A really great question appeared on paper 4 in the trial exam. it mentioned tartaric acid, or 2,3-dihydroxybutanedioic acid to give it its IUPAC name.
Only ONE out of 21 students got the question right, hence, it seems, only one out of 21 understood it.
The question stated: It has two dissociation constants K1 and K2 for which the pKa values are 2.99 and 4.40. suggest equations showing the two dissociation that give rise to these pKa values.
Two main mistakes were made.
1) pKa is not the same as pH.
It is invalid to say the pKa 2.99 must be the structure that lost two protons because the 2.99 value is “more acidic” than the 4.40 value. Students that thought this also went on to say that the structure at 4.40 must therefore only have lost one proton, and drew the associated structures.
PKa is an acid dissociation constant. You were taught how to construct a Ka equation (and therefore a pKa expession) for any acid. The general form was HA(aq) <-> H+(aq) + A-(aq), and hence Ka = ( [H+(aq)] x [A-(aq)] ) / [HA(aq)]
The tartaric acid contains two acidic H’s. So lets modify the above equation slightly to allow us to see more clearly both dissociations.
H2A(aq) <-> H+(aq) + HA-(aq) is the first dissociation. Ka = ( [H+(aq)] x [HA-(aq)] ) / [H2A(aq)]
HA-(aq) <-> H+ (aq) + A(2-)(aq) is the second dissociation. Ka = ( [H+(aq)] x [A(2-)aq)] ) / [HA-(aq)]
the pKa values must relate to these two dissociations, so which pKa value corresponds to which dissociation?
Notice in each case, only one proton is shown dissociating for the particular starting species. Never are we showing or talking about the total number of H+ ions produced, neither are we talking about the -log of [H+], i.e. pH.
It will be harder for the second dissociation to occur because a H+ will be lost from a negative species, HA-(aq) so it’s dissociation constant, pKa, must be higher. (higher pKa values mean the process is harder). So the second dissociation has the pKa value of 4.40, and therefore the first dissociation will have the lower pKa value of 2.99.
It is true that the acidity of the solution after the second dissociation will be higher than if only the first dissociation occurred, but we are not talking about the pH of the solution.
2) the alcohol OH groups were shown dissociating. Aliphatic OH’s are not considered ‘acidic’ at A-level (phenols of course are different, and are recognized as being acidic). In general, if the pKa of a compound is higher than that of water then they are not considered as acids. The pKa for alcohols under standard conditions is about 16, while that for water is about 15.8. If you find that 15.8 value strange, and think it should be 14, then you are still confusing pKa and pH. pKa, Is not the same as pH. The pH for water is about 14, but the pKa is 15.8.
As always, for any acid, we will use water here,
chemical equation: H2O(l) <-> H+(aq) + [OH-(aq)
from which we write Ka = ( [H+(aq)] x [OH-(aq)] ) / [H2O]
The concentration of water is almost a constant value, 1000g in 1 L. Mr of H2O is 18g, which corresponds to 1000/18 moles per L, so the concentration of water is 55.6 moles per dm3 of water. pH of H2O is 7, so [H+(aq)] is 1×10-7 mol dm-3 and the [OH-(aq)] is the same as the [H+(aq)], so [H+(aq)] x [OH-(aq)] = (1×10-7) x (1×10-7) which = 1×10-14. therefore, 1×10-14 / 55.6 = 15.8.
In summary, for water (under standard conditions), the pH is 14, and it’s pKa is 15.8
pKa and pH are connected ( / related) to each other yes, but not the same.