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Current Question (ID: 8484)

Question:

$\text{The ionization constant of propanoic acid is } 1.32 \times 10^{-5}\text{. The degree of ionization of 0.05M acid solution will be:}$

Options:

1. $\alpha = 0.63 \times 10^{-2}$
2. $\alpha = 1.63 \times 10^{-4}$
3. $\alpha = 1.63 \times 10^{-2}$
4. $\alpha = 0.05 \times 10^{-2}$

Solution:

$\text{Hint: Propanoic acid is a weak acid.}$ $\text{Explanation:}$ $\text{Let the degree of ionization of propanoic acid be } \alpha\text{.}$ $\text{Then, representing propionic acid as HA, we have:}$ $\text{HA} + (.05 - 0.0\alpha) \approx .05\text{H}_2\text{O} \leftrightarrow \text{H}_3\text{O}^+ + \text{A}^-$ $\text{At equilibrium: } [\text{H}_3\text{O}^+] = 0.05\alpha, [\text{A}^-] = 0.05\alpha$ $K_\alpha = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$ $= \frac{(0.05\alpha)(0.05\alpha)}{0.05} = 0.05\alpha^2$ $\alpha = \sqrt{\frac{K_\alpha}{0.05}} = \sqrt{\frac{1.32 \times 10^{-5}}{0.05}} = 1.63 \times 10^{-2}$

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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
}