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Current Question (ID: 8543)

Question:

$\text{The } \text{K}_{sp} \text{ of } \text{Ag}_2\text{CrO}_4\text{, } \text{AgCl, } \text{AgBr and } \text{AgI are, respectively, } 1.1\text{x}10^{-12}\text{, } 1.8\text{x}10^{-10}\text{, } 5.0\text{x}10^{-13}\text{, and } 8.3\text{x}10^{-17}. \text{ The salt precipitates that last if the } \text{AgNO}_3 \text{ solution is added to the solution containing equal moles of } \text{NaCl, } \text{NaBr, } \text{NaI and } \text{Na}_2\text{CrO}_4 \text{ is -}$

Options:

1. $\text{AgI}$
2. $\text{AgCl}$
3. $\text{AgBr}$
4. $\text{Ag}_2\text{CrO}_4$

Solution:

$\text{Hint: Compound having less solubility value will precipitate first.} \n\n\text{Calculate the solubility value for } \text{Ag}_2\text{CrO}_4\text{, } \text{AgCl, } \text{AgBr and } \text{AgI respectively as follows:} \n\n\text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^{+} + \text{CrO}_4^{2-} \n\n\text{Solubility product} \n\n\text{K}_{sp} = (2\text{s})^2 \times \text{S} = 4\text{s}^3 \n\n\text{K}_{sp} = (1.1 \times 10^{-12}) \n\n\text{S} = \sqrt[3]{\frac{\text{K}_{sp}}{4}} = 0.65 \times 10^{-4} \n\n\text{AgCl } \rightleftharpoons \text{Ag}^{+} + \text{Cl}^{-} \n\n\text{K}_{sp} = \text{S} \times \text{S } (\text{K}_{sp} = 1.8 \times 10^{-10}) \n\n\text{S} = \sqrt{\text{K}_{sp}} = 1.34 \times 10^{-5} \text{ (Correction from original image, Ksp for AgCl is } 1.8 \times 10^{-10}\text{, not } 5 \times 10^{-13}\text{ for AgBr)} \n\n\text{AgI } \rightleftharpoons \text{Ag}^{+} + \text{I}^{-} \n\n\text{K}_{sp} = \text{S} \times \text{S } (\text{K}_{sp} = 8.3 \times 10^{-17}) \n\n\text{S} = \sqrt{\text{K}_{sp}} = 0.9 \times 10^{-8} \n\n\text{AgBr } \rightleftharpoons \text{Ag}^{+} + \text{Br}^{-} \n\n\text{K}_{sp} = \text{S} \times \text{S } (\text{K}_{sp} = 5.0 \times 10^{-13}) \n\n\text{S} = \sqrt{\text{K}_{sp}} = 7.07 \times 10^{-7} \n\n\text{Comparing the solubilities:} \n\n\text{S(Ag}_2\text{CrO}_4) = 0.65 \times 10^{-4} = 6.5 \times 10^{-5} \n\n\text{S(AgCl)} = 1.34 \times 10^{-5} \n\n\text{S(AgI)} = 0.9 \times 10^{-8} = 9 \times 10^{-9} \n\n\text{S(AgBr)} = 7.07 \times 10^{-7} \n\n\text{The order of solubility from lowest to highest is:} \n\n\text{AgI} < \text{AgCl} < \text{AgBr} < \text{Ag}_2\text{CrO}_4 \n\n\text{Since the solubility of } \text{Ag}_2\text{CrO}_4 \text{ is highest, it will precipitate last.}$

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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
}