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Current Question (ID: 21149)

Question:

$\text{pH of 0.005 M calcium acetate (} pK_a \text{ of CH}_3\text{COOH} = 4.74) \text{ is:}$

Options:

1. $7.04$
2. $9.37$
3. $9.26$
4. $8.37$

Solution:

$\text{Hint: Use the formula, that is, } \text{pH} = 7 + \frac{pK_a}{2} + \frac{\log C}{2}$ $\text{Step 1:}$ $(\text{CH}_3\text{COO})_2\text{Ca} \rightarrow \text{Ca}^{2+} + 2\text{CH}_3\text{COO}^-$ $\text{The initial concentration of } (\text{CH}_3\text{COO})_2\text{Ca} \text{ is 0.005 M. 1 mole } (\text{CH}_3\text{COO})_2\text{Ca} \text{ generated 2 moles CH}_3\text{COO}^- \text{. Hence, the amount of CH}_3\text{COO}^- \text{ generated is 0.01 M}$ $\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-$ $\text{Step 2:}$ $\text{The formula for calculating pH of the solution is as follows:}$ $\text{pH} = 7 + \frac{pK_a}{2} + \frac{\log C}{2}$ $\text{pH} = 7 + \frac{4.74}{2} + \frac{\log 0.01}{2}$ $\text{pH} = 7 + 2.37 - 1$ $\text{pH} = 8.37$

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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
}