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Current Question (ID: 21067)

Question:

$\text{An acidic buffer is obtained on mixing:}$ $1.\ 100\ \text{mL of}\ 0.1\ \text{M}\ \text{CH}_3\text{COOH and}\ 100\ \text{mL of}\ 0.1\ \text{M}\ \text{NaOH.}$ $2.\ 100\ \text{mL of}\ 0.1\ \text{M}\ \text{HCl and}\ 200\ \text{mL of}\ 0.1\ \text{M}\ \text{NaCl.}$ $3.\ 100\ \text{mL of}\ 0.1\ \text{M}\ \text{HCl and}\ 200\ \text{mL of}\ 0.1\ \text{M}\ \text{CH}_3\text{COONa.}$ $4.\ 100\ \text{mL of}\ 0.1\ \text{M}\ \text{CH}_3\text{COOH and}\ 200\ \text{mL of}\ 0.1\ \text{M}\ \text{NaOH.}$

Options:

1. $100\ \text{mL of}\ 0.1\ \text{M}\ \text{CH}_3\text{COOH and}\ 100\ \text{mL of}\ 0.1\ \text{M}\ \text{NaOH.}$
2. $100\ \text{mL of}\ 0.1\ \text{M}\ \text{HCl and}\ 200\ \text{mL of}\ 0.1\ \text{M}\ \text{NaCl.}$
3. $100\ \text{mL of}\ 0.1\ \text{M}\ \text{HCl and}\ 200\ \text{mL of}\ 0.1\ \text{M}\ \text{CH}_3\text{COONa.}$
4. $100\ \text{mL of}\ 0.1\ \text{M}\ \text{CH}_3\text{COOH and}\ 200\ \text{mL of}\ 0.1\ \text{M}\ \text{NaOH.}$

Solution:

$1.\ 100\ \text{mL of}\ 0.1\ \text{M}\ \text{HCl and}\ 200\ \text{mL of}\ 0.1\ \text{M}\ \text{CH}_3\text{COONa will form a buffer.}$ $\text{The initial and final millimoles of reactant and product in the reaction is as follows:}$ $\text{HCl} + \text{CH}_3\text{COONa} \rightarrow \text{CH}_3\text{COOH} + \text{NaCl}$ $10\ \quad 20\ \quad -\ \quad -$ $-\ \quad 10\ \quad 10\ \quad -$ $\text{So finally we get a mixture of}$ $\text{CH}_3\text{COOH} + \text{CH}_3\text{COONa that will work like an acidic buffer solution}$

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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
}