Import Question JSON

Current Question (ID: 8557)

Question:

$\text{H}_2\text{S gas when passed through a solution of cations containing HCl precipitates the cations of the second group in qualitative analysis but not those belonging to the fourth group. It is because:}$

Options:

1. $\text{Presence of HCl decreases the sulphide ion concentration}$
2. $\text{Presence of HCl increases the sulphide ion concentration}$
3. $\text{Solubility product of group II sulphides is more than that of group IV sulphides}$
4. $\text{Sulphides of group IV cations are unstable in HCl}$

Solution:

$\text{Hint: Sulphide of second group cation has high less solubility product value than sulphide of fourth group cation}$ $\text{In qualitative analysis of cations of second group, H}_2\text{S gas is passed in presence of HCl due to common ion effect, lower concentration of sulphide ions is obtained which is sufficient for the precipitation of second group cations in the form of their sulphides due to lower value of their solubility product } (K_{sp}).$ $\text{Here, fourth group cations are not precipitated because it require more sulphide ions for exceeding their ionic product to their solubility products which is not obtained here due to common ion effect.}$

Import JSON File

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
}