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 0.2712 M hydrochloric acid solution.

In three titrations, the volumes of the sodium hydroxide solution required to reach the equivalence point were 8.03 mL, 8.18 mL, and 7.94 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 21.63 mL, 21.45 mL, and 21.71 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 + NaOH ------------------> NaCl + H2O

for neutralization reaction,

M1V1 = M2V2

say, M1V1 are related to NaOH and M2V2 are of HCl

given that, M2 = 0.2712 M, V2 = 10mL, V1 = (8.03 + 8.18 +7.94)/3= 8.05 mL

M1 = 0.2712 x 10 / 8.05

     = 0.33 M


NaOH + CH3COOH -------------------> CH3COONa + NaOH

M1V1 = M2V2
say, M1V1 are related to NaOH and M2V2 are of CH3COOH

GIVEN THAT, M1 = 0.33 M, V1 = (21.63+21.45+21.71)/3 = 21.59mL, V2 = 5.0mL, M2= to be determined

M2 = (0.33 x 21.59)/ 5
   = 1.42 M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.