I need the answers for the calculation portion and the discussion portion. If th
ID: 1034427 • Letter: I
Question
I need the answers for the calculation portion and the discussion portion.If the vinegar contains 5% acetic acid, the amount of vinegar which is needed to reach and point with 25 mL of 0.1849 M NaOH is... Trial #1: Initial: 0 mL NaOH Ending: 25 mL NaOH Mass of flask: 108.19g Mass of flask + solution : 113.65 g
Trial #2: Initial: 25 mL NaOH Ending: 56.7 mL NaOH Mass of flask: 108.10g Mass of flask + solution: 113.65g
Household ammonia is 3% ammonia, The amount of household ammonia which is needed to reach an endpoint with 25 mL of 0.221 M HCl is... Trial 1: Initial: 0 mL HCl Ending: 13.5 mL HCl Mass of flask: 108.02g Mass of flask + ammonia = 111.17 g
Trial 2: Initial: 13.5 mL HCl Ending: 27.8 mL HCl Mass of flask: 108.11 g Mass of flask + ammonia : 111.26g
5e Percent ammonia in household ammonia 1. Dispose of the rest of the NaOH solution in the buret in the sink. Wash the buret v distilled water. Obtain approximately 60 mL of the standard solution of 0.200 M hydrochloric acid. Pour approximately 4-5 mL of the solution into the buret to rinse out any remaining distilled water. Fill the buret to approximately the 0 mL mark and open the stopcock letting the solution flow through the tip removing any air bubbles. Note the beginning volume of HCI in your buret to the nearest 0.1 mL 2. It is assumed that household ammonia contains 3% ammonia. Calculate how much ammonia must be weighed out so that the end point requires 25 mL of your hydrochloric acid solution for the methyl red end point. Weigh the household ammonia in an Erlenmeyer flask and add a few drops of methyl red indicator 3. Titrate the household ammonia sample with HCI 4. Repeat the titration with another sample of household ammonia. CALCULATE 1. Percent acetic acid in vinegar from the two titrations. 2. Average percent acetic acid in vinegar. 3. Percent ammonia in household ammonia from the two titrations. 4. Average percent ammonia in household ammonia. DISCUSSION Discuss the consistency of the values of percentage by mass for both solutions, and compare their values to those shown on their container. Suggest possible reasons for any significant differences
Explanation / Answer
Percent acetic acid in vinegar
CH3COOH + NaOH --> CH3COONa + H2O
Trial#1 : volume vinegar = 5.46 g
volume 0.1849 M NaOH used = 25 ml
moles NaOH = 0.1849 M x 25 ml = 4.6225 mmol
moles CH3COOH present = 4.6225 mmol
mass CH3COOH acid present = 4.6225 mmol x 60.05 g/mol/1000 = 0.28 g
1a. percent acetic acid = 0.28 g x 100/5.49 g = 5.06%
Trial#2 : volume vinegar = 5.55 g
volume 0.1849 M NaOH used = 31.7 ml
moles NaOH = 0.1849 M x 31.7 ml = 5.86133 mmol
moles CH3COOH present = 5.86133 mmol
mass CH3COOH acid present = 5.86133 mmol x 60.05 g/mol/1000 = 0.352 g
1b. percent acetic acid = 0.353 g x 100/5.55 g = 6.34%
2. Average percent acetic acid in vinegar = 5.70%
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Household NH3
NH3 + HCl ---> NH4Cl
Trial#1, mass NH3 solution = 3.15 g
Volume 0.221 M HCl used = 13.5 ml
moles HCl used = 0.221 M x 13.5 ml = 2.9835 mmol
moles NH3 in solution = 2.9835 mmol
mass NH3 = 2.9835 mmol x 17 g/mol/1000 = 0.051 g
3a) percent NH3 = 0.051 x 100/3.15 g = 1.61%
Trial#2, mass NH3 solution = 3.15 g
Volume 0.221 M HCl used = 14.3 ml
moles HCl used = 0.221 M x 14.3 ml = 3.1603 mmol
moles NH3 in solution = 3.1603 mmol
mass NH3 = 3.1603 mmol x 17 g/mol/1000 = 0.054 g
3a) percent NH3 = 0.054 x 100/3.15 g = 1.70%
4) average percent NH3 = 1.655%