Example: Titration of 10.00 mL 0.100 mol/L H3PO4(aq) with 0.100 mol/L NaOH(aq)
Reactions:
1 H3PO4(aq) + 1 NaOH(aq) H2O + NaH2PO4(aq)
1 NaH2PO4(aq) + 1 NaOH(aq) H2O + Na2HPO4(aq)
1 Na2HPO4(aq) + 1 NaOH(aq) H2O + Na3PO4(aq)
At the start we have only H3PO4 in the erlenmeyer flask. This weak acid is partially ionized. The pH can be calculated.
Upon adding strong base, this will react with the weak acid H3PO4 to form NaH2PO4, containing the conjugate base of the weak acid. So we have a buffer solution. The pH can be calculated.
When the first equivalence point EP is reached, the only compound that is present in the erlenmeyer flask is NaH2PO4: an ampholyte.
If we add more strong base from the burette, the second reaction will occur. NaH2PO4 is converted into
Na2HPO4. Both compounds form a buffer.
When the second equivalence point EP is reached, the only compound that is present in the erlenmeyer flask is Na2HPO4: an ampholyte.
If we add more strong base from the burette, the third reaction will occur. Na2HPO4 is converted into
Na3PO4. Both compounds form a buffer.
When the third equivalence point EP is reached, there is no clear inflection point. The phosphate anion can hydrolyse according to:
PO43–(aq) + H2O(l) HPO42–(aq) + OH–(aq)
The corresponding equilibrium constant (2.1 x 10-2) is not that small. This means that the equilibrium position of this reaction is not sufficiently shifted to the left and that at EP3 a HPO42–/PO43–-buffer is still present.
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