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Current Question (ID: 8498)

Question:
$\text{The p}K_a \text{ of HCN is 9.30. The pH of a solution prepared by mixing 2.5 moles of KCN and 2.5 moles of HCN in water and making up the total volume to 500 ml will be:}$
Options:
  • 1. $9.30$
  • 2. $8.30$
  • 3. $7.30$
  • 4. $10.30$
Solution:
$\text{Hint: Use Henderson-Hasselbalch Equation.}$ $\text{Step 1:}$ $\text{In the Henderson's equation,}$ $\text{pH} = \text{p}K_a + \log \frac{[\text{salt}]}{[\text{acid}]}$ $\text{Calculate the concentration of HCN and KCN respectively as follows:}$ $[\text{HCN}] = \frac{\text{Number of moles}}{\text{volume of solution (L)}}$ $= \frac{2.5 \text{ mol}}{500 \times \frac{1}{1000}}$ $= 5 \text{ M}$ $\text{Similarly, }[\text{KCN}] = \frac{2.5 \text{ mol}}{0.5 \text{ L}} = 5 \text{ M}$ $\text{Since both concentrations are equal:}$ $\frac{[\text{KCN}]}{[\text{HCN}]} = \frac{5}{5} = 1$ $\text{pH} = 9.30 + \log(1) = 9.30 + 0 = 9.30$

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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
}