When 20 ml of 0.2 M – DCl solution is mixed with 80 ml of 0.1 M – NaOD solution, pD of the resulting — Ionic Equilibrium Chemistry Question
Question
When 20 ml of 0.2 M – DCl solution is mixed with 80 ml of 0.1 M – NaOD solution, pD of the resulting solution becomes 13.6. The ionic product of heavy water, $\text{D}_2\text{O}$, is
💡 Solution & Explanation
Moles of $D^+$ from DCl = $20 \times 0.2 = 4 \text{ mmol}$. Moles of $OD^-$ from NaOD = $80 \times 0.1 = 8 \text{ mmol}$. Excess $OD^- = 8 - 4 = 4 \text{ mmol}$. Total volume = $100 \text{ ml}$. Final $[OD^-] = 4 / 100 = 0.04 \text{ M} = 4 \times 10^{-2} \text{ M}$. Given $pD = 13.6$, we have $[D^+] = 10^{-13.6}$. Since $\log 2.5 \approx 0.4$, $10^{0.4} \approx 2.5$, meaning $[D^+] \approx 2.5 \times 10^{-14} \text{ M}$. The ionic product $K_w = [D^+][OD^-] = (2.5 \times 10^{-14}) \times (4 \times 10^{-2}) = 10 \times 10^{-16} = 10^{-15}$. Therefore, correct answer is A.