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Current Question (ID: 8516)

Question:

$\text{K}_\text{a} \text{ for HCN is } 5 \times 10^{-10} \text{ at } 25^{\circ}\text{C}. \text{ For maintaining a constant pH of } 9, \text{ the volume of } 5 \text{ M KCN solution required to be added to } 10 \text{ mL of } 2 \text{ M HCN solution is-}$

Options:

1. $2 \text{ mL}$
2. $3 \text{ mL}$
3. $4.2 \text{ mL}$
4. $5.6 \text{ mL}$

Solution:

\text{Hint: Use Henderson-Hasselbalch Equation} \text{Step 1:} \text{Calculate the molarity of KCN and HCN are as follows:} \text{Let X is the volume of KCN solution to be added.} (10 + X) \text{ mL} = \text{Total volume of solution after mixing the two solutions.} \text{Molarity (M}_2\text{) of HCN in the final solution:} M_2 = \frac{M_1 V_1}{V_2} = \frac{2\text{M} \times 10 \text{ mL}}{(10 + X) \text{ mL}} = \frac{20\text{M}}{(10+X)} \text{Molarity (M'}_2\text{) of KCN in the final solution:} M'_2 = \frac{M_1 V_1}{V_2} = \frac{5\text{M} \times X \text{ mL}}{(10 + X) \text{ mL}} = \frac{5X\text{M}}{(10 + X)} \text{Step 2:} \text{Calculate the volume of } 5\text{M KCN solution is as follows:} \text{Now using Henderson equation, we get} \text{pH} = -\log (K_a) + \log \frac{[\text{KCN}]}{[\text{HCN}]} 9.00 = -\log (5 \times 10^{-10}) + \log \frac{\frac{5X}{(10+X)}\text{M}}{\frac{20}{(10+X)}\text{M}} 9.00 = 9.30 + \log \left(\frac{X}{4}\right) \log \left(\frac{X}{4}\right) = -0.30 \frac{X}{4} = 0.50 X = 2.0 \text{ mL}

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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
}