Solubility and K_sp

While we have only discussed solubility in terms of "soluble" or "not soluble" in this web site, scientists have again assigned a quantitative measurement to what has so far been a qualitative observation. This measurement once again involves our old friend, K the equilibrium constant.

Normally, we measure equilibrium constants by calculating the concentration of the products over the concentration of the reactants. However, in a dissociation reaction such as the following,

  NaCl(s) Na⁺(aq) + Cl^-(aq)

the reactant, NaCl, is a solid. Solids are never included in equilibrium constants, so the new constant is simply the concentrations of the products multiplied together. So, for the dissociation reaction:

  A_xB_y(s) X A⁺(aq) + Y B^-(aq)

the solubility equilibrium constant, called the solubility product, is:

  K_sp = [A⁺]^x[B^-]^y

If you know the solubility product of a certain solid and how it dissociates (which and how many ions are produced), you can calculate the concentrations of all of the product ions; further, you can mathematically predict what certain changes, such as adding more of one product, will affect the equilibrium. In short, the solubility product lets you do with dissolution reactions what you can already do with other equilibrium reactions.

For example, AuI₃ dissociates as shown below:

  AuI₃ Au³⁺ + 3 I^-    K_sp = 1.0 x 10^-46

What is the concentration of iodide (I^-) if some AuI₃ is placed in water when the system reaches equilibrium? Each AuI₃ that dissociates will give an aluminum ion and three hydroxide ions. Let's call the concentration of Au ion "x" and the concentration of iodide "3x," because iodide concentration will be three times larger. Now, plug these values into the solubility product expression:

  1.0 x 10^-46 = [Au³⁺][I^-]³ = [x][3x]³

Multiplying out the cubed term gives 27x³, which becomes 27x⁴ when multiplied by the Al ion concentration. The final expression with the solubility product becomes:

  1.0 x 10^-46 = 27x⁴

Some easy division and 4th-rooting (with your handy scientific calculator) gives a value for x:

  x = 1.387 x 10^-12

Multiplying this value by three gives us the iodide concentration:

  [I^-] = 4.162 x 10^-12

| |

home site map/index background information chemical reactions states of matter acid/base reactions

thermodynamics redox reactions atoms and molecules organic chemistry nuclear chemistry

help features reference thinkquest