How many milliliters of 0.20 M NaOH must be added to 75 mL of 0.050 M HCl to make a neutral solution?

0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o

Lady bird [3.3K]

Answer:

10.57% of N in acetanilide

Explanation:

All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:

2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻

The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:

Moles Na₂CO₃ anf Moles H₂SO₄ in excess:

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

Moles H₂SO₄ added:

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

Moles that react:

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

Moles of NH₃ = Moles N:

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

mass N and mass percent:

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

<h3>10.57% of N in acetanilide</h3>

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