For a certain diprotic acid, the pH at one half the volume to the first equivalence point is 4.15 and the pH at one half the volume between equivalence points is 7.22. Given this information, what are the values for Ka1 and Ka2

The values for

&

Explanation:

               Dissociation of diprotic acid is as follows

                        ----------------(i)

                         ---------------(ii)

At the first equivalence point 50% of the acidic proton is lost , then it has the other 50% of its conjugate base.

  From equation (i) we can write

              ∵ At equivalence point [Conjugate base = Acid]  

   From equation (ii) we can also write

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer