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Current Question (ID: 8513)

Question:

$\text{A solution of benzoic acid (a weak monobasic acid) is titrated with NaOH. The pH of the solution is 4.2 when half of the acid is neutralized. The dissociation constant of the acid will be:}$

Options:

1. $3.2 \times 10^{-5}$
2. $6.42 \times 10^{-4}$
3. $6.31 \times 10^{-5}$
4. $8.7 \times 10^{-8}$

Solution:

$\text{Hint: At half neutralization reaction; }[\text{Salt}]_0 = [\text{Acid}]_0$ $\text{Explanation: For an acidic buffer; }\text{pH} = \text{pK}_a + \log \frac{[\text{Salt}]_0}{[\text{Acid}]_0}$ $\text{Since at half neutralization, }[\text{Salt}]_0 = [\text{Acid}]_0\text{, the log term becomes zero:}$ $\text{pH} = \text{pK}_a$ $4.2 = \text{pK}_a$ $\text{Therefore, }K_a = 10^{-\text{pK}_a} = 10^{-4.2} = 6.31 \times 10^{-5}$

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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
}