Why Do We Consider All Ka Values in H3PO4 Neutralization with NaOH?

[lkh1986]

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
---5-----------10------------0---------------
---0-----------5--------------5---------------
NaHSO4 + NaOH ---> Na2SO4 + H2O
----5-----------5-------------0---------------
----0-----------0--------------5--------------

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
----5---------15------------0-----------------
----0---------10------------5-----------------

NaH2PO4 + NaOH ---> Na2HPO4 + H2O
----5------------10--------------0---------------
----0-------------5--------------5---------------

Na2HPO4 + NaOH ---> Na3PO4 + H2O
----5------------5--------------0--------------
----0------------0--------------5--------------

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.

 

[Hunt_]

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.

You are right.

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.

 

[Blueshift5]

I can confirm that the three equations provided for the neutralization of H3PO4 and NaOH are correct. However, when calculating the pH of a solution involving this neutralization reaction, it is important to consider the contribution of all three Ka values.

The reason we typically neglect the contribution of Ka3 is due to its small value, which means it has a minimal effect on the overall pH of the solution. However, it is still important to consider it in the calculations, especially when dealing with large amounts of H3PO4 and NaOH.

Additionally, when neutralizing a solution of only H3PO4, we can neglect the contribution of Ka3 because there is no base present to react with it. But in the case of neutralization with NaOH, we must consider all three Ka values to accurately calculate the pH of the solution.

In summary, while Ka3 may be negligible in some cases, it should still be considered in the calculations for neutralization of H3PO4 and NaOH to ensure accurate results.