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 0.2679 M hydrochloric acid solution.

In three titrations, the volumes of the sodium hydroxide solution required to reach the equivalence point were 9.44 mL, 9.59 mL, and 9.35 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 21.59 mL, 21.41 mL, and 21.67 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

Part 1:
Molarity of HCl (known) x volume of HCl (10 mL) = molarity of NaOH (Unknown) x volume of NaOH (Average of three titrations)

So
(0.2679 x 10) / ((9.44+9.59+9.35)/3) = Molarity of HCl
Molarity of HCl = 0.2832 M

Part 2:
Molarity of vinegar (unknown) x volume of vinegar (5 mL) = Molarity of NaOH (previous answer) x volume of NaOH (average of three titrations)

So
(0.2832 x ((21.59+21.41+21.67)/3) / 5 = Molarity of vinegar
Molarity of vinegar = 1.22097 or 1.221M

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