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g If 50.0 mL of a 0.75 M acetic acid solution is titrated with 1.0 M sodium hydroxide, what is the pH after 10.0 mL of NaOH have been added

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Answer:

pH = 2.66

Explanation:

  • Acetic Acid + NaOH → Sodium Acetate + H₂O

First we calculate the number of moles of each reactant, using the given volumes and concentrations:

  • 0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid
  • 1.0 M NaOH * 10.0 mL = 10 mmol NaOH

We calculate how many acetic acid moles remain after the reaction:

  • 37.5 mmol - 10 mmol = 27.5 mmol acetic acid

We now calculate the molar concentration of acetic acid after the reaction:

27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M

Then we calculate [H⁺], using the following formula for weak acid solutions:

  • [H⁺] = [tex]\sqrt{C*Ka}=\sqrt{0.458M*1.76x10^{-5}}[/tex]
  • [H⁺] = 0.0028

Finally we calculate the pH:

  • pH = -log[H⁺]
  • pH = 2.66

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