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Current Question (ID: 8490)

Question:
$\text{The pH value of a 0.01 M solution of an organic acid is 4.15. The pK}_a \text{ of the acid will be:}$
Options:
  • 1. $8.95$
  • 2. $1.43$
  • 3. $3.56$
  • 4. $6.30$
Solution:
$\text{Hint: } K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$ $\text{STEP 1: Calculate [H}^+\text{] of the acid}$ $\text{Let the organic acid be HA}$ $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$ $[\text{HA}] = 0.01\text{M and pH} = 4.15$ $\text{So, } -\log[\text{H}^+] = 4.15 \Rightarrow [\text{H}^+] = 7.08 \times 10^{-5}\text{M}$ $\text{Step 2: Calculate K}_a \text{ using the general expression}$ $K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = \frac{(7.08 \times 10^{-5})^2}{0.01}$ $K_a = 5.01 \times 10^{-7}$ $\text{Step 3: Calculate pK}_a$ $\text{pK}_a = -\log K_a$ $\text{pK}_a = -\log(5.01 \times 10^{-7}) = 6.30$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}