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Current Question (ID: 21100)

Question:

$\text{An aqueous solution contains 0.10 M } \text{H}_2\text{S} \text{ and 0.20 M HCl. If the equilibrium constant for the formation of } \text{HS}^- \text{ from } \text{H}_2\text{S} \text{ is } 1.0 \times 10^{-7}$ $\text{and that of } \text{S}^{2-} \text{ from } \text{HS}^- \text{ ions is } 1.2 \times 10^{-13} \text{ then the concentration of } \text{S}^{2-} \text{ ions in aqueous solution will be:}$

Options:

1. $5 \times 10^{-8} \text{ M}$
2. $3 \times 10^{-20} \text{ M}$
3. $6 \times 10^{-21} \text{ M}$
4. $5 \times 10^{-19} \text{ M}$

Solution:

$\text{Step 1:}$ $\text{The desired equation is as follows:}$ $\text{H}_2\text{S} \rightleftharpoons 2\text{H}^+ + \text{S}^{2-}$ $\text{The two equations are as follows:}$ $\text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^- \quad \text{K}_{(1)} = 1.0 \times 10^{-7}$ $\text{HS}^- \rightleftharpoons \text{H}^+ + \text{S}^{2-} \quad \text{K}_{(2)} = 1.2 \times 10^{-13}$ $\text{H}_2\text{S} \rightleftharpoons 2\text{H}^+ + \text{S}^{2-}$ $\text{K}_{\text{eq.}} = 1 \times 10^{-7} \times 1.2 \times 10^{-13}$ $= 1.2 \times 10^{-20}$ $\text{Step 2:}$ $\text{Calculate the concentration of } \text{S}^{2-} \text{ as follows:}$ $\frac{[\text{H}^+]^2 [\text{S}^{2-}]}{[\text{H}_2\text{S}]}$ $= 1.2 \times 10^{-20} = \frac{(0.2)^2 \times [\text{S}^{2-}]}{0.1}$ $[\text{S}^{2-}] = \frac{0.1 \times 1.2 \times 10^{-20}}{0.04}$ $= 3 \times 10^{-20} \text{ M}$

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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
}