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Topic: Derivation of Fraction of Titration Equation from Charge Balance  (Read 7366 times)

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Offline haz658

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   In the sixth edition of Daniel C. Harris's Quantitative Chemical Analysis, chapter 12 page 247, Harris derives an equation for the fraction of titration completed when titrating a weak acid with a strong base, specifically NaOH. He accomplished this from the charge balance equation where
                   [H+] + [Na+] = [A-] + [OH-]
Through replacement of the Na+ and A- terms, he obtains the following.
            
Through rearrangement of terms, he derives the fraction of titration.
            
I attempted to derive this equation from the one above, but instead obtained this.
  
Can someone please offer some advice how to properly derive this equation? If you would like me to elaborate regarding the steps I took to obtain my answer, I will gladly do so.

Offline Borek

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Re: Derivation of Fraction of Titration Equation from Charge Balance
« Reply #1 on: August 16, 2010, 04:13:35 PM »
How is the titration fraction defined? Seems to me like initial concentration of A- is negative. Are you sure nominator on the right doesn't contain multiplication instead of subtraction?
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Offline haz658

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Re: Derivation of Fraction of Titration Equation from Charge Balance
« Reply #2 on: August 16, 2010, 07:04:33 PM »
 The titration fraction is the fraction/percentage of titration performed with respect to completion (the point at which the number of moles of base added is equal to that of acid present in the original solution, or vice versa). I apologize if my equation appears sloppy. Any time alphaA- is present in the equations, the negative sign represents the charge of the species A-, the conjugate base of the weak acid HA, and the alpha represents the fraction of that species present in solution (as opposed to the HA form). I checked my algebra, and I am fairly certain that my rearrangement of terms in the original equation, where expressions for sodium cation and hydroxide anion have been substituted in, is correct. It would seem that I am missing some sort of manipulation of the present variables in order to obtain the solution given. Thanks for your response. I am not confident that I will be able to figure this out without any external assistance.

Offline Borek

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Re: Derivation of Fraction of Titration Equation from Charge Balance
« Reply #3 on: August 17, 2010, 03:49:30 AM »
OK, I see what you mean, tex formula formatting is sloppy.

My guess is that you are on the right track, although not seeing the text and not knowing the book I don't see how he did it (yet ;) ).

Just rearranging what you wrote I got



To get rid of volumes you can try to substitute



and solve for Φ - but I can be wrong.
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Offline haz658

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Re: Derivation of Fraction of Titration Equation from Charge Balance
« Reply #4 on: August 17, 2010, 12:52:38 PM »
   You did it! Thank you so much for your help. :) It is always frustrating when these manipulations are needed, because I always struggle to find the correct one that will achieve the desired result and I can never find it or foresee how to arrive at the final result from the point I have reached.
        If you were curious, I made the conversion for Vb/Va to (CbVb/CaVa) x (Ca/Cb), as you suggested. The steps follow.
$$ \frac {C_{b} V_{b}} {C_{a} V_{a}} = \alpha + \frac {OH^{-} - H^{+}} {C_{a}} \left (1 + \frac {C_{b} V_{b}} {C_{a} V_{a}} \frac {C_{a}} {C_{b}} \right ) /$$
$$ \frac {C_{b} V_{b}} {C_{a} V_{a}} = \alpha + \frac {OH^{-} - H^{+}} {C_{a}} + \frac {OH^{-} - H^{+}} {C_{b}} \frac {C_{b} V_{b}} {C_{a} V_{a}} /$$




Thanks again for your wisdom. Fantastic.  :)

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