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vazorg [7]
3 years ago
7

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
Chemistry
1 answer:
V125BC [204]3 years ago
8 0

Answer:

pH = 2.66

Explanation:

  • Acetic Acid + NaOH → Sodium Acetate + H₂O

First we <u>calculate the number of moles of each reactant</u>, using the <em>given volumes and concentrations</em>:

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

We<u> calculate how many acetic acid moles remain after the reaction</u>:

  • 37.5 mmol - 10 mmol = 27.5 mmol acetic acid

We now <u>calculate the molar concentration of acetic acid after the reaction</u>:

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

Then we <u>calculate [H⁺]</u>, using the<em> following formula for weak acid solutions</em>:

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

Finally we <u>calculate the pH</u>:

  • pH = -log[H⁺]
  • pH = 2.66
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