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H3PO4 Stepwise Neutralization 
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#1
Oct1707, 08:34 AM

P: 99

For the triprotic acid, phosphoric acid, H3PO4, there are Ka1, Ka2,
and Ka3. And we know from the book that the value of Ka3 is too small, so we neglect the [H+] contribution from it. Normally, we encounter questions involving pH calculations of neutralization of a diprotic acid and a base, for example, H2SO4 and NaOH. Let say there are 5 mmol of H2SO4 and 10 mmol of NaOH. The stepwise reactions are: H2SO4 + NaOH > NaHSO4 + H2O 5100 055 NaHSO4 + NaOH > Na2SO4 + H2O 550 005 However, if the reaction is between H3PO4 and NaOH, let say there are 5 mmol H3PO4 and 15 mmol NaOH. The stepwise reactions are: H3PO4 + NaOH > NaH2PO4 + H2O 5150 0105 NaH2PO4 + NaOH > Na2HPO4 + H2O 5100 055 Na2HPO4 + NaOH > Na3PO4 + H2O 550 005 So, are the 3 equations above correct? I mean, can we use the 3 equations above for calculating the pH value, as what we would do for the case of the neutralization involving a diprotic acid and NaOH? If so, why do we neglect the [H+] from H3PO4? Or is it that we ONLY neglect the [H+] from Ka3 when we calculate the pH of a solution containing ONLY H3PO4, but we should somehow "include" and "consider" the Ka3 when we wish to find the pH value of a solution involving neutralization of H3PO4 and NaOH, since this neutralization reaction is based on the stoichiometric ratio. Thanks. 


#2
Oct1807, 06:12 PM

P: 26

You can prove that H3PO4 third dissocitation is negligible as long as H3PO4 is the only source of hydronium ions in solution. ( either generally or u can take a numerical example and try it out yourself . ) But notice that the reactions u posted : H3PO4 + NaOH > NaH2PO4 + H2O NaH2PO4 + NaOH > Na2HPO4 + H2O Na2HPO4 + NaOH > Na3PO4 + H2O are not entirely quantitative. that is , the advancement decreases from one reaction to the other. For each rxn , k = ka/kw . Notice the 3rd rxn is much weaker than the 1st reaction , which is pretty much 'complete'. You can find the ka's and calculate the K of each reaction to convince urself. Finding the pH here is different from that of sulfuric acid. 


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