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Current Question (ID: 8675)

Question:

$\text{The softening of hard water by synthetic ion-exchange resins is based on -}$

Options:

1. $\text{Increasing the lattice energy.}$
2. $\text{Physical separation.}$
3. $\text{Lowering the activation energy.}$
4. $\text{Exchange of cations and anions.}$

Solution:

$\text{HINT: Ions in the water are exchanged for other ions fixed to the resin.}$ $\text{Explanation:}$ $\text{The process of treating the permanent hardness of water using synthetic}$ $\text{resins is based on the exchange of cations (e.g., }\text{Na}^+, \text{Ca}^{2+}, \text{Mg}^{2+}\text{ etc)}$ $\text{and anions (e.g., }\text{Cl}^-, \text{SO}_4^{2-}, \text{HCO}_3^-\text{ etc) present in water by }\text{H}^+\text{ and}$ $\text{OH}^-\text{ ions respectively.}$ $\text{Synthetic resins are of two types: 1) Cation exchange resins 2) Anion}$ $\text{exchange resins}$ $\text{Cation exchange resins are large organic molecules that contain the}$ $\text{- }\text{SO}_3\text{H group. The resin is firstly changed to RNa (from }\text{RSO}_3\text{H) by}$ $\text{treating it with }\text{NaCl}.\text{ This resin then exchanges }\text{Na}^+\text{ ions with }\text{Ca}^{2+}\text{ and}$ $\text{Mg}^{2+}\text{ ions, thereby making the water soft.}$ $2\text{RNa}_{\text{(s)}} + \text{M}^{2+}_{\text{(aq)}} \rightarrow \text{R}_2\text{M}_{\text{(s)}} + 2\text{Na}^{+}_{\text{(aq)}}$ $\text{There are cation exchange resins in }\text{H}^+\text{ form. The resins exchange }\text{H}^+\text{ ions}$ $\text{for }\text{Na}^+, \text{Ca}^{2+},\text{ and }\text{Mg}^{2+}\text{ ions.}$ $2\text{RH}_{\text{(s)}} + \text{M}^{2+}_{\text{(aq)}} \rightleftharpoons \text{MR}_{2\text{(s)}} + 2\text{H}^{+}_{\text{(aq)}}$ $\text{Anion exchange resins exchange }\text{OH}^-\text{ions for anions like }\text{Cl}^-,\text{HCO}_3^-,\text{ and}$ $\text{SO}_4^{2-}\text{ present in water.}$ $\text{RNH}_{2\text{(r)}} + \text{H}_2\text{O}_{\text{(l)}} \rightleftharpoons \text{RNH}_3^{+}\text{OH}_{\text{(r)}}^-$ $\downarrow + \text{X}_{\text{(aq)}}^-$ $\text{RNH}_3^{+}\text{X}_{\text{(r)}}^- + \text{OH}_{\text{(aq)}}^-$ $\text{During the complete process, water first passes through the cation exchange}$ $\text{process. The water obtained after this process is free from mineral cations}$ $\text{and is acidic in nature.}$ $\text{This acidic water is then passed through the anion exchange process where}$ $\text{OH}^-\text{ ions neutralize the }\text{H}^+\text{ ions and de-ionize the water obtained.}$

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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
}