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Current Question (ID: 8458)

Question:

$\text{The pH of a solution made by mixing } 50 \text{ mL of } 0.01 \text{ M barium hydroxide solution with } 50 \text{ mL of } \text{H}_2\text{O} \text{ is}$

Options:

1. $3.0$
2. $6.0$
3. $12.0$
4. $15.0$

Solution:

$\textbf{Hint: } \text{Find the molarity after dilution, then find concentration of } \text{OH}^-$ $\textbf{Step 1:}$ $\text{The volume of solution is doubled by the addition of } 50 \text{ mL of water, therefore, the molarity of dilute solution is as follows:}$ $M_1V_1 = M_2V_2$ $0.01 \times 50 = M_2 \times 100$ $M_2 = \frac{0.01 \times 50}{100}$ $M_2 = 0.005 \text{ M}$ $\textbf{Step 2:}$ $\text{The concentration of } \text{OH}^- \text{ ion is as follows:}$ $\text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^-$ $\begin{array}{ccc} \text{Initial conc.} & 0.005 & 0 & 0 \\ \text{Final conc.} & 0 & 0.005 & 2 \times 0.005 = 0.01 \end{array}$ $[\text{OH}^-] = 0.01 = 1 \times 10^{-2}$ $\textbf{Step 3:}$ $\text{Calculate the pH of the solution is as follows:}$ $\therefore [\text{OH}^-] = 0.01 = 1 \times 10^{-2}$ $\text{pOH} = 2$ $\therefore \text{pH} = 14 - 2 = 12$

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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
}