Question

calculate the volume of 0.500 M NaOH which must be added to 2.25g of phosphoric acid (formula weight 98.00g/mol) to yield a buffer of pH= 6.70 when diluted to one liter. Given acid dissociation for H3PO4 are : pka1= 2.14, pka2= 7.20, pka3 = 12.10

Answer 1

Titration of the phosphoric acid H3PO4 is an interesting case. Although often listed together with strong mineral acids (hydrochloric, nitric and sulfuric) phosphoric acid is relatively weak, with pKa1=2.14, pKa2=7.20 and pKa3=12.10. That means titration curve contains only two inflection points and phosphoric acid can be titrated either as a monoprotic acid or as a diprotic acid. In the first case acid has to be titrated against indicator changing color around pH 4.7 (for example methyl orange), in the second case - against indicator changing color around pH 9.6 (for example thymolphthalein). Phenolphthalein can't be used, as it starts to change color around pH 8.2, when phosphoric acid is titrated in about 95%.

A simple buffer system that can be described using the Henderson–Hasselbalch equation:

pH=pKa−log10c(H2A− )c(HA2− )

As both concentrations c(H2A− ) and c(HX2AX− ) are equivalent, the logarithmic expression gets zero and your final pH equals the second pKa value of phosphoric acid, which is 7.20

H3PO4+NaOH⟶NaH2PO4+H2O

NaH2PO4+NaOH⟶Na2HPO4+H2O

Using the following formulae : Vx Nx = V N (1)

0.500 M NaOH = 0.500 N NaOH (*)
0.2 M H3PO4 = 0.6 N H3PO4 (*)

Vx= 0.6/0.500*

Relate moles of H₃PO₄ with the moles of NaOH with the helps of coefficients in front of NaOH and H₃PO₄ seen in balanced equations.

               moles ↔ Litre use definition of molarity (moles/litrs)

Start with what is given for H₃PO₄

0.015 lit of H₃PO₄ x 0.2 moles of H₃PO₄ / lit of H₃PO₄ x 3 moles of NaOH/1mole of H₃PO₄ x lit of NaOH/0.1 mol of NaOH = 0.09 lit or 90 ml.