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Current Question (ID: 8478)

Question:

$\text{A mixture of 10 mL of 0.2 M Ca(OH)}_2 \text{ and 25 mL of 0.1 M HCl is prepared. The pH of the resultant mixture would be:}$

Options:

1. $1.90$
2. $13.42$
3. $1.47$
4. $12.63$

Solution:

$\text{Hint: Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O}$ $\text{Step 1: Calculate the number of moles of OH}^- \text{ and H}^+$ $\text{Total number of moles of Calcium hydroxide} = \frac{(10 \times 0.2)}{(1000)} = 0.002 \text{ moles}$ $\text{Total number of moles of HCl} = \frac{(25 \times 0.1)}{(1000)} = 0.0025 \text{ moles}$ $\text{Step 2: Calculate the number of moles of calcium hydroxide unreacted}$ $\text{Reaction: Ca(OH)}_2 + 2\text{HCl} \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O}$ $\text{1 mole of Ca(OH)}_2 \text{ reacts with 2 moles of HCl. Thus, 0.0025 moles of HCl will react with 0.00125 moles of Ca(OH)}_2$ $\text{Total number of moles of Ca(OH)}_2\text{unreacted} = 0.002 - 0.00125 = 0.00075$ $\text{Step 3: Calculate pH}$ $\text{Total volume of solution} = 35\text{mL}$ $\text{Thus molarity of solution} = \frac{(0.00075 \times 1000)}{(35)} = 0.00214\text{M}$ $\text{Thus,}$ $[\text{OH}^-] = 2 \times 0.0214 = 0.0428$ $\text{and pOH} = -\log(0.0428) = 1.37$ $\Rightarrow \text{pH} = 14 - 1.37 = 12.63$

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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
}