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Current Question (ID: 8492)

Question:

$\text{A 0.001 M aniline solution has a pH of:}$ $(K_b = 4.27 \times 10^{-10})$

Options:

1. $6.19$
2. $7.81$
3. $8.34$
4. $9.81$

Solution:

$\text{Hint: Aniline is a weak base.}$ $\text{Explanation:}$ $\text{Step 1: Calculate } \alpha$ $\text{Given: } K_b = 4.27 \times 10^{-10} \text{ and concentration (c) = 0.001M}$ $\text{We know that,}$ $K_b = c\alpha^2$ $\Rightarrow 4.27 \times 10^{-10} = 0.001 \times \alpha^2$ $\Rightarrow \alpha = 6.53 \times 10^{-4}$ $\text{Step 2: Calculate pH}$ $[\text{OH}^-] = c\alpha$ $\Rightarrow [\text{OH}^-] = 0.065 \times 10^{-5}$ $\text{pOH} = -\log[\text{OH}^-]$ $\Rightarrow \text{pOH} = -\log(0.065 \times 10^{-5}) = 6.19$ $\text{Thus, pH} = 14 - \text{pOH} = 7.81$

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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
}