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Current Question (ID: 8531)

Question:

$\text{The solubility of } \text{Ni (OH)}_2 \text{ in } 0.1 \text{ M NaOH is:}\text{K}_{\text{sp}} (\text{Ni (OH)}_2) = 2 \times 10^{-15}$

Options:

1. $2 \times 10^{-8} \text{ M.}$
2. $1 \times 10^{-13} \text{ M}$
3. $1 \times 10^{8} \text{ M}$
4. $2 \times 10^{-13}$

Solution:

\text{Hint: } K_{\text{sp}} = [\text{Ni}^{2+}][\text{OH}^{-}]^2 \text{Explanation:} \text{STEP 1: } K_{\text{sp}} \text{ for } \text{Ni(OH)}_2 = 2 \times 10^{-15} [\text{OH}^{-}] = 0.1 \text{ (From NaOH)} \text{The equilibrium for } \text{Ni(OH)}_2 \text{ will be represented as:} \text{Ni(OH)}_2 \rightleftharpoons \text{Ni}^{2+} + 2\text{OH}^{-} x \quad \quad \quad (2x + 0.1) \text{Thus, } K_{\text{sp}} = [\text{Ni}^{2+}][\text{OH}^{-}]^2 \text{STEP 2:} K_{\text{sp}} = x(2x + 0.1)^2 \text{Here we can neglect } 2x \text{ as } K_{\text{sp}} \text{ is small so } 2x \ll 0.1 \text{Or, } K_{\text{sp}} = x(0.1)^2 x = \frac{2 \times 10^{-15}}{10^{-2}} = 2 \times 10^{-13} \text{So, the solubility of } \text{Ni(OH)}_2 \text{ in } 0.1 \text{ M NaOH is } 2 \times 10^{-13}

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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
}