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 a0.2829 M hydrochloric acid solution.

In three titrations, the volumes of the sodium hydroxide solution required to reach the equivalence point were 8.67 mL, 8.82 mL, and 8.58 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 21.57 mL, 21.39 mL, and21.65 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

HCl is a mono basic/protic acid. So its molarity is equal to its normality

NaOH is a monoacidic base.So its molarity is equal to its normality

CH3 COOH a mono basic/protic acid. So its molarity is equal to its normality

CALCULATION OF NORMALITY AND HENCE THE MOLARITY OF NaOH SOLUTION

V1N1 = V2N2

V1 = Volume of NaOH = Average of the the values 8.67 mL, 8.82 mL, and 8.58 mL.

= 8.67 mL +, 8.82 mL, +and 8.58 mL / 3= 8.69 mL

N1 = Normality of NaOH = ?

V2 = Volume of HCl= 10 mL

N2 = Normality of HCl = 0.2829 N

V1N1 = V2N2

N1 = V2N2 / N 1

   = 10 x 0.2829 / 8.69 = 0.3255 N = 0.3255 M

Molarity of NaOH   = 0.3255 M

CALCULATION OF NORMALITY AND HENCE THE MOLARITY OF CH3 COOH SOLUTION (Vinegar)

V1N1 = V2N2

V1 = Volume of CH3 COOH = 5 mL.

N1 = Normality of CH3 COOH = ?

V2 = Volume of NaOH = Average of the the values 21.57 mL, 21.39 mL, and 21.65 mL  

         = 21.57 mL,+ 21.39 mL, + 21.65 mL / 3 = 21.54 mL

N2 = Normality of NaOH = 0.3255 N (from previous titration)

V1N1 = V2N2

N1 = V2N2 / N1

   = 21.54 x 0.3255/5 = 1.4022 N = 1.4022M

Molarity of CH3COOH = = 1.4022M

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