Acid+Neutrilization

=__Acid Neutralization__= =Facilities, Personal Protection and Equipment=
 * Carry out neutralizations in a well-ventilated fume hood
 * Use the sash or a safety shield for protection against vigorous reactions
 * Wear an apron, splash-proof goggles and a full-face shield and nitrile gloves (other glove material may not provide proper protection). Long gloves or gauntlets protect forearms from splashes.
 * A five-gallon polyethylene bucket is recommended for neutralizing 1 to 10 liters.
 * A large container is needed for addition of cold water / ice and base, and to safely stir the reaction.
 * Perform all steps SLOWLY.
 * **Caution**: Vapors and heat are generated.
 * Have ice on hand to cool an dilute the solution.
 * pH paper

Procedure
Can be used with Acetic, Formic, Hydrochloric, Nitric, Perchloric, Phosphoric, Sulfuric, and Trichloroacetic Acids
 * 1) Estimate the how much base you will need for neutrilization, and the volume vessel you will need to carry out the following. 5 gallon buckets can be handy for this.
 * 2) Pour amount of acid specified above slowly into water to make 3 N or less.
 * 3) Neutralize by adding 6 N sodium hydroxide solution, stirring continually.
 * 4) As heat builds up, add ice to dilute and cool solution.
 * 5) Monitor pH change with a suitable indicator or check periodically with pH paper (e.g., to go from pH 2 to pH 12 takes only 30 ml of 6 N NaOH in 15 L of solution).
 * 6) When pH > 2 is reached, the solution may be washed down the sanitary sewer with tap water rinse to clear the drain trap.

Example Neutrilization
You have a solution containing:
 * 300 mL of ammonium hydroxide
 * 1.5 L of hydrochloric acid
 * 250 mL of sulfuric acid
 * 400 mL 20% trichloroacetic acid solution

How many grams of sodium hydroxide will you need to neutralize all of these items?

__Step One__: Calculate how many moles of acid protons you have. hydrochloric acid: 1500 mL ÷ 83.0 mL/mole = 18 moles sulfuric acid: 250 mL ÷ 27.2 mL/mole = 9 moles trichloroacetic acid: 400 mL ÷ 871 mL/mole = 0.5 moles Total Acid: 28 moles

__Step Two__: Calculate how many moles of base you have: Ammonium hydroxide: 300 mL ÷ 67 mL/mole = 5 moles base

__Step Three__: Subtract moles of base from moles of acid: 28 moles acid - 5 moles base = 23 moles base needed

__Step Four__: Calculate the grams of sodium hydroxide you would need to complete the neutralization: 23 moles base × 40 g/mole NaOH = 920 grams NaOH

You need 920 grams sodium hydroxide to complete the neutralization.