A sodium hydroxide solution was prepared according to the procedure outlined in

Question

A sodium hydroxide solution was prepared according to the procedure outlined in this week's experiment. This sodium hydroxide solution was standardized by titration against exactly 10 mL of a 0.2867 M hydrochloric acid solution.

In three titrations, the volumes of the sodium hydroxide solution required to reach the equivalence point were 8.23 mL, 8.38 mL, and 8.14 mL.

What is the molarity of your sodium hydroxide solution?

Molarity of NaOH = M

This sodium hydroxide solution was then used to determine the amount of acetic acid in vinegar. Three titrations of the sodium hydroxide solution against exactly 5.0 mL of vinegar using phenolphthalein indicator required 22.13 mL, 21.95 mL, and 22.21 mL of the sodium hydroxide solution.

What is the molarity of the acetic acid in vinegar?

Molarity of acetic acid in vinegar = M

Explanation / Answer

(1) NaOH + HCl => NaCl + H2O

Average volume of NaOH = (8.23 + 8.38 + 8.14)/3 = 8.25 mL = 0.00825 L


Moles of NaOH = moles of HCl = volume x concentration of HCl

= 10/1000 x 0.2867 = 0.002867 mol


Molarity of NaOH = moles/volume of NaOH

= 0.002867/0.00825

= 0.3475 M = 0.348 M


(2) NaOH + CH3COOH => CH3COONa + H2O

Average volume of NaOH = (22.13 + 21.95 + 22.21)/3 = 22.10 mL = 0.02210 L


Moles of acetic acid = moles of NaOH = volume x concentraton of NaoH

= 0.02210 x 0.3475 = 0.007679 mol


Molarity of acetic acid = moles/volume of acetic acid

= 0.007679/0.005

= 1.536 M = 1.54 M

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