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Current Question (ID: 8512)

Question:

$\text{The mixture that will produce a buffer solution when mixed in equal volumes is:}$

Options:

1. $0.1 \text{ mol dm}^{-3} \text{ NH}_4\text{OH and } 0.1 \text{ mol dm}^{-3} \text{ HCl}$
2. $0.05 \text{ mol dm}^{-3} \text{ NH}_4\text{OH and } 0.1 \text{ mol dm}^{-3} \text{ HCl}$
3. $0.1 \text{ mol dm}^{-3} \text{ NH}_4\text{OH and } 0.05 \text{ mol dm}^{-3} \text{ HCl}$
4. $0.1 \text{ mol dm}^{-3} \text{ CH}_3\text{COONa and } 0.1 \text{ mol dm}^{-3} \text{ NaOH}$

Solution:

$\text{Hint: Weak base and its salt with strong acid forms buffer when both are present together in solution.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{When the concentration of }\text{NH}_4\text{OH}\text{ (a weak base) is higher than the strong acid (HCl), a mixture of a weak base and its conjugate acid is obtained, which acts as a basic buffer.}$ $\text{Step 2:}$ $\text{The reaction is as follows:}$ $\text{NH}_4\text{OH } 0.05 \text{ M} + \text{HCl } 0.05\text{M} \rightarrow \text{NH}_4\text{Cl } 0.05\text{ M} + \text{H}_2\text{O}$ $\text{Initially:}\quad\quad\quad 0.1 \quad\quad\quad 0.05\text{M} \quad\quad\quad 0$ $\text{After:}\quad\quad\quad\quad\quad 0.05 \quad\quad\quad 0 \quad\quad\quad\quad 0.05$ $\text{In the solution, the weak base and its conjugate acid are left. It is an example of a buffer. Hence, option third is the correct answer.}$

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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
}