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Current Question (ID: 8472)

Question:
$\text{The value of the pH of 0.01 mol dm}^{-3} \text{CH}_3\text{COOH } (K_a = 1.74 \times 10^{-5}) \text{ is:}$
Options:
  • 1. 3.4
  • 2. 3.6
  • 3. 3.9
  • 4. 3.0
Solution:
$\text{Hint: } [\text{H}^+] = \sqrt{K_a \cdot C}$ $\text{Step 1:}$ $\text{Calculate the concentration of H}^+ \text{ ion as follows:}$ $\text{Given that, } (K_a = 1.74 \times 10^{-5})$ $\text{Concentration of CH}_3\text{COOH} = 0.01 \text{ mol dm}^{-3}$ $[\text{H}^+] = \sqrt{K_a \cdot C}$ $= \sqrt{1.74 \times 10^{-5} \times 0.01} = 4.17 \times 10^{-4}$ $\text{Step 2:}$ $\text{pH} = -\log [\text{H}^+]$ $= -\log (4.17 \times 10^{-4})$ $= 3.4$

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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
}