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Current Question (ID: 21069)

Question:
$100 \text{ mL of } 0.1 \text{ M HCl is taken in a beaker and to it } 100 \text{ mL of } 0.1 \text{ M NaOH is added in steps of } 2 \text{ mL and the pH is continuously measured. Which of the following graphs correctly depicts the change in pH?}$
Options:
  • 1. $\text{Graph 1}$
  • 2. $\text{Graph 2}$
  • 3. $\text{Graph 3}$
  • 4. $\text{Graph 4}$
Solution:
$\text{In this problem, we are adding a strong base NaOH to a strong acid (HCl) and observing the change in pH.}$ $1. \text{ Initially, the solution is acidic because it contains } 0.1 \text{ M HCl, which gives a low pH.}$ $2. \text{ As NaOH is added, it starts to neutralize the HCl. Since both NaOH and HCl are strong, they completely dissociate, leading to a sharp increase in pH as the equivalence point (where moles of HCl equal moles of NaOH) is approached.}$ $3. \text{ At the equivalence point, the pH will be around 7 because we are neutralizing a strong acid with a strong base.}$ $4. \text{ After the equivalence point, any further addition of NaOH will make the solution basic, causing the pH to increase further.}$ $\text{Therefore, the graph that correctly depicts the change in pH should show:}$ $- \text{A starting low pH.}$ $- \text{A steep rise as it approaches pH 7.}$ $- \text{A leveling off after passing pH 7 as it becomes basic.}$ $\text{The correct answer is graph 2, which shows an initial low pH, a steep rise at the equivalence point, and a leveling off afterward.}$

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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
}