Question

Calculate the pH during the titration of 50.00 mL of 0.1000 M phosphoric [H3PO4; Ka1=7.1 x 10^(-3), Ka2=6.3 x 10^(-8), Ka3=4.2 x 10^(-13)] after adding 33.42 mL of 0.1000 M NaOH.

Answer 1

This is a triprotic acid

H3PO4 <-> H2PO4- + H+

H2PO4- <-> H+ + HPO4-2

HPO4-2 <-> H+ + PO4-3

so...

1 mol of H+ : 1 mol of OH-

find mmol of OH-

mmol of base = MV = 33.42*0.1 = 3.342 mmol of OH- were used

mmol of phosphoric acid = MV = 50*0.1 = 5 mmol of H3PO4 are present

so...

H3PO4 + OH- <-> H2PO4- + H2O

mmol of H3PO4 left after reaction = 5-3.342 = 1.658 mmol of H3PO4

mmol of H2PO4- formed after reaction = 0+3.342 mmol of H2PO4-

so...

this will form a buffer, since ther eis acid + conjugate base

use buffer equation as follows:

pH = pKa + log(H2PO4-/H3PO4)

pKa = -log(Ka1) = -log(7.1*10^-3) = 2.1487

note that we must use pKa1, since this is the FIRST ionization (i.e. the 1st reaction contains H2PO4- and H3PO4)

pH = pKa + log(H2PO4-/H3PO4)

ph = 2.1487 + log(3.342/1.658 ) = 2.453121

the pH resulting will be 2.453121