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Consider a 125 mL buffer solution at 25°C that contains 0.500 mol of hypochlorous acid (HOCl) and 0.500 mol of sodium hypochlorite (NaOCl). What will be the pH of this buffer solution after adding 0.341 mol of HCl? The Ka of hypochlorous acid is 2.9 x 10-8 . Assume the change in volume of the buffer solution is negligible

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Answer:

pH = 6.82

Explanation:

To solve this problem we can use the Henderson-Hasselbach equation:

  • pH = pKa + log[tex]\frac{[NaOCl]}{[HOCl]}[/tex]

We're given all the required data to calculate the original pH of the buffer before 0.341 mol of HCl are added:

  • pKa = -log(Ka) = -log(2.9x10⁻⁸) = 7.54
  • [HOCl] = [NaOCl] = 0.500 mol / 0.125 L = 4 M
  • pH = 7.54 + log [tex]\frac{4}{4}[/tex]
  • pH = 7.54

By adding HCl, we simultaneously increase the number of HOCl and decrease NaOCl:

  • pH = 7.54 + log[tex]\frac{[NaOCl-HCl]}{[HOCl+HCl]}[/tex]
  • pH = 7.54 + log [tex]\frac{(0.500mol-0.341mol)/0.125L}{(0.500mol+0.341mol)/0.125L}[/tex]
  • pH = 6.82

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