Equilibrium constant of (Tritium is an isotope of H) differ from those of at 298 K. Let at 298 K, pu — Ionic Equilibrium Chemistry Question
Question
Equilibrium constant of $\text{T}_2\text{O}$ (Tritium is an isotope of H) differ from those of $\text{H}_2\text{O}$ at 298 K. Let at 298 K, pure $\text{T}_2\text{O}$ 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)
💡 Solution & Explanation
For pure $\text{T}_2\text{O}$, $pT = 7.60$, so $pOT = 7.60$, giving $pK_w = pT + pOT = 15.20$. Moles of TCl = $100 \times 0.4 = 40 \text{ mmol}$. Moles of NaOT = $400 \times 0.2 = 80 \text{ mmol}$. Excess NaOT = $80 - 40 = 40 \text{ mmol}$. Total volume = $500 \text{ ml}$. $[OT^-] = 40 \text{ mmol} / 500 \text{ ml} = 0.08 \text{ M} = 8 \times 10^{-2} \text{ M}$. $pOT = -\log(8 \times 10^{-2}) = 2 - 3\log 2 = 2 - 0.9 = 1.1$. Then, $pT = pK_w - pOT = 15.20 - 1.1 = 14.1$. Therefore, correct answer is D.