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Current Question (ID: 21095)

Question:

$\text{The molar solubility of Cd(OH)}_2 \text{ is } 1.84 \times 10^{-5} \text{ M in water. The expected solubility of Cd(OH)}_2 \text{ in a buffer solution of pH } = 12 \text{ is:}$

Options:

1. $2.49 \times 10^{-10} \text{ M}$
2. $1.84 \times 10^{-9} \text{ M}$
3. $6.23 \times 10^{-11} \text{ M}$
4. $1.49 \times 10^{-9} \text{ M}$

Solution:

$\text{Hint: Use the formula, that is, } K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2$ $\text{Step 1:}$ $\text{First, calculate the value of } K_{sp} \text{ as follows:}$ $\text{The reaction is as follows:}$ $\text{Cd(OH)}_2 \rightleftharpoons \text{Cd}^{2+} + 2\text{OH}^-$ $\text{The } K_{sp} \text{ formula is as follows:}$ $K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2$ $\text{The molar solubility value is } 1.84 \times 10^{-5} \text{ M}$ $K_{sp} = s \times (2s)^2$ $= 4 \times (1.84 \times 10^{-5})^3$ $\text{Step 2:}$ $\text{Calculate the molar solubility when pH is 12.}$ $\text{pOH = 14 - pH; = 2}$ $\text{pOH = -log[OH}^-\text{]}$ $[\text{OH}^-] = 10^{-2} \text{ M}$ $K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2$ $[\text{Cd}^{2+}] = \frac{K_{sp}}{[\text{OH}^-]^2}$ $K_{sp} = \frac{4 \times (1.84 \times 10^{-5})^3}{(10^{-2})^2} = 24.9 \times 10^{-11}$ $= 2.49 \times 10^{-10}$

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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
}