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Current Question (ID: 21156)

Question:

$\text{A 25 mL buffer solution is prepared by mixing CH}_3\text{COOH of concentration 0.1 M and CH}_3\text{COONa of concentration 0.01 M. If the pH of the solution is 5, then the } pK_a \text{ of CH}_3\text{COOH is:}$ $1. \text{ Four (4)}$ $2. \text{ Five (5)}$ $3. \text{ Six (6)}$ $4. \text{ Seven (7)}$

Options:

1. $\text{Four (4)}$
2. $\text{Five (5)}$
3. $\text{Six (6)}$
4. $\text{Seven (7)}$

Solution:

$\text{Hint: } pH = pK_a + \log \left( \frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]} \right)$ $\text{Step 1:}$ $\text{The formula of Henderson Hasselbach equation is as follows:}$ $pH = pK_a + \log \left( \frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]} \right)$ $\text{The given values are as follows:}$ $[\text{CH}_3\text{COOH}] = 0.1$ $[\text{CH}_3\text{COONa}] = 0.01$ $\text{Step 2:}$ $\text{Calculate the value of } pK_a \text{ as follows:}$ $pH = pK_a + \log \left( \frac{[\text{CH}_3\text{COONa}]}{[\text{CH}_3\text{COOH}]} \right)$ $= pH - \log \left( \frac{0.01}{0.1} \right)$ $= pH - \log [10^{-1}] = pH + \log 10$ $pK_a = 5 + 1 = 6$ $\text{Thus, option third is the correct answer.}$

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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
}