Question

calculate the pH of 2.0 ×10^-4 M H3X solution assuming the first dissociation to be 100% second to be 50% and third to be negligible

​

Answer 1

Answer:

Explanation:

[OH-] = 8 x 10^-3

In equilibrium concentration is C∝ where ∝ is the degree of dissociation

OH-] = (8x 10^-3) x 0.5

        = 4 x 10^-3 moles

pOH = -log 4 x 10^-3   = 2.4

   pH = 14 - 2.4

   pH= 11.6