Equilibrium constant of T2O (Tritium is an isotope of H) differ from those of at 298 K. Let at 298 K, pure T2O has pT (like pH) 7.60. What is the pT of a solution prepared by adding 100 ml of 0.4 M -

Question

Equilibrium constant of T2O (Tritium is an isotope of H) differ from those of $H_2O$ at 298 K. Let at 298 K, pure T2O has pT (like pH) 7.60. What is the pT of a solution prepared by adding 100 ml of 0.4 M - TCl to 400 ml of 0.2 M - NaOT? (log 2 = 0.3)

Prashant Kotian · Accepted Answer

Correct answer: D. For pure T2O, pT = 7.60. Since pure T2O is neutral, [T+] = [OT-] = 10^-7.6 M. Thus, KT2O = [T+][OT-] = (10^-7.6)^2 = 10^-15.2. This means pKT2O = 15.2.
Moles of T+ added = 100 ml × 0.4 M = 40 mmol.
Moles of OT- added = 400 ml × 0.2 M = 80 mmol.
Reaction: T+ + OT- -> T2O.
Excess OT- = 80 - 40 = 40 mmol.
Total volume = 500 ml.
[OT-] = 40 mmol / 500 ml = 0.08 M = 8 × 10^-2 M.
pOT = -log[OT-] = -log(8 × 10^-2) = 2 - 3 log 2 = 2 - 0.9 = 1.1.
pT = pKT2O - pOT = 15.2 - 1.1 = 14.1.

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