To determine the weight of ammonium sulfate originally present in the solution, we need to analyze the chemical reactions involved and the stoichiometry of the neutralization process. Let's break this down step by step.
Understanding the Reaction
Ammonium sulfate, represented as (NH₄)₂SO₄, reacts with sodium hydroxide (NaOH) to produce ammonia (NH₃), water (H₂O), and sodium sulfate (Na₂SO₄). The reaction can be summarized as follows:
- (NH₄)₂SO₄ + 2 NaOH → 2 NH₃ + Na₂SO₄ + 2 H₂O
When the solution is boiled, the ammonia is expelled, leaving behind sodium sulfate and any excess sodium hydroxide. The remaining NaOH in the solution is then neutralized by hydrochloric acid (HCl).
Neutralization of Excess Base
The neutralization reaction between NaOH and HCl is straightforward:
From the problem, we know that 40 mL of 0.1 M HCl was required to neutralize the excess NaOH. First, let's calculate the moles of HCl used:
Calculating Moles of HCl
Using the formula for moles:
- Moles of HCl = Molarity × Volume (in liters)
Converting 40 mL to liters gives us 0.040 L. Therefore:
- Moles of HCl = 0.1 mol/L × 0.040 L = 0.004 moles
Relating HCl to NaOH
Since the reaction between NaOH and HCl is a 1:1 ratio, the moles of NaOH that were in excess will also be 0.004 moles. Now, we need to find out how many moles of NaOH were initially present in the solution.
Calculating Initial Moles of NaOH
The initial volume of NaOH used was 100 mL of 1 M NaOH:
- Moles of NaOH = 1 mol/L × 0.100 L = 0.1 moles
Now, we can find the moles of NaOH that reacted with ammonium sulfate:
- Moles of NaOH that reacted = Initial moles - Excess moles
- Moles of NaOH that reacted = 0.1 moles - 0.004 moles = 0.096 moles
Calculating Moles of Ammonium Sulfate
From the balanced equation, we see that 2 moles of NaOH react with 1 mole of ammonium sulfate. Therefore, the moles of ammonium sulfate can be calculated as follows:
- Moles of (NH₄)₂SO₄ = Moles of NaOH that reacted / 2
- Moles of (NH₄)₂SO₄ = 0.096 moles / 2 = 0.048 moles
Finding the Weight of Ammonium Sulfate
Now that we have the moles of ammonium sulfate, we can calculate its weight. The molar mass of ammonium sulfate is approximately:
- N: 14.01 g/mol × 2 = 28.02 g/mol
- H: 1.01 g/mol × 8 = 8.08 g/mol
- S: 32.07 g/mol × 1 = 32.07 g/mol
- O: 16.00 g/mol × 4 = 64.00 g/mol
Adding these together gives:
- Molar mass of (NH₄)₂SO₄ = 28.02 + 8.08 + 32.07 + 64.00 = 132.17 g/mol
Finally, we can find the weight of the ammonium sulfate:
- Weight = Moles × Molar Mass
- Weight = 0.048 moles × 132.17 g/mol ≈ 6.35 g
Thus, the weight of ammonium sulfate originally present in the solution is approximately 6.35 grams.
