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Current Question (ID: 21124)

Question:

$\text{The first and second dissociation constants of an acid } \text{H}_2\text{A} \text{ are } 1.0 \times 10^{-5} \text{ and } 5.0 \times 10^{-10} \text{ respectively.}$ $\text{The overall dissociation constant of the acid will be:}$

Options:

1. $5.0 \times 10^{-5}$
2. $5.0 \times 10^{15}$
3. $5.0 \times 10^{-15}$
4. $0.2 \times 10^{5}$

Solution:

$\text{Hint: Multiply the two dissociation constant values so that the overall dissociation constant is obtained}$ $\text{The dissociation equations of the acid are as follows:}$ $\text{H}_2\text{A} \rightarrow \text{H}^+ + \text{HA}^- \quad \ldots (K_1) = 1.0 \times 10^{-5}$ $\text{HA}^- \rightarrow \text{H}^+ + \text{A}^{2-} \quad \ldots (K_2) = 5.0 \times 10^{-10}$ $\text{Combine the two equations and multiply the dissociation constants, so that the overall dissociation constant of the acid is obtained.}$ $\text{H}_2\text{A} \rightarrow 2\text{H}^+ + \text{A}^{2-} \quad \ldots (K_3) = K_1 \times K_2$ $\text{Overall dissociation constant } (K_3) = 1.0 \times 10^{-5} \times 5.0 \times 10^{-10}$ $= 5.0 \times 10^{-15}$ $\text{Hence, option third is the correct answer.}$

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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
}