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Current Question (ID: 21119)

Question:

$\text{At 25°C, the solubility product of Mg(OH)}_2 \text{ is } 1.0 \times 10^{-11}. \text{ At which pH, will Mg}^{2+} \text{ ions start precipitating in the form of Mg(OH)}_2 \text{ from a solution of } 0.001 \text{ M Mg}^{2+} \text{ ions?}$

Options:

1. $8$
2. $9$
3. $10$
4. $11$

Solution:

$\text{Hint: } K_{sp} = [\text{Mg}^{2+}][\text{OH}^{-}]^2$ $\text{Solubility of Mg(OH)}_2 = 1.0 \times 10^{-11}$ $\text{Solution molarity is } 0.001$ $\text{Mg(OH)}_2 (s) \rightarrow \text{Mg}^{2+} + 2\text{OH}^{-}$ $\text{Moles of Mg(OH)}_2 (s) = \text{moles of } [\text{Mg}^{2+}]$ $K_{sp} = [\text{Mg}^{2+}] \times [\text{OH}^{-}]^2$ $1 \times 10^{-11} = 0.001 \times [\text{OH}^{-}]^2$ $[\text{OH}^{-}] = 10^{-4} \text{ M}$ $[\text{H}^{+}] = \frac{10^{-14}}{[\text{OH}^{-}]}$ $= \frac{10^{-14}}{10^{-4}}$ $= 10^{-10} \text{ M}$ $\text{pH} = -\log[\text{H}^{+}]$ $\text{putting values}$ $\text{pH} = 10$

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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
}