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Current Question (ID: 8514)

Question:

$\text{The pH of a solution containing 0.1 mol of }\text{CH}_3\text{COOH}\text{, 0.2 mol of }\text{CH}_3\text{COONa}\text{, and 0.05 mol of NaOH in 1 L of solution is-}$ $(\text{p}K_a \text{ of }\text{CH}_3\text{COOH} = 4.74 \text{ and log 5} = 0.7)$

Options:

1. $4.56$
2. $5.44$
3. $5.04$
4. $3.74$

Solution:

$\text{Hint: A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa.}$ $\text{Step 1:}$ $\text{When }\text{CH}_3\text{COOH}\text{, }\text{CH}_3\text{COONa}\text{ and NaOH are mixed together, }\text{CH}_3\text{COOH}\text{ and NaOH react with each other. The reaction is as follows:}$ $\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}$ $\text{Initial conc.:}\quad 0.1 \quad\quad\quad 0.05 \quad\quad\quad - \quad\quad\quad -$ $\text{After reaction:}\quad 0.05 \quad\quad\quad - \quad\quad\quad 0.05 \quad\quad\quad -$ $\text{After reaction is completed, the total concentration of acid is 0.05 M and total concentration of }\text{CH}_3\text{COONa}\text{ is}$ $0.2 + 0.05 = 0.25$ $\text{Step 2:}$ $\text{Calculate the pH of the solution as follows:}$ $\text{pH} = \text{p}K_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}$ $= 4.74 + \log\frac{0.25}{0.05}$ $= 4.74 + \log 5$ $= 5.44$

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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
}