so ksp is reaalllyy small--i wont be much of a help if you were looking for ksp to prove what i said;;

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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.

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

CaF2 Ca(2+) +2F(-) Ksp=(x)(4x^2) where x=solubility Therefore, Ksp=3.7 x 10^-11

Solubility product constant, Ksp, uses concentrations of soluble (dissolved) substances. A solid is not dissolved.

The Ksp for BaS04, which is barium sulfate, is 1.1 x 10^-10. Ksp is the solubility product. It is the product of the solubility of the ions in moles per liter.

due to the solubility product constant(ksp)

NaCl dissolve so easily that it is not even given a solubility product constant, as this value ( also known as the Ksp) tells us the solubility for compounds that don't readily dissolve.

Yes, AgBr is insoluble in water.You can use the Ksp (the solubility product) to determine the solubility.Ksp = 5.0 x 10^-13 and the equilibrium equation is AgBr(s) Ag+ + Br-This video explains it in great detail:www .

To determine the stability of an ion compound in water: look up the product solubility constant (Ksp) for the compound to be dissolved; write the chemical equation and modify the Ksp equation.

Solubility: the total amount of a solute dissolved in a solvent, at a given temperature, pressure and at saturation.Ex.: Sodium chloride - 359 g/L at room temperature.Solubility product: the concentration of the first ion multiplied by the concentration of the second ion, released by the dissociation of an ionic compound, at equilibrium. The appreciation is made with the solubility product constant:Ksp = [A].[B]Ex.: Lithium fluorideKsp = [Li+].[F-] = 3,8.10-3

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.

ksp= [Ca2+][Cl-]^2 = (x)((2x)^2) Ksp =4x^3 where x= the amount soluble of one mole of product

Ksp or solubility product is meaured for aqueous solutions of salts, for acids is Ka , for bases is Kb and for water is Kw.

Particular solubility. Depending on the properties of both solvent and solute, all combinations have a solubility constant, Ksp.

Equilibrium in the case of a generally insoluble salt takes the form of "Ksp," which means solubility product constant. Ksp = [Ca+2]3[PO4-3]2

the higher the Ksp value the more soluble a compound is.

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-]

Ion product < Ksp Unsaturated solution Ion product = Ksp Saturated solution Ion product > Ksp Supersaturated solution

Ksp= [Products]^mole ratio so Ksp=[Ag+][NO3] Since you didnt provide any numerical values, that's as far as you can go. You can look for known Ksp values at certain temperatures in some AP Chemistry books or online.

Radium sulfate (formula RaSO4) will dissolve at 0.00021 g/100g of water. To determine this: Get the solubility product constant Ksp of the values. This will indicate how much solid gets dissolve in the solution. Also consider the effect on solubility (for example, common ions).

Type your answer here... For AuCl, Ksp = 2.0 x 10-13 at 25°C. What is the solubility of AuCl at this temperature?

If the ion product concentration is greater than the Ksp value a precipitate will form. If it equals the Ksp the solution is saturated and no precipitate forms.

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