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Current Question (ID: 8494)

Question:

$\text{The percentage ionization of 0.02 M dimethylamine solution if it also contains 0.1 M NaOH solution }(K_b \text{ of dimethylamine} = 5.4 \times 10^{-4})\text{ will be:}$

Options:

1. $0.54\%$
2. $0.05\%$
3. $5.40\%$
4. $54.00\%$

Solution:

$\text{Hint: NaOH is a strong base so it will undergo complete ionization}$ $\text{Explanation:}$ $\text{Step 1: Write down the given data}$ $K_b = 5.4 \times 10^{-4} \text{ and concentration (dimethylamine)} = 0.02 \text{ M}$ $\text{concentration (NaOH)} = 0.1 \text{ M}$ $\text{Step 2: Calculate \% of ionization}$ $\text{If 0.1 M of NaOH is added to the solution, then NaOH (being a strong base) undergoes complete ionization.}$ $\text{NaOH}_{(aq)} \rightleftharpoons \text{Na}^+_{(aq)} + \text{OH}^-_{(aq)}$ $\text{Conc.} \quad 0.1\text{M} \quad 0.1\text{M}$ $\text{and,}$ $(\text{CH}_3)_2\text{NH} + \text{H}_2\text{O} \rightleftharpoons (\text{CH}_3)_2\text{NH}_2^+ + \text{OH}^-$ $\text{Conc.} \quad 0.02-x \quad x \quad 0.1\text{M (Common ion)}$ $\text{Then, }[(\text{CH}_3)_2\text{NH}_2^+] = x$ $K_b = \frac{[(\text{CH}_3)_2\text{NH}_2^+][\text{OH}^-]}{[(\text{CH}_3)_2\text{NH}]} = \frac{x \times 0.1}{0.02}$ $x = 1.08 \times 10^{-4}$ $\text{The percentage of dimethylamine ionized is } \frac{1.08 \times 10^{-4}}{0.02} \times 100 = 0.54\%$

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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
}