PLEASE HELP... A. REPORT Concentration Acid-Base Titration 20 2. Molarity of Aci

Question

PLEASE HELP...

A. REPORT Concentration Acid-Base Titration 20 2. Molarity of Acid in vinegur Acetic (M) of NaoH volunme s C 3, Initial NaoH in buret level Trial 1 Trial 2 4, Final NaOH level in burct Trial 3 5, Volume (mL) of NaoH used 6. Average volume (mL) 7. Average volume in liters (L) 8. Moles of calcularions) in fitradion mole NaOH 9. Moles of HGH neutralized by Naot 10. 102 of (Show calculations) mole H 11. Grams of HC,H10, (Show caleulations) 2. Percent (m/v) HC1H30, in vinegar (Shiny calculationr)

Explanation / Answer

Volume of NaOH used

Trail 1= 11 ml,   

Trail 2= 23-11= 12 ml,   

Trail 3= 33-23= 10 ml

Average volume = (11+12+10)/3

= 11 ml

1000 ml =1.0 L

So

11 ml -1.0 L /1000 ml= 0.011 L

Chemical reaction of NaOH and CH3COOH vinegar is as follows:

CH3COOH + NaOH = CH3COONa +H2O

Here molarity of NaOH = 0.5 M

Number of moles of NaOH = volume in L * molarity

= 0.011 *0.5

= 0.0055 moles NaOH

Now calculate the moles of CH3COOH as follows:

0.0055 moles NaOH * 1 Mole CH3COOH / 1 Mole NaOH

= 0.0055 Mole CH3COOH

According to the problem the volume of CH3COOH = 5 ml OR 0.005 L

Molarity = number of moles / volume in L

= 0.0055/ 0.005

= 1.1 M

Amount of CH3COOH = number of moles * molar masss

= 0.0055 Mole CH3COOH *60.05 g/ mole

= 0.33 g CH3COOH

% of CH3COOH (m/v)

0.33 g / 5 .0 ml *100

= 6.6 % CH3COOH

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