Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: confused???? on May 12, 2008, 01:13:46 PM
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how do you write the Bronsted Lowry ionic equation for the reaction that occurs when NaOH(aq) is added to a buffer solution containing acetic acid and potassium acetate??
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What happens in that solution?
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how do you write the Bronsted Lowry ionic equation for the reaction that occurs when NaOH(aq) is added to a buffer solution containing acetic acid and potassium acetate??
Key features of buffer
Salt of weak acid - strongly dissociated. Look at ionic products.
Weak acid - slightly dissociated - an equilibrium state. Look at the reactant and the ionic products.
Understand the interaction of the products from both reactions and which products are in large concentration and which are small when weak acid mixed with solution of its potassium salt.
Then think about adding [H+] and apply Le Chatelier to understand where the H+ "goes"
Then think about adding [OH-] and apply Le Chatelier to understand where the OH- "goes"
Clive
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so would the equations look like this?
NaOH + OH- (arrow) H3O + NaO
KCH3Coo + H2O (arrow) H3O + KCH2COO
I think this is right if not could you please tell me what I am doing wrong
Thanks
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After mixing you have solution that contains:
potassium acetate - salt
sodium hydroxide - strong base
acetic acid - weak acid
Do you see a pair of compounds that should react immediatley?
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so would the equations look like this?
NaOH + OH- (arrow) H3O + NaO
KCH3Coo + H2O (arrow) H3O + KCH2COO
I think this is right if not could you please tell me what I am doing wrong
Thanks
I may be giving too much away here .
Maybe if you look at the reactions in the order of my questions it might help sort things out. (I'm writing the equations slightly unconventionally to emphasise ions)
Salt of weak acid - strongly dissociated. Look at ionic products.
CH3COO-K+ --> CH3COO- + K+
Weak acid - slightly dissociated (ignoring water to simplify)- an equilibrium state. Look at the reactant and the ionic products.
CH3COOH <-> CH3COO- + H+
Understand the interaction of the products from both reactions and which products are in large concentration and which are small when weak acid mixed with solution of its potassium salt.
This is the clincher
Note that CH3COO- is a product of both reactions but the equilibrium is different for both.
Final hint: the salt dissociation dominates
The standard argument goes (look up other examples in your test book):
Now .
1. What happens to the equilibrium of both reactions under the condition of a lot of CH3COO-? (Think Le Chatelier)
2. What happens when you add H+? (Le Chat again)
3. What happens when you add OH-? (Le Chat again !! )
I have seen arguments that use the equilibrium constant to explain shift in equilibrium position - and I don't understand them
Clive
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I have seen arguments that use the equilibrium constant to explain shift in equilibrium position - and I don't understand them
Please elaborate - where is the problem? Any argument that refers to the equilbrium constant is just a more strict approach to the LeChatelier's principle.
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I have seen arguments that use the equilibrium constant to explain shift in equilibrium position - and I don't understand them
Please elaborate - where is the problem? Any argument that refers to the equilbrium constant is just a more strict approach to the LeChatelier's principle.
Say
2A + D <--> C
then Kc = [C]
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[A]2[D]
If concentration of C is increased independently of the other reactants then my textbook talks about "concentration term" or "reaction quotient" (without defining them) having to change and "return" to Kc
This seems to me to be no different from Le Chatelier's principle with numbers. This "Kc approach" doesn't explain the behaviour of the system - it just gives you a method, no better than Le Chat to determine which direction the reaction will "go" to re-establish equilibrium.
It doesn't explain whether the concentrations of all participants adjust to new levels to give the original Kc value.
In other words why is Kc re-established?
It just seems a bit "empty" as an explanation.
Clive
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If concentration of C is increased independently of the other reactants then my textbook talks about "concentration term" or "reaction quotient" (without defining them) having to change and "return" to Kc
Concentration is just concentration, reaction quotient is exactly the same formula you used for Kc. Difference is, reaction quotient (denoted often by Q) doesn't have to be equal to Kc - if it is not, reaction will proceed till Q=Kc.
This seems to me to be no different from Le Chatelier's principle with numbers. This "Kc approach" doesn't explain the behaviour of the system - it just gives you a method, no better than Le Chat to determine which direction the reaction will "go" to re-establish equilibrium.
It is no different - but it deals with numbers so it is much better! Le Chatelier's principle will tell you "reaction will shift to the left". KC tells you exactly HOW FAR it will shift.
It doesn't explain whether the concentrations of all participants adjust to new levels to give the original Kc value.
In other words why is Kc re-established?
LCP doesn't tell you "why" either. For that you need thermodynamical approach and chemical potential. LCP is just a simple, qualitative rule of thumb, distilled from the the more complete theory.
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This seems to me to be no different from Le Chatelier's principle with numbers. This "Kc approach" doesn't explain the behaviour of the system - it just gives you a method, no better than Le Chat to determine which direction the reaction will "go" to re-establish equilibrium.
It is no different - but it deals with numbers so it is much better! Le Chatelier's principle will tell you "reaction will shift to the left". KC tells you exactly HOW FAR it will shift.
Aha !! ** Sound of scales falling from eyes **
Thanks
Clive
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how do you write the Bronsted Lowry ionic equation for the reaction that occurs when NaOH(aq) is added to a buffer solution containing acetic acid and potassium acetate??
How is the answer to the original question getting along?
Clive
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Thanks for all the help I think I got it now!!
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Thanks for all the help I think I got it now!!
So can you post what you believe to be the correct approach for future reference?
Clive