Predicting whether a precipitate will form

Consider the slightly soluble compound MX for which K_s is known. 
For a solution of MX at equilibrium in the presence of undissolved solid, the product of concentrations in Q_s is equal to K_s, the solubility constant for the compound.  For solutions that are not at equilibrium Q_s is NOT equal to K_s.
 

MX(s) M+(aq) + X–(aq) Qs = [M+][X–]

Thus comparing the value of Q_s with K_s for a solution containing the cation and anion of an sparingly soluble solid allows one to establish

• if the system is at equilibrium (Q_s = K_s) AND
• if Q_s is not equal to K_s to predict whether the amount of precipitate will be increased when equilibrium is reached

If the calculated Q_s is
 

• greater than K_s then the reaction that results in a lower concentration of the ions in solution (precipitation) is faster until Q_s becomes equal to K_s (and more precipitate forms)
   
• less than K_s then the reaction that increases the concentration of the ions in solution is faster until Q_s equals K_s.  If there is precipiate present it will dissolve.  If there is no precipitate present, no change will be observed.

Thus one can predict if mixing two solutions will give rise to a precipitate by calculating Qs in the mixed solution. The ion concentrations required are calculated by
 

1. calculating the amount in moles of each substance mixed using n = cV (c and V are the concentration and volume of the solution containing that before mixing).
2. dividing the result of 1 by the total volume in litres of the solution after mixing.