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Current Question (ID: 8554)

Question:
$\text{In an acidic Buffer solution (CH}_3\text{COOH + CH}_3\text{COONa), the species mainly present in the solution are:}$ $\text{(Ignore negligible amount)}$
Options:
  • 1. $\text{CH}_3\text{COOH, CH}_3\text{COO}^-, \text{CH}_3\text{COONa, H}^+$
  • 2. $\text{CH}_3\text{COO}^-, \text{Na}^+, \text{CH}_3\text{COOH}$
  • 3. $\text{CH}_3\text{COONa, CH}_3\text{COO}^-, \text{H}^+$
  • 4. $\text{CH}_3\text{COO}^-, \text{Na}^+, \text{H}^+, \text{CH}_3\text{COONa}$
Solution:
$\text{Hint: Sodium acetate is a strong salt and acetic acid is a weak acid.}$ $\text{Explanation:}$ $\text{CH}_3\text{COONa is a strong electrolyte and it is completely dissociated while ionisation of CH}_3\text{COOH is suppressed due to common ion effect of CH}_3\text{COO}^- \text{ ion. The reaction is as follows:}$ $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$ $\text{When sodium acetate is added then concentration of acetate ion increases and reaction shift toward backward direction. Hence, only CH}_3\text{COO}^-, \text{Na}^+, \text{ and CH}_3\text{COOH are present.}$

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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
}