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Current Question (ID: 8539)

Question:

\text{The maximum concentration of equimolar solutions, of ferrous sulphate and sodium sulphide, so that when mixed in equal volumes, there is no precipitation of iron sulphide, will be:} \text{(For iron sulphide, } K_{sp} = 6.3 \times 10^{-18}\text{).}

Options:

1. $5.02 \times 10^{-9} \text{ M}$
2. $5.02 \times 10^{9} \text{ M}$
3. $2.25 \times 10^{-13} \text{ M}$
4. $\text{Can't predict}$

Solution:

\text{Hint: Precipitation occurs when ionic product > solubility.} \text{Step 1: Let the maximum concentration of each solution be } x \text{ mol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., } \frac{x}{2}. \text{Thus,} [FeSO_4] = [Na_2S] = \frac{x}{2} \text{ M} \text{So, } [Fe^{2+}] = [FeSO_4] = \frac{x}{2} \text{ M} \text{Step 2: Calculate x using Solubility product} \text{Reaction: } FeS \rightleftharpoons Fe^{2+}_{(aq)} + S^{2-}_{(aq)} K_{sp} = [Fe^{2+}][S^{2-}] 6.3 \times 10^{-18} = \frac{x}{2} \times \frac{x}{2} 6.3 \times 10^{-18} = \frac{x^2}{4} x^2 = 4 \times 6.3 \times 10^{-18} = 25.2 \times 10^{-18} x = \sqrt{25.2 \times 10^{-18}} = \sqrt{25.2} \times 10^{-9} \approx 5.02 \times 10^{-9} \text{If the concentrations of both solutions are equal to or less than } 5.02 \times 10^{-9} \text{ M, then there will be no precipitation.}

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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
}