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Current Question (ID: 8486)

Question:

$\text{The ionization constant of phenol is } 1.0 \times 10^{-10}\text{. The concentration of phenolate ion in 0.05 M solution of phenol will be:}$

Options:

1. $4.2 \times 10^{-4} \text{ M}$
2. $3.6 \times 10^{-5} \text{ M}$
3. $7.8 \times 10^{-6} \text{ M}$
4. $2.2 \times 10^{-6} \text{ M}$

Solution:

$\text{Hint: } K_a = \frac{[\text{C}_6\text{H}_5\text{O}^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{OH}]}$ $\text{Explanation:}$ $\text{Step 1: Write down the equilibrium}$ $\text{C}_6\text{H}_5\text{OH} + \text{H}_2\text{O} \rightleftharpoons \text{C}_6\text{H}_5\text{O}^- + \text{H}_3\text{O}^+$ $\text{Initial conc.}\quad\quad 0.05\quad\quad\quad\quad 0\quad\quad\quad 0$ $\text{At equilibrium}\quad\quad 0.05-x\quad\quad\quad x\quad\quad\quad x$ $K_a = \frac{[\text{C}_6\text{H}_5\text{O}^-][\text{H}_3\text{O}^+]}{[\text{C}_6\text{H}_5\text{OH}]}$ $K_a = \frac{x \times x}{0.05-x}$ $\text{Step 2: Calculate phenolate ion concentration.}$ $\text{As the value of the ionization constant is very less, x will be very small.}$ $\text{Thus, we can ignore x in the denominator.}$ $\therefore x = \sqrt{1 \times 10^{-10} \times 0.05}$ $= \sqrt{5 \times 10^{-12}}$ $= 2.2 \times 10^{-6}\text{ M} = [\text{H}_3\text{O}^+]$ $\text{Since } [\text{H}_3\text{O}^+] = [\text{C}_6\text{H}_5\text{O}^-], [\text{C}_6\text{H}_5\text{O}^-] = 2.2 \times 10^{-6} \text{ M}$

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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
}