What is solubility product constant?

Answer 1

At saturated solution of (partly) insoluble salt (imaginable: A2B) the following Equilibrium exists:

(A2B)s --> 2A+ + B2-

The thermodynamic Equilibrium constant Ksol equals: Ksol = [A+]2*[B2-] , the square hooks symbolise the ion concentration in mole/L. Note also the power in superscript [A]power which is equal to the number of ions in the (standardised) reaction equation. Further it should be noted that the amount of solid salt (A2B)s is not in the nominator of this Equilibrium equation, Ksol = [A+]2*[B2-] and that's why it is called (Equilibrium) solubility product.

Answer 2

Solubility is normally measured as the amount of a substance that can dissolve in a solvent and is measured in units of (amount of substance)/(volume of solvent).

Barium nitrate, for example, has a solubility of 0.105 moles/litre in water, or it can also be expressed as 27.3g/litre.

The solubility product is an equilibrium constant for the process:

salt (solid) <===> cations (aq) + ions (aq)

As the activity of the solid is taken to have a value of 1, the solubility product is the result of multiplying the concentrations of the dissolved ions at the solid's maximum solubility.

The solubility product (Ksp) can be thought of as a more general expression of the solubility of a substance. To obtain the solubility product, we multiply the concentrations (mol/litre)* of ALL the ions that result from the substance dissolving when they are at their maximum concentration.

In the case of barium nitrate, we have:

Ba(NO3)2 ---> Ba2+ + NO3- + NO3-

So the solubility product , Ksp = [Ba2+][NO3-][NO3-] = [Ba2+][NO3-]2

The experimentally determined value of the solubility product for Barium Nitrate is 4.64×10-3.

For barium nitrate dissolving in pure water, [NO3-]= 2[Ba2+], so we can write

Ksp = [Ba2+][NO3-]2 = 4[Ba2+]3 = 4.64×10-3

Solving this, [Ba2+] = 0.105 mol/litre

As [Ba2+] equals the concentration of barium nitrate that is dissolved, this is the solubility of barium nitrate in pure water.

The usefulness of the solubility product is that we can use it to look at the effects of ions from other sources than the salt. For example, if we try to dissolve barium nitrate in 10 molar nitric acid, we use the same expression as before:

Ksp = [Ba2+][NO3-]2 = 4.64×10-3

But this time [NO3-] = 10 mol/litre (approx), as that is already in solution, so:

100 [Ba2+] = 4.64×10-3

[Ba2+] = 4.64×10-5

The solubility of barium nitrate in 10M nitric acid is 4.64×10-5 Mol/litre, MUCH less than its solubility in pure water, and not something that you could determine from the solubility alone.

*Strictly, we should use activity rather than concentration.

Answer 3

This is product beween the ions concentration.