What causes a saturated solution to form?

1 Answer
Feb 6, 2017

Answer:

#"The intrinsic solubility of the solute."#

Explanation:

By definition, a saturated solution defines a condition where the solute, the material in solution, is in equilibrium with UNDISSOLVED solute, i.e. a solution where the following equilibrium operates:

#MX_2(s) rightleftharpoons M^+ + 2X^-#

A temperature is usually specified, because a hot solution can (generally) dissolve more solute than a cold one. Where sparingly soluble salts are employed, often a #"solubility product"#, #K_"sp"# is reported, which is another equilibrium expression, such that,

#[M^(+)][X^-]^2=K_"sp"# (and here, since #[X^-]=[2M^+]#, if we wished to assess the solubility #S# of the salt, we would write, #K_"sp"=Sxx2S^2=4S^3#, and eventually that #S=""^(3)sqrt((K_"sp")/4)#.........

Solubility products are extensively tabulated for sparingly soluble, and precious metal salts. Why? Because suppose you had to dispose of a solution of lead or mercury salts. You don't want to throw too much of these salts (in solution) down the drain. Likewise, if you wanted to isolate gold or rhodium or platinum salts, you would want to make sure that your mother solution did not contain too much of the precious metal before you disposed of it.

Please note that the definition of a saturated solution as one where the solution #"all the solute that it can"#, is #"INADEQUATE"# and #"INCORRECT"#. Sometimes a #"supersaturated solution"# can be formed, such that, given the scenario above,

#[M^(+)][X^-]^2>K_"sp"#

These solutions are metastable, and can often be restored to equilibrium, to saturation, by scratching the sides of the glass, or adding a seed crystal.