![]() ![]() As the above hopefully shows, the common ion effect is governed by Le Chatelier’s principle.This would produce more OH - ions to rebalance those that were lost when Pb(OH) 2 (s) started forming. The same could be done with the OH - ions dissolved adding Pb(NO 3) 2 to the solution would cause Pb(OH) 2 (s) to precipitate and the equilibrium will shift to the right. Remember that Na like NO 3 - \, is a spectator ion and will not form a precipitate! The Ca(OH) 2 equilibrium will respond by shifting to the right to produce more Ca 2 ions. If we added sodium carbonate, Na 2CO 3, we would begin to precipitate CaCO 3 (s) while reducing the Ca 2 (aq) concentration. Using a solubility table, we can see that compounds containing sodium ions (Na ) will dissolve in water, and that calcium carbonate, CaCO 3, has low solubility. To do this, we need to add a compound that will reduce the amount of Ca 2 or OH - ions in solution by precipitating one of the ions out of solution. To increase the solubility of Ca(OH) 2, we need to do the opposite the equilibrium must shift to the right and favor the dissolving reaction. ![]() When this happens, we will have decreased the solubility of Ca(OH) 2 as more of it is in precipitate form now. To maintain the K sp concentration of Ca(OH) 2 at equilibrium, the equilibrium must shift to the left (favoring the crystallization reaction) and makes more Ca(OH) 2 (s). But the extra Ca 2 (aq) will now disturb the Ca(OH) 2 equilibrium: Using a solubility table and our Predicting the solubility of salts lesson recall that compounds with a nitrate (NO 3 -) anion are highly soluble in water, If we added some calcium nitrate (Ca(NO 3) 2) to the solution, the following happens:Ĭa(NO 3) 2 (s) \, → \, Ca 2 (aq) 2NO 3 - (aq)Īs it is highly soluble, this is not an equilibrium, it is a straightforward dissolving process. Applying Le Chatelier’s principle, increasing or will cause the system to shift away from them, which is towards more Ca(OH) 2 (s). To do this, we add a compound that will dissolve to produce more Ca 2 or OH - \, ions in solution. To decrease the solubility of Ca(OH) 2, we need the equilibrium to shift to the left and favor the precipitate. For example, the salt calcium hydroxide, Ca(OH) 2, when saturated has the equilibrium:Ĭa(OH) 2 (s) ⇌ \, \rightleftharpoons \, ⇌ Ca 2 (aq) 2OH - (aq).This is important in predicting how the solubility will change. Like any process at equilibrium, the common ion effect is governed by Le Chatelier’s principle. The common ion effect is a way to change the solubility of a compound by adding a soluble salt that has an ion in common with the compound you are trying to change the solubility of.Even though K sp cannot change, there are ways to change solubility in a given solution without changing the temperature! This is useful for when we want to reduce or increase the solubility of some compounds. ![]() The reverse reaction is the crystallization process that changes the substance from aqueous back to the solid state, as a precipitate: The forward reaction is the dissolving process that changes the substance from solid to aqueous state: ![]() This can be broken up into the individual forward and reverse reactions. M mX x (s) ⇌ \, \rightleftharpoons \, ⇌ M x (aq) X m- (aq) Whenever we talk about a compound with low solubility or a saturated solution, we should always write the equilibrium that has been created.How pH and thermodynamics influence the solubility of salts.How to use a solubility table to suggest ways to increase the solubility of a saturated solution.How to use the common ion effect to decrease the solubility of a saturated solution. ![]()
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