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Current Question (ID: 8451)

Question:

$\text{The conjugate bases of Bronsted acids H}_2\text{O and HF are respectively:}$

Options:

1. $\text{H}_3\text{O}^+ \text{ and H}_2\text{F}^+\text{, respectively.}$
2. $\text{OH}^- \text{ and H}_2\text{F}^+\text{, respectively.}$
3. $\text{H}_3\text{O}^+ \text{ and F}^-\text{, respectively.}$
4. $\text{OH}^- \text{ and F}^-\text{, respectively.}$

Solution:

\text{Hint: Brønsted acid loses } \text{H}^+ \text{ ion} \text{A Brønsted-Lowry acid is any species that is capable of donating a proton and it forms an anion as conjugate base. The reactions are as follows:} \text{H}_2\text{O} \leftrightarrow \text{H}^+ + \text{OH}^- \text{HF} \leftrightarrow \text{H}^+ + \text{F}^- \text{The conjugate bases of } \text{H}_2\text{O} \text{ and HF are } \text{OH}^- \text{ and } \text{F}^- \text{, respectively.}

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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
}