The distinctive odor of vinegar is due to acetic acid, CH3COOH, which reacts wit

Question

The distinctive odor of vinegar is due to acetic acid, CH3COOH, which reacts with sodium hydroxide in the following fashion:
CH3COOH(aq)+NaOH(aq)H2O(l)+NaC2H3O2(aq)

Part A

If 4.00 mL of vinegar needs 43.5 mL of 0.115M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 2.00 qt sample of this vinegar?

2.)If 25.8 mL of AgNO3 is needed to precipitate all the Cl ions in a 0.795-mg sample of KCl(forming AgCl), what is the molarity of the AgNO3 solution?

3.)If 45.3 mL of 0.100 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

The distinctive odor of vinegar is due to acetic acid, CH3COOH, which reacts with sodium hydroxide in the following fashion:
CH3COOH(aq)+NaOH(aq)H2O(l)+NaC2H3O2(aq)

Part A

If 4.00 mL of vinegar needs 43.5 mL of 0.115M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 2.00 qt sample of this vinegar?

2.)If 25.8 mL of AgNO3 is needed to precipitate all the Cl ions in a 0.795-mg sample of KCl(forming AgCl), what is the molarity of the AgNO3 solution?

3.)If 45.3 mL of 0.100 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

Explanation / Answer

PART A

43.5 mL of 0.115M NaOH = 43.5 x 0.115 / 1000 = 0.005 Moles

Moles of acetic acid reacted = 0.005 Moles

Mass of acetic acid = 0.005 x 60.05 = 0.3004 gm

0.3004 gm of acetic acid present in 4 ml

2quarts US liquid = 1.893 liter or 1893 ml

Mass of acetic acid in 2 qt = 0.3004 x 1893 /4 = 142.164 gm acetic acid

PART B

0.795-mg sample of KCl = 0.795 / 74.55 = 0.01066 milli moles

Milli moles of AgCl reacted =   0.01066 milli moles

Molarity of the solution =   0.01066 x 1000 /25.8 = 0.4133 mM or 4.133 x 10-3 M

PART C

45.3 mL of 0.100 M HCl = 45.3 x 0.1 /1000 = 0.00453 Moles

Moles of KOH Reacted = 0.00453

Mass of KOH = 0.00453 x 56.1 = 0.254 gm

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