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When 0.1 mol of CH 3 NH 2 (ionisation constant, K b = 5 × 10 -4 ) is mixed with 0.08 mol HCl and the volume is made up to 1 litre. Find the [H + ] ofthe resulting solution. Given log2 = 0.3

Options

  1. A8 × 10 -2
  2. B2 × 10 -11
  3. C1.23 × 10 -4
  4. D8 × 10 -11

Correct answer

D. 8 × 10 -11

Step-by-step solution

CH 3 NH 2 + HCl ⇌ CH 3 NH 3 + Cl - Initial moles 0.1 0.08 0 Resulting solution contains [salt] = 0.08 [base] = 0.1 - 0.08 = 0.02 Applying pOH = pK b + log salt base pK b = -log K b = -log 5 × 10 -4 = 4 - log 5 = 3.30 pOH = 3.30 + log 0.08 0.02 = 3.30 + 0.60 = 3.90 pH + pOH = 14, pH = 14 - 3.902 = 10.1 Applying pH = -log [H + ] 10.1 = − log [ H + ] So H + = 8 × 10 − 11 [H + ] ≫ 8 × 10 -11

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