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Current Question (ID: 8481)

Question:

$\text{The concentration of CH}_3\text{COOH that will have the same [H}^+\text{] as obtained from } 10^{-2} \text{ M HCOOH, is-}$ $\text{(Ka(CH}_3\text{COOH}) = 10^{-5}, \text{Ka(HCOOH}) = 10^{-4}\text{)}$

Options:

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

Solution:

$\text{Hint: } C_1\alpha_1 = C_2\alpha_2$ $\text{Step 1:}$ $\text{HCOOH} = K_{a1} = 10^{-4}, C_1 = 10^{-2}$ $\text{CH}_3\text{COOH} = K_{a2} = 10^{-5}, C_2 = ?$ $C_1\alpha_1 = C_2\alpha_2$ $\text{Step 2:}$ $\Rightarrow C_1\sqrt{\frac{Ka_1}{C_1}} = C_2 \times \sqrt{\frac{Ka_2}{C_2}}$ $\Rightarrow \sqrt{C_1 Ka_1} = \sqrt{C_2 Ka_2} \Rightarrow C_1 Ka_1 = C_2 Ka_2$ $\Rightarrow C_2 \times 10^{-5} = 10^{-2} \times 10^{-4}$ $\Rightarrow C_2 = 10^{-1} \text{ M}$

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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
}