Chatelier’s principle that if we increase the concentration of any one of the ions, it should combine with the ion of its opposite charge and some of the salt will be precipitated till once again K sp = Q sp . Similarly, if the concentration of one of the ions is decreased, more salt will dissolve to increase the concentration of both the ions till once again K sp = Q sp . This is applicable even to soluble salts like sodium chloride except that due to higher concentrations of the ions, we use their activities instead of their molarities in the expression for Q sp . Thus if we take a saturated solution of sodium chloride and pass HCl gas through it, then sodium chloride is precipitated due to increased concentration (activity) of chloride ion available from the dissociation of HCl.
Sodium chloride thus obtained is of very high purity and we can get rid of impurities like sodium and magnesium sulphates. The common ion effect is also used for almost complete precipitation of a particular ion as its sparingly soluble salt, with very low value of solubility product for gravimetric estimation. Thus we can precipitate silver ion as silver chloride, ferric ion as its hydroxide (or hydrated ferric oxide) and barium ion as its sulphate for quantitative estimations. Dissolution of S mol/L of Ni(OH) provides S mol/L of Ni + and 2S mol/L of OH – , but the total concentration of OH – = ( .
+ 2S) mol/L because the solution already contains . mol/L of OH – from NaOH. K sp = . × – = [Ni + ] [OH – ] = (S) ( .
+ 2S) As K sp is small, 2S << . , thus, ( . + 2S) ≈ . Hence, .
× – = S ( . ) S = . × – M = [Ni + ] Problem . Calculate the molar solubility of Ni(OH) in .
M NaOH. The ionic product of Ni(OH) is . × – . Let the solubility of Ni(OH)