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Dorslek
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So it seems I've been seeing two different ways of interpreting buffer solution reactions. One to me seems to work with Le Chatelier's Principle and the other doesn't.
I am dropping a strong base, NaOH, into a weak acid/base buffer solution.
View A:
HA + H2O <--> (H3O+) + (A-)
Reasoning: The (OH-) of the NaOH will end up consuming the (H3O+) and turning into H2O thus the pH isn't really effected. In response to the declining concentration of (H3O+), the equilibrium will shift to the right to produce more of the products and less of the reactants.
View B:
HA + (OH-) <--> H2O + (A-)
Reasoning: The (OH-) will start consuming HA on the reactant side thus the pH is not changed as (OH-) is not produced.
My confusion stems in how these two are connected. View A makes complete sense to me however View B is used often as well. View A seems to deal with the (OH-) consuming the product versus View B where (OH-) is consuming the reactant.
The shifts in equilibrium also appear to be different. In View A the equilibrium is from L -> R to make up for a decreasing (H3O+) concentration to produce more of (H3O+) and (A-). In View B, there would be less reactant so it would seem then that the equilibrium shift would be from R -> L to produce more HA.
Thanks for the help, appreciate it
I am dropping a strong base, NaOH, into a weak acid/base buffer solution.
View A:
HA + H2O <--> (H3O+) + (A-)
Reasoning: The (OH-) of the NaOH will end up consuming the (H3O+) and turning into H2O thus the pH isn't really effected. In response to the declining concentration of (H3O+), the equilibrium will shift to the right to produce more of the products and less of the reactants.
View B:
HA + (OH-) <--> H2O + (A-)
Reasoning: The (OH-) will start consuming HA on the reactant side thus the pH is not changed as (OH-) is not produced.
My confusion stems in how these two are connected. View A makes complete sense to me however View B is used often as well. View A seems to deal with the (OH-) consuming the product versus View B where (OH-) is consuming the reactant.
The shifts in equilibrium also appear to be different. In View A the equilibrium is from L -> R to make up for a decreasing (H3O+) concentration to produce more of (H3O+) and (A-). In View B, there would be less reactant so it would seem then that the equilibrium shift would be from R -> L to produce more HA.
Thanks for the help, appreciate it
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