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Current Question (ID: 8510)

Question:

$\text{If a solution of 0.1 M NH}_4\text{OH and 0.1 M NH}_4\text{Cl has pH 9.25, then the pK}_b \text{ of NH}_4\text{OH will be:}$

Options:

1. $9.25$
2. $4.75$
3. $3.75$
4. $8.25$

Solution:

$\text{HINT: Henderson - Hasselbalch equation.}$ $\text{Explanation:}$ $\text{Step 1:}$ $\text{Solution of NH}_4\text{OH} + \text{NH}_4\text{Cl act as basic buffer.}$ $\text{The equation is as follows:}$ $\text{pOH} = \text{pK}_b + \log \frac{[\text{Salt}]}{[\text{Base}]}$ $\text{Calculate the value of pOH is as follows:}$ $\text{pOH} + \text{pH} = 14$ $\text{pOH} = 14 - 9.25$ $\text{pOH} = 4.75$ $\text{Step 2:}$ $\text{Calculate the value of pK}_b \text{ is as follows:}$ $\text{pK}_b + \log \frac{[\text{Salt}]}{[\text{base}]} = \text{pOH}$ $4.75 - \log \left(\frac{0.1}{0.1}\right) = \text{pK}_b$ $4.75 - 0 = \text{pK}_b$ $\text{pK}_b = 4.75$

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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
}