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Current Question (ID: 8527)

Question:

$\text{The ionization constant of nitrous acid is } 4.5 \times 10^{-4}. \text{ The pH of a } 0.04 \text{ M sodium nitrite solution will be:}$

Options:

1. $2.56$
2. $6.14$
3. $7.96$
4. $11.74$

Solution:

Hint: NaNO2 is the salt of strong base (NaOH) and weak acid (HNO2) Explanation: Step 1: Calculate Kh NO2- + H2O ⇌ HNO2 + OH- Kh = [HNO2][OH-] / [NO2-] Kh = Kw / Ka = (10^-14) / (4.5 × 10^-4) = 2.22 × 10^-11 Step 2: Calculate [OH-] using Kh If x moles undergo hydrolysis: [NO2-] = 0.04 - x ≈ 0.04 [HNO2] = x, [OH-] = x Kh = x^2 / 0.04 = 2.22 × 10^-11 x = [OH-] = 9.43 × 10^-7 M Step 3: Calculate pH [H+] = 10^-14 / (9.43 × 10^-7) = 1.06 × 10^-8 M pH = -log[H+] = 7.97

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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
}