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Current Question (ID: 8479)

Question:

$\text{When 0.1 mol of } \text{CH}_3\text{NH}_2 (K_b = 5 \times 10^{-4}) \text{ is mixed with 0.08 mol of HCl and diluted to 1 L, the } \text{H}^+ \text{ ion concentration in the solution will be:}$

Options:

1. $8 \times 10^{-11} \text{ M}$
2. $6 \times 10^{-5} \text{ M}$
3. $1.6 \times 10^{-11} \text{ M}$
4. $8 \times 10^{-2} \text{ M}$

Solution:

$\text{Hint: Use a formula to find pH of Buffer}$ $\text{Explanation:}$ $\text{Step 1: The reaction occurring is:}$ $\text{CH}_3\text{NH}_2 + \text{HCl} \rightarrow \text{CH}_3\text{NH}_3^+ \text{Cl}^-$ $\text{Initial: } 0.1 \quad 0.08 \quad 0$ $\text{Final: } 0.02 \quad 0 \quad 0.08$ $\text{Thus, it is a basic buffer solution.}$ $\text{The expression for the hydroxide ion concentration is}$ $[\text{OH}^-] = K_b \times \frac{[\text{Base}]}{[\text{Conjugate acid}]}$ $\text{Step 2: Substitute values in the above expression.}$ $[\text{OH}^-] = 5 \times 10^{-4} \times \frac{0.02}{0.08} = 1.25 \times 10^{-4}$ $\text{But } [\text{H}^+][\text{OH}^-] = K_w = 10^{-14}$ $\therefore [\text{H}^+] = \frac{10^{-14}}{[\text{OH}^-]} = \frac{10^{-14}}{1.25 \times 10^{-4}} = 8 \times 10^{-11} \text{ M}$

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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
}