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Current Question (ID: 8488)

Question:

$\text{The degree of ionization of 0.05 M acetic acid is:}$ $(\text{pK}_a = 4.74)$

Options:

1. $1.91 \times 10^{-2}$
2. $2.46 \times 10^{2}$
3. $1.93 \times 10^{2}$
4. $2.34 \times 10^{-3}$

Solution:

$\text{Hint: }\text{pK}_a = -\log K_a$ $\text{Explanation:}$ $\text{Step 1: Write down the given data}$ $\text{Concentration (c) = 0.05 M and }\text{pK}_a = 4.74$ $\Rightarrow \text{pK}_a = -\log K_a$ $\Rightarrow K_a = -\text{antilog pK}_a = 1.82 \times 10^{-5}$ $\text{We have to calculate }\alpha$ $\text{Step 2: Find out }\alpha$ $\text{We know that, for weak acid;}$ $\text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}^+$ $\text{Initial: }0.05 \quad - \quad -$ $\text{At equilibrium: }0.05 - 0.05\alpha \quad 0.05\alpha \quad 0.05\alpha$ $K_a = \frac{(0.05\alpha)^2}{0.05-0.05\alpha}$ $1.82 \times 10^{-5} = \frac{0.05\alpha^2}{1-\alpha}$ $\text{Consider }1 - \alpha \approx 1$ $\Rightarrow \alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.05}}$ $\Rightarrow \alpha = 1.92 \times 10^{-2}$

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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
}