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Buffer solution + equilibrium + Organic compound

  • Thread starter myvow
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For this question,why Ka =(ch3coo-)(h+)/(ch3cooh) instead of Ka =(ch3coo +x)(x)/(ch3cooh-x) ?
20141228_7bca95847216b430816fm7mRBFmSHq75.png
 
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  • #2
epenguin
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Borek
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why Ka =(ch3coo-)(h+)/(ch3cooh) instead of Ka =(ch3coo +x)(x)/(ch3cooh-x)
First equation uses equilibrium concentrations, second uses initial concentrations modified by the extent of the dissociation, to calculate equilibrium concentrations. Do you understand what these means?
 
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epenguin
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For this question,why Ka =(ch3coo-)(h+)/(ch3cooh) instead of Ka =(ch3coo +x)(x)/(ch3cooh-x) ?
20141228_7bca95847216b430816fm7mRBFmSHq75.png
You have now modified your initial post. I do not understand what you have in mind by your x.
You might have been confused by the CH3COONa. That is a conventional formula, it really just represents the composition of sodium acetate but in aqueous solution and for that matter in crystal sodium acetate is all Na+ and CH3COO-. When you add to this "HCl" - I.e. H+ + Cl- - a fraction of the acetate ions CH3COO- become protonated to form CH3COOH and then the calculations you are asked about involve that fraction, which actually determines the pH as you can see from the given formulae.
 
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