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Current Question (ID: 8542)

Question:

$\text{When equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together } (K_{sp} \text{ (cupric iodate)} = 7.4 \times 10^{-8}), \text{ from the following, the correct observation would be:}$

Options:

1. $\text{Precipitation will occur}$
2. $\text{Precipitation will not occur}$
3. $\text{There are 50\% chances of precipitation}$
4. $\text{Data is not sufficient to make any conclusive observation}$

Solution:

$\text{Hint: If ionic product < } K_{sp}\text{; precipitation will not occur.}$ $\text{Explanation:}$ $\text{When equal volumes are mixed, the concentrations are diluted by half. The molar concentrations after mixing become 0.001 M each.}$ $\text{Then, NaIO}_3 \rightarrow \text{Na}^+ + \text{IO}_3^- \text{; } [\text{IO}_3^-] = 0.001\text{M}$ $\text{and Cu(ClO}_3)_2 \rightarrow \text{Cu}^{2+} + 2\text{ClO}_3^- \text{; } [\text{Cu}^{2+}] = 0.001\text{M}$ $\text{Now, the solubility equilibrium of copper iodate can be written as:}$ $\text{Cu(IO}_3)_2 \rightarrow \text{Cu}^{2+} + 2\text{IO}_3^-$ $\text{Ionic product of copper iodate } = [\text{Cu}^{2+}][\text{IO}_3^-]^2 = (0.001)(0.001)^2 = 10^{-9}$ $\text{Since ionic product < } K_{sp} \text{; precipitation will not occur.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}