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Current Question (ID: 8483)

Question:

$\text{The ionization constant of acetic acid is } 1.74 \times 10^{-5}\text{. The pH of acetic acid in its 0.05 M solution will be:}$

Options:

1. $7.81$
2. $3.03$
3. $8.54$
4. $1.45$

Solution:

$\text{Hint: Use pH = -log[H}^+\text{]}$ $\text{Explanation:}$ $\text{Step 1: Calculate degree of dissociation,}$ $\alpha = \sqrt{\frac{K_a}{c}}$ $c = 0.05 \text{ M}$ $K_a = 1.74 \times 10^{-5}$ $\text{Then,}$ $\alpha = \sqrt{\frac{1.74 \times 10^{-5}}{0.05}}$ $\alpha = \sqrt{34.8 \times 10^{-5}}$ $\alpha = \sqrt{3.48 \times 10^{-4}}$ $\alpha = 1.86 \times 10^{-2}$ $\text{Step 2: Calculate pH of the solution}$ $\text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+$ $\text{Thus, concentration of CH}_3\text{COO}^- = c\alpha$ $= 0.05 \times 1.86 \times 10^{-2}$ $= 0.093 \times 10^{-2}$ $= 0.00093$ $\text{Since } [\text{OAc}^-] = [\text{H}^+]\text{,}$ $[\text{H}^+] = 0.00093 = 9.3 \times 10^{-3}$ $\text{pH} = -\log[\text{H}^+]$ $= -\log(9.3 \times 10^{-3})$ $\therefore \text{pH} = 3.03$ $\text{Hence, the concentration of acetate ion in the solution is 0.00093 M, and its pH is 3.03.}$

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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
}