Question

When we mix together, from separate sources, the ions of a slightly soluble ionic salt , the salt will precipitate if ${{{Q}}_{{{sp}}}}..........{{{K}}_{{{sp}}}}$ , and will continue to precipitate until ${{{Q}}_{{{sp}}}}..........{{{K}}_{{{sp}}}}$?
(A) is greater than; equals
(B) is less than; is greater than
(C) is less than; equals
(D) equals; is less than
(E) equals; is greater than

Answer 1

We have to calculate the ionic product first. Then we should know the conditions of precipitation when it depends on the ionic product. The relation of ionic product and solubility product is very important for predicting precipitation.

Complete answer:
First we will take a reaction and try to understand what an ionic product is. The reaction is Calcium carbonate dissolved in water to form aqueous solution:
${CaCo_3}_{(aq)}$ $\rightarrow$ ${Ca^{2+}}_{aq}$ + ${{Co_3}^{2-}}_{(aq)}$
Here on dissolving in water the ions dissociates and ionic product [Q] is as given below
${{Q = [C}}{{{a}}^{{{2 + }}}}{{] [CO}}_{{3}}^{{{2 - }}}{{]}}$
The ${{{K}}_{{{sp}}}}$ of the product will be given already for example for ${{CaC}}{{{O}}_{{3}}}{{ , }}{{{K}}_{{{sp}}}}{{ = 4}}{{.8 \times 1}}{{{0}}^{{{ - 9}}}}$
Now when instead of water the calcium nitrate is dissolved in sodium carbonate then formation of calcium carbonate will be there, so will the calcium carbonate precipitate or not.
So the conditions are :
When the ionic product at equilibrium is greater than the solubility product the precipitate will start forming.
${{Q > }}{{{K}}_{{{sp}}}}$ precipitate starts forming
When the ionic product at equilibrium is equal to the solubility product the precipitate can be formed till that point.
${{Q = }}{{{K}}_{{{sp}}}}$ precipitate forms till it is equal.
When the ionic product is less than the solubility product the precipitate will not form.
${{Q < }}{{{K}}_{{{sp}}}}$ precipitate does not form.

Hence, the answer to the above question is option A.

Note: For insoluble salts the conditions are different. In insoluble salt the ionic product should be greater than ${{1}}{{{0}}^{{{ - 10}}}}$ for the salt to precipitate and continue to precipitate till ${{1}}{{{0}}^{{{ - 10}}}}$. If the value becomes less then no precipitation.It is to be noted that these conditions are only applied for sparingly soluble salt, it does not apply upon completely soluble or insoluble salts. Solubility products tell us how much moles of salt is soluble in one litre of solvent.