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Current Question (ID: 8489)

Question:

$\text{The first ionization constant of H}_2\text{S is } 9.1 \times 10^{-8}\text{. The concentration of HS}^- \text{ ion in its 0.1 M solution will be:}$

Options:

1. $12.3 \times 10^{-7} \text{ M}$
2. $11.4 \times 10^{-6} \text{ M}$
3. $3.5 \times 10^{-4} \text{ M}$
4. $9.54 \times 10^{-5} \text{ M}$

Solution:

$\text{Hint: } K_{a_1} = \frac{[\text{H}^+][\text{HS}^-]}{[\text{H}_2\text{S}]}$ $\text{Explanation:}$ $\text{Step 1: Write the equilibrium concentration}$ $\text{Let the concentration of HS}^- \text{ be x M.}$ $\text{H}_2\text{S} \rightleftharpoons \text{H}^+ + \text{HS}^-$ $C_i\text{: }\quad 0.1\quad\quad 0\quad\quad 0$ $C_f\text{: }\quad 0.1-x\quad x\quad\quad x$ $\text{Then, } K_{a_1} = \frac{[\text{H}^+][\text{HS}^-]}{[\text{H}_2\text{S}]}$ $\text{Step 2: Calculate HS}^- \text{ concentration}$ $9.1 \times 10^{-8} = \frac{(x)(x)}{0.1-x}$ $(9.1 \times 10^{-8})(0.1-x) = x^2$ $\text{Taking 0.1 - x ≈ 0.1 M, we have } (9.1 \times 10^{-8})(0.1) = x^2$ $9.1 \times 10^{-9} = x^2$ $x = \sqrt{9.1 \times 10^{-9}}$ $= 9.54 \times 10^{-5}\text{ M}$ $\Rightarrow [\text{HS}^-] = 9.54 \times 10^{-5}\text{ M}$

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
}