Question

A.) Vinegar is a solution of acetic acid, HCH3COO, in water. Because only the first hydrogen atom ionizes, acetic acid is a monoprotic acid. A 25.00 mL sample of vinegar was analyzed using 0.002450 M NaOH solution. The analysis required 8.93 mL of the sodium hydroxide solution. Write a balanced equation for the reaction of acetic acid with sodium hydroxide. How many mg of acid are present in the sample of vinegar? What is the mass percent acetic acid in the vinegar solution. (You may assume the density of vinegar is exactly 1 g/mL.)

B.) Copper (II) sulfide is oxidized by molecular oxygen to produce gaseous sulfur trioxide and solid copper (II) oxide. Write a balanced chemical equation for this reaction. Identify which atoms are being oxidized and which are reduced. Write the two balanced half reactions that make up the reaction and prove that your balanced equation is correct. Write a paragraph analyzing the half reactions and the balanced equation. What stands out for you? What surprises you? What makes this equation challenging? Look at the half reactions and the balanced equation to answer this part.

Answer 1

1. Write a balanced equation for the reaction of acetic acid with sodium hydroxide.

Answer: CH₃COOH (aq) + NaOH (aq) --> CH₃COONa (aq) + H₂O (l)

2. How many mg of acid are present in the sample of vinegar?

Answer: Since the acid is monoprotic and also base is monobasic, there will be 1:1 mole ratio

moles NaOH = 0.002450 M x 8.93 mL x 10^-3 L= 0.0000218785 mol

moles acetic acid = 0.0000218785 mol

mass of acetic acid = moles acetic acid x molar mass of acetic acid = 0.0000218785 mol x 60.05 g/mol = 0.00131380392 grams, since 1 gram = 1000 mg hence

mg of acid are present in the sample of vinegar = 1.3138 mg

3. What is the mass percent acetic acid in the vinegar solution

mass % = (mass of acetic acid in sample / mass of vinegar solution titrated) x 100%

mass % = (0.0013138 g / 25 g) x 100% = 0.005255 %

mass percent acetic acid in the vinegar solution = 5.2 x 10^-3 %

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Balanced equation: CuS + 2O₂ → CuO + SO₃

"S" atom is being oxidized & "O" atoms are being reduced.

half reactions:

oxidation half cell: S^2- --> S⁶⁺ + 8 e^-

reduction half cell: 2O₂ + 8e^- --> 4 O^2-

S⁶⁺ + 4 O^2- --> SO₃ and Cu²⁺ + O^2- --> CuO

What stands out for you? What surprises you?

Answer: copper being intact in terms of oxidation state does not changed.

What makes this equation challenging?

Answer: Anion in CuS (S^2-) becomes cation in SO₃ (S⁶⁺) and copper being intact, this will make challenging during coefficient of oxygen.

Hope this helped you!

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