Question

Consider the titration of 25.00 mL of 0.300 M H3 PO4 with 0.300 M NaOH. 1.What is the pH of the titration medium after the addition of the following volumes of NaOH titrant: F. 50.00 mL: G. 60.00 mL: H. 62.50 mL: I. 75.00 mL: J. 85.00 mL

Can writer should how to find Ka1,Ka2, and Ka3?

Answer 1

Hi , friend i can solve only three among five . i wanted to skip this but i thought it will help somewhat

F ) 50 mL NaOH added

this second equivalence point . at this point pH = 1/2 [pKa2 + pKa3 ]

pH = 1/2 (7.21 + 12.32)

pH = 9.76

H ) 62.50 mL

it is third half-equivalence point . here pH = pKa3

pH = 12.32

J )   85 mL

75 mL NaOH consumed upto 3rd equivalence point . remaining base volume = 10 mL

base millimoles = 10 x 0.3 = 3

[OH-] = 3 / (25 +85) = 0.0273 M

pOH = -log [OH-] = 1.56

pH + pOH = 14

pH = 12.40