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
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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
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