The ionic product of heavy water ($D_2O$) at 298 K is $1.1 \times 10^{-15}$. What is the pD of a neutral solution of heavy water at this temperature?

## CORE INFORMATION\n✓ **Answer**: B\n## CONCEPT FRAMEWORK\n **Primary Concept**: Ionic product of water and pD\n **Related Topics**: Acid-base equilibria, pH, pOH\n **Prerequisites**: Understanding of logarithms and the concept of ionic product.\n## SOLUTION PATHWAY\n**Given Information**:\n Ionic product of $D_2O$, $K_w(D_2O) = 1.1 \times 10^{-15}$\n Temperature = 298 K\n**Method & Steps**:\n 1. Determine the concentration of $D^+$ and $OD^-$ ions in a neutral solution of heavy water.\n Reasoning: In a neutral solution, the concentration of $D^+$ ions is equal to the concentration of $OD^-$ ions.\n Formula/Rule: $K_w(D_2O) = [D^+][OD^-]$ where $[D^+]$ and $[OD^-]$ are the concentrations of deuterium ions and deuteroxide ions respectively.\n 2. Calculate the concentration of $D^+$ ions.\n Working: Since the solution is neutral, $[D^+] = [OD^-]$. Therefore, $[D^+]^2 = K_w(D_2O) = 1.1 \times 10^{-15}$. Taking the square root, we get $[D^+] = \sqrt{1.1 \times 10^{-15}} \approx 1.05 \times 10^{-7.5} \approx 3.32 \times 10^{-8}$ M\n Key Point: Remember to take the square root of the ionic product to find the concentration of $D^+$ ions in a neutral solution.\n 3. Calculate the pD of the solution.\n Formula/Rule: $pD = -\log_{10}[D^+]$ \n Working: $pD = -\log_{10}(3.32 \times 10^{-8}) \approx 7.48$\n Result: The pD of a neutral solution of heavy water at 298 K is approximately 7.48.\n Verification: The pD value is slightly higher than 7, which is expected since the ionic product of heavy water is slightly lower than that of ordinary water.\n ## OPTION ANALYSIS\nA :x: 7.00 is the pD of a neutral solution if the ionic product were $10^{-14}$, which is not the case here.\nB ✓: 7.48 is the calculated pD value.\nC :x: 7.96 is significantly higher than the expected value.\nD :x: 6.52 is significantly lower than the expected value.\n ## LEARNING AIDS\n **Common Mistakes**:\n 1.Forgetting to take the square root of the ionic product when calculating the concentration of $D^+$ ions in a neutral solution: This leads to an incorrect pD value.\n How to avoid: Clearly understand that in a neutral solution, $[D^+] = [OD^-]$.\n 2.Incorrectly using the pH formula instead of the pD formula: This would give a misleading result.\n How to avoid: Remember pD is defined as $-\log_{10}[D^+]$.\n **Quick Tips**:\n 1.Remember that for neutral solutions, $pD = \frac{1}{2} pK_w(D_2O)$.\n 2.Understand the analogy between pH and pD; they both represent the acidity/neutrality/alkalinity of the solution.

NEET Chemistry Hydrogen MCQs

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A metal hydride crystallizes in a face-centered cubic lattice with hydrogen atoms occupying tetrahedral voids. If the edge length of the unit cell is 'a', what is the distance between two nearest hydrogen atoms?

a/2

a/√2

a/4

√3a/4

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📚 Subject:Chemistry

📑 Chapter:Hydrogen

🎯Topic:Hydrogen

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :3 mins

🔄Ncert Concepts :Hydrogen, Crystal Structure, Solid State

Which of the following isotopic combinations of dihydrogen exhibits the largest rotational constant?

📚 Subject:Chemistry

📑 Chapter:Hydrogen

🎯Topic:H2O2

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :2 mins

🔄Ncert Concepts :Hydrogen Peroxide, Volume Strength, Molarity

The oxidation state of oxygen in $H_2O_2$ is:

-2

-1

Question Tags

📚 Subject:Chemistry

📑 Chapter:Hydrogen

🎯Topic:H2O2

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :1 mins

🔄Ncert Concepts :Hydrogen Peroxide, Oxidation State, Redox Reactions

The ionic product of heavy water ( $D_2O$ ) at 298 K is $1.1 imes 10^{-15}$ . What is the pD of a neutral solution of heavy water at this temperature?