At , the absolute dissociation constant () of heavy water () is . If its density is exactly , calcul — Ionic Equilibrium Chemistry Question
Question
At $35^\circ C$, the absolute dissociation constant ($K_d$) of heavy water ($D_2O$) is $4.0 \times 10^{-15}$. If its density is exactly $1.04 \text{ g/mL}$, calculate the ionic product ($K_w$) of $D_2O$ at $35^\circ C$. If your answer is $y \times 10^{-13}$, enter the value of $y$.
Answer: 2.08
💡 Solution & Explanation
The molar mass of $D_2O$ is $20 \text{ g/mol}$. Its molarity is $[D_2O] = \frac{1040 \text{ g/L}}{20 \text{ g/mol}} = 52 \text{ M}$. The ionic product $K_w = K_d \times [D_2O]$. Substituting the values, $K_w = (4.0 \times 10^{-15}) \times 52 = 208 \times 10^{-15} = 2.08 \times 10^{-13}$. Thus, $y = 2.08$.
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