How will formation of an insoluble product shift the equilibrium?

[bamtori]

How...

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[Average Guy]

I'll try my best to explain it (if anyone sees a mistake here or can explain it better, post and I'll re-edit my comment). When a precipitate forms it is just a solid forming in solution. Solids are NEVER in the equation for equilibrium. Since that is the case, the formation of that solid essentially results in the loss of product (it's not exactly lost, just not in the equilibrium equation because it's a solid and not part of the solution). So think of it as removing the precipitate from the resulting solution. And of course when applying Le Chatelier's principle the loss of product means the creation of more product and so the equilibrium will shift to the right of the equation.

 

[BigCountry]

Here's my take, equilibrium expressions (Keq or Ksp) have equations products/reactants. However only aqueous and gases are included, no solids. Something that is insoluble is basically like saying a solid. Solids are not in the equation, therefore it has NO EFFECT.

 

[dentalstudentprospect2]

When looking at chemical equilibrium, we have to use K which is a function of concentration. Formation of a solid essentially takes species out of solution thus lowering its concentration. The reaction will correct itself by shifting to restore the balance of concentrations. So if you form a solid product, you lose the concentration of product species and the reaction must shift to the right in order to maintain the Keq value.