The ionic product of heavy water, , is at . The PD value of pure heavy water at is — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

For heavy water, the ionic product is $K_w = [\text{D}^+][\text{OD}^-] = 1.0 \times 10^{-16}$. Taking the negative logarithm, $pD + pOD = 16$. In pure heavy water, the concentration of deuterium ions equals the concentration of deuteroxide ions ($[\text{D}^+] = [\text{OD}^-]$), which implies $pD = pOD$. Therefore, $2pD = 16 \implies pD = 8$. Therefore, correct answer is 8.