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Current Question (ID: 21101)

Question:

$\text{An aqueous solution contains an unknown concentration of } \text{Ba}^{2+}. \text{When 50 mL of a 1 M solution of } \text{Na}_2\text{SO}_4 \text{ is added, BaSO}_4 \text{ just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO}_4 \text{ is } 1 \times 10^{-10}. \text{ What is the original concentration of Ba}^{2+}?$

Options:

1. $5 \times 10^{-9} \text{ M}$
2. $2 \times 10^{-9} \text{ M}$
3. $1 \times 10^{-9} \text{ M}$
4. $1.0 \times 10^{-10} \text{ M}$

Solution:

$\text{Hint: } K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]$ $\text{Step 1:}$ $\text{Calculate the new concentration of } \text{Na}_2\text{SO}_4 \text{ as follows:}$ $M_1V_1 = M_2V_2$ $1 \times 50 = M_2 \times 500$ $M_2 = 0.1 \text{ M}$ $\text{Thus, the concentration of } \text{Na}_2\text{SO}_4 \text{ is 0.1 M. 1 mole of } \text{Na}_2\text{SO}_4 \text{ dissociates and gives 1 mole } \text{SO}_4^{2-}. \text{ Thus, the concentration of } \text{SO}_4^{2-} \text{ ion is 0.1 M.}$ $\text{Step 2:}$ $\text{Calculate the concentration of } \text{Ba}^{2+} \text{ ion using } K_{sp} \text{ formula as follows:}$ $\text{BaSO}_4 \rightleftharpoons \text{Ba}^{2+} + \text{SO}_4^{2-}$ $K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]$ $1 \times 10^{-10} = [\text{Ba}^{2+}] [0.1]$ $[\text{Ba}^{2+}] = 1 \times 10^{-9} \text{ M}$ $\text{Hence, answer is option third.}$

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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
}