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Topic: acid/base pH calculation  (Read 7688 times)

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PJB

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acid/base pH calculation
« on: May 27, 2005, 11:26:48 AM »


“Calculate the H3O+ concentration and pH of the solution that results when 22.0mL of 0.15M acetic acid is mixed with 22.0 mL of 0.15 M NaOH”.

Ok…. I think that this will eventually be an I.C.E. problem  but first I need to get to a net equation?  The book talks about adding the acid to water, then the base to water and going from there.  First of all, where do they get the water and second of all, is it alright to just add NaOH + CH3CO2H --> H2O + CH3CO2-  or is that ignoring steps and going to cause problems later on?

Another question, do I use the reciprocal of Kb for CH3CO2 since it is the resulting conjugate base of acetic acid?  Am I properly ignoring Na, since it doesn’t affect the pH, or do I need to have it in the equation?  

Oh, what I would do to have  an instructor for this “wonderful” little correspondence course!   :)  You smart chemistry people are all very much appreciated by this “chemistry challenged” student!
Pam



Offline AWK

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Re:acid/base pH calculation
« Reply #1 on: May 27, 2005, 12:29:03 PM »
Yes, this is a simple hydrolysis problem. You should remember about dilution of solution twice!
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Offline Borek

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Re:acid/base pH calculation
« Reply #2 on: May 27, 2005, 12:50:07 PM »
“Calculate the H3O+ concentration and pH of the solution that results when 22.0mL of 0.15M acetic acid is mixed with 22.0 mL of 0.15 M NaOH”.

First thing - volumes ond concentration are identical. That means you get just solution of sodium acetate and you must concentrate on hydrolysis (as already pointed out by AWK).

Quote
Ok…. I think that this will eventually be an I.C.E. problem  but first I need to get to a net equation?  The book talks about adding the acid to water, then the base to water and going from there.  First of all, where do they get the water and second of all, is it alright to just add NaOH + CH3CO2H --> H2O + CH3CO2-  or is that ignoring steps and going to cause problems later on?

All is solved in water, there is plenty of water around :)

You have lost Na+ on the right side of the equation. Besides, you should either write everything in ionic form, or not. So it is either

NaOH + CH3CO2H --> H2O + CH3CO2Na

or

Na+ + OH- + CH3CO2H --> H2O + CH3CO2- + Na+

Why CH3CO2H? Because it is a weak acid and it is dissociated only partially. NOw, look at both sides - there is Na+ floating here and there. Cancel it out and you will get net ionic reaction for the problem.

Quote
Another question, do I use the reciprocal of Kb for CH3CO2 since it is the resulting conjugate base of acetic acid?

Be more specific, I have no idea what you are writing about.

Quote
Am I properly ignoring Na, since it doesn’t affect the pH, or do I need to have it in the equation?

You may safely ignore it. See above.
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PJB

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Re:acid/base pH calculation
« Reply #3 on: May 27, 2005, 09:33:39 PM »
Sorry about the vague Kb reciprocal...  I just kind of popped that in out of nowhere!  My question refers to the statement in my text that "the equilibrium constant for the acid-base reaction is the reciprocal of the K for ionization of the conjugate base or acid."  I think it's that whole statement I don't quite understand.  Do I just need the k for the conjugate base "if the solution is basic"so ignore the k for reactants?

Hope that clarifies, its hard sometimes for me to ask an intelligent question!  This chem II stuff is just not my cup of tea.  I'm also taking organic chemistry and so far I'm feeling better about it, than I am about Chem II.

Thanks for your *delete me*
Pam

Offline Borek

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Re:acid/base pH calculation
« Reply #4 on: May 28, 2005, 05:48:44 AM »
OK, I know what you are asking about ;)

Lest say we have a weak acid HA.

Dissociation:

HA <-> H+ + A-

Ka = [H+][A-]/[HA]

A- is conjugate base. Hydrolysis:

A- + H2O <-> HA + OH-

Kb for conjugate base is

Kb = [HA][OH-]/[A-]

(water concentration is treated as constant and is 'hidden' in the Kb value)

If you multiply both constants, [HA] and [A-] cancels out:

Ka*Kb = [H+][A-]/[HA] * [HA][OH-]/[A-] = [H+][OH-]

What is left is nothing else than Kw (water autodissociation constant), so

Ka*Kb = Kw

What your textbook states is that

Ka = Kw/Kb

or

Kb = Kw/Ka

hence the reciprocals mentioned.
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