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Current Question (ID: 8495)

Question:

$\text{The pH of a 0.05 M aqueous solution of diethylamine is 12. Its }K_b\text{ value will be:}$

Options:

1. $2 \times 10^{-3}$
2. $2.5 \times 10^{-3}$
3. $3 \times 10^{-3}$
4. $4.5 \times 10^{-3}$

Solution:

$\text{Hint: Find the concentration of OH}^-\text{ and use this value to find }K_b$$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{Initial conc.}$ $\text{Equilibrium conc.}$ $(\text{C}_2\text{H}_5)_2\text{NH} + \text{H}_2\text{O} \rightarrow (\text{C}_2\text{H}_5)_2\text{NH}_2^+ + \text{OH}^-$ $\quad\quad\quad 1 \quad\quad\quad\quad 0 \quad\quad\quad\quad 0$ $\quad\quad (1-\alpha) \quad\quad\quad \alpha \quad\quad\quad\quad \alpha$ $[\text{OH}^-] = C\alpha$ $\text{where C is conc. of base and C = 0.05 M}$ $\text{pH} = 12 \therefore \text{pOH} = 2$ $\text{pOH} = -\log[\text{OH}^-]$ $[\text{OH}^-] = 10^{-2} \text{ M}$ $C\alpha = 10^{-2}$ $\text{or } 0.05 \times \alpha = 10^{-2} \quad (\because C = 0.05)$ $\therefore \alpha = 0.2$ $\text{Step 2:}$ $\text{Now for base, }K_b = \frac{C\alpha^2}{(1-\alpha)} = \frac{0.05 \times (0.2)^2}{(1-0.2)}$ $= \frac{0.05 \times 0.04}{0.8} = 2.5 \times 10^{-3}$ $\text{Hence, option second is correct.}$

Import JSON File

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
}