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Current Question (ID: 8528)

Question:
$\text{A } 0.02 \text{ M solution of pyridinium hydrochloride has a pH of } 3.44. \text{ The ionization constant of pyridine will be:}$
Options:
  • 1. $6.6 \times 10^{-6}$
  • 2. $1.51 \times 10^{-9}$
  • 3. $2.36 \times 10^{-7}$
  • 4. $3.43 \times 10^{-9}$
Solution:
\text{Hint: } K_b = \frac{K_w}{K_h} \text{Explanation:} \text{Given, pH = } 3.44 \text{ and concentration = } 0.02 \text{ M} \text{pH = } 3.44 \Rightarrow -\text{log}[\text{H}^+] = 3.44 [\text{H}^+] = 3.63 \times 10^{-4} K_h = \frac{(3.63 \times 10^{-4})^2}{0.02} = 6.6 \times 10^{-6} K_b = \frac{K_w}{K_h} = \frac{10^{-14}}{6.6 \times 10^{-6}} = 1.51 \times 10^{-9}

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}