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Ammonia Titration Laboratory Activity The Problem:

Ammonia Titration Laboratory Activity The Problem: Determine the molarity (M) and percent basicity (by mass) of ammonium hydroxide by acid/base titration. Information: 1. Ammonia in water forms NH 4 OH . (@ prep table) 2. Do not titrate ammonia samples of more than

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Ammonia Titration Laboratory Activity The Problem:

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  1. Ammonia Titration Laboratory Activity • The Problem: • Determine the molarity (M)andpercent basicity (by mass) of ammonium hydroxide by acid/base • titration. • Information: • 1. Ammonia in water forms NH4OH. (@ prep table) • 2. Do not titrate ammonia samples of more than • 10.0ml at a time. • 3. The burette at your lab station contains 1.2 M HCl. • 4. Do no more than three trials.

  2. Requirements and Goals of this Activity: • 1. Work with your lab partner to summarize the • procedure, which must be checked before you • begin. • 2. Record all data in a table. • 3. Show all work for the calculations. • 4. Assignment counts as a checked lab activity • (vs formal lab)

  3. Titration of Ammonium Hydroxide Solution • Procedure: • ●Mass an empty 125 ml Erlenmeyer flask. • ●Measure 10.0 ml of ammonia and add to flask. • ●Mass the flask with ammonia. • ●Add 2-3 drops of phenolphthalein. (turns pink) • ●Titrate with 1.2 M HCL until pink color • disappears (i.e. turns clear).

  4. Trial 1 Trial 2 MEF & NH4OH ____0.01__ g _________ g MEF _________ g _________ g MNH4OH _________ g _________ g NH4OH (100 ml buret) Vf____0.1__ mL ________ mL Vi _________ mL ________ mL Actual V used _________ mL ________ mL HCl (50 ml buret) Vf____0.01__ mL ________ mL Vi _________ mL ________ mL Actual V used _________ mL ________ mL

  5. Calculations: NH4OH + HCl NH4Cl + H2O (6.75-8.25) (.00750 L HCl) (1.2 mol / L) = .0090 moles HCl (.0090 mol HCl )(1 mol NH4OH) = .0090mol NH4OH (1 mol HCl) .0090 moles NH4OH = 0.9 M NH4OH .0100 L solution (0.81 – 0.99) (.0090 moles NH4OH) (35.0 g ) = 0.32 g NH4OH (1 mol NH4OH) (0.29 – 0.35) N = 14.0 5 H = 5.0 O = 16.0 35.00.32 g NH4OH x 100% = 3.2 % 9.98 g solution (2.9 – 3.5) (9.48-10.48)

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