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Current Question (ID: 18237)

Question:

$\text{One liter of 0.5 M KCl solution is electrolyzed for one minute in a current of 1.608 mA. Considering 100 \% efficiency, the pH of the resulting solution will be:}$

Options:

1. $7$
2. $9$
3. $8$
4. $10$

Solution:

$\text{HINT: Use , W/E = It/96500, then find current efficiency.}$ $\text{STEP 1:}$ $\text{Electrolysis of KCl will favour the evolution of H}_2 \text{ and Cl}_2.}$ $2\text{KCl} + \text{H}_2\text{O} = 2\text{KOH} + \text{Cl}_2 + \text{H}_2$ $\text{2 KCl will give 2KOH}$ $\text{so solution will have OH}^- \text{ and K}^+$ $\text{amount of charge will be = It = 1.608 \times 10^{-3} \times 1 \times 60 \approx 0.09650}$ $\text{STEP 2:}$ $\text{Since 96500 coulomb will deposite on OH}^-$ $\text{so value can be calculated for .09650 } \approx 10^{-6}$ $\text{that will give pOH value = 6}$ $\text{pH = 14 - pOH = 14 - 6 = 8}$

Import JSON File

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
}