chemistry Le Châtelier’s principle help?
A buffered solution is made by dissolving hydrofluoric acid and sodium fluoride in water. The equilibrium equation is HF(aq) + H2O(l) <--><--> H3O+(aq) + F-(aq)
Use Le Châtelier’s principle to describe what happens to properties of the solution when a strong base is added to it.
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HF <==> H+ + F-
If OH- is added to the solution, then H+ will react and cause a shift to the right. As H+ react with OH-, the H+ concentration will drop and the equilibrium will shift to the right in an attempt to bring the H+ concentration back closer to what it was (which is why the pH stays nearly constant), while decreasing the concentration of HF and increasing the concentration of F-.
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I fear that Rangan is off the beaten track. A catalyst is a substance that speeds up a chemical reaction by providing an alternate pathway with a lower activation energy. The added OH- ion is NOT a catalyst.
A catalyst speeds up BOTH the forward and reverse reactions for a system at equilibrium. By definition, for a system at equilibrium, the rate of the forward reaction is equal to the rate of reverse reaction. You can't speed up only half an equilibrium.
Is it something to do with the acid (hydrofluoric) dissociating in the presence of the base to create more water molecules and thus more moles on the LHS of the equation. If this is the case then there will be a higher proportion of products in the solution as the reaction moves to oppose the change.
I'm not sure if that is correct though, sorry!
Good luck :)
A strong base will work as a catalyst. now when you add a catalyst in a le chatlier's principal, (no matter if the pressure is low or high) the catalyst will increase the rate of forward reaction only if the /_\ n (del n) is greater than zero. AND it will produce H3o randomly