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Current Question (ID: 21115)

Question:

$\text{In aqueous solution, the ionization constants for carbonic acid are}$ $K_1 = 4.2 \times 10^{-7}$ $\text{and}$ $K_2 = 4.8 \times 10^{-11}$ $\text{The correct statement for a saturated 0.034 M solution of the carbonic acid is:}$

Options:

1. $\text{The concentration of } \text{H}^+ \text{ is double that of } \text{CO}_3^{2-}$
2. $\text{The concentration of } \text{CO}_3^{2-} \text{ is 0.034 M.}$
3. $\text{The concentration of } \text{CO}_3^{2-} \text{ is greater than that of } \text{HCO}_3^{-}$
4. $\text{The concentration of } \text{H}^+ \text{ and } \text{HCO}_3^{-} \text{ are approximately equal.}$

Solution:

$\text{Hint: The second dissociation is almost negligible as compared to first}$ $\text{Step 1:}$ $\text{The two reactions are as follows:}$ $\text{A} \rightarrow \text{H}_2\text{CO}_3(\text{aq.}) \rightleftharpoons \text{H}^+(\text{aq.}) + \text{HCO}_3^{-}(\text{aq.}) \text{, } K_1 = 4.2 \times 10^{-7}$ $\text{B} \rightarrow \text{HCO}_3^{-}(\text{aq.}) \rightleftharpoons \text{H}^+(\text{aq.}) + \text{CO}_3^{2-}(\text{aq.}) \text{, } K_2 = 4.8 \times 10^{-11}$ $\text{The value of } K_2 \ll K_1$ $\text{The concentration of } \text{H}^+ \text{ ion is } [\text{H}^+]_A \text{. The } \text{HCO}_3^{-} \text{ dissociates in a negligible amount. Hence, the total concentration of } \text{H}^+ \text{ ion is almost equal to } [\text{H}^+]_A.$ $\text{Step 2:}$ $\text{As, } K_2 < K_1$ $[\text{H}^+] \text{ comes from dissociation of } \text{H}_2\text{CO}_3.$ $[\text{H}^+]_{\text{total}} \approx [\text{H}^+]_A \text{ and from first equilibrium.}$ $[\text{H}^+]_A \approx [\text{HCO}_3^{-}] \approx [\text{H}^+]_{\text{total}}$ $\text{Thus, we can say that the concentration of } \text{H}^+ \text{ and } \text{HCO}_3^{-} \text{ are approximately equal.}$ $\text{Hence, option fourth is the correct answer}$

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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
}