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Current Question (ID: 8493)

Question:

$\text{The self-ionization constant for pure formic acid, } K = [\text{HCOOH}_2^+][\text{HCOO}^-] \text{ has been estimated as } (10)^{-4} \text{ at room temperature. The percentage of formic acid molecules in pure formic acid that are converted to formate ions is}$ $\text{(Given: } d_{\text{HCOOH}} = 1.22 \text{ g/cc})$

Options:

1. $0.0185\%$
2. $0.0073\%$
3. $0.074\%$
4. $0.037\%$

Solution:

$\text{Hint: Percentage dissociation} = \frac{\text{final concentration of } (\text{HCOO})^-}{\text{Initial concentration of HCOOH}} \times 100$ $\text{Explanation}$ $\text{Step 1:}$ $\text{Calculate the final concentration of } (\text{HCOO})^- \text{ as follows:}$ $2\text{HCOOH} \rightleftharpoons (\text{HCOO})^- + \text{HCOOH}_2^+$ $K_c = [(\text{HCOO})^-][\text{HCOOH}_2^+]$ $10^{-4} = x^2$ $x = (10)^{-2}$ $\text{The concentration of } (\text{HCOO})^- \text{ is } (10)^{-2}$ $\text{Step 2:}$ $\text{Calculate the percentage of formic acid is converted to formate ion is as follows:}$ $\text{Weight of 1L formic acid in g} = 1.22 \times 10^3$ $\text{Mole} = \frac{1.22 \times (10)^3}{46} = 26.5 \text{ M} = \text{initial concentration of acid}$ $\text{Percentage dissociation} = \frac{\text{final concentration of } (\text{HCOO})^-}{\text{Initial concentration of HCOOH}} \times 100$ $= \frac{(10)^{-2}}{26.5} \times 100 = 0.037\%$

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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
}