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Chemistry II

Chemistry II. Acid-Base PH and Titrations. Neutralization Reactions. Acid + Base salt + water NaOH + H 2 SO 4 Na 2 SO 4 + H 2 O. 2. 2 . Base with OH -. Acid with H +. Salt with metal from the base and the negative part of the acid.

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Chemistry II

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  1. Chemistry II Acid-Base PH and Titrations

  2. Neutralization Reactions • Acid + Base salt + water • NaOH + H2SO4 Na2SO4 + H2O 2 2 Base with OH- Acid with H+ Salt with metal from the base and the negative part of the acid Since 1 mole of and acid neutralizes 1 mole of a base, we can use this to find out unknown concentrations by a process of Titration.

  3. Titration Lab • Burette 0.0 ml 0.10 M KOH 50.0 ml 0.10 M KOH Stopcock Ma Va = Mb Vb Erlenmeyer flask Unknown Concentration of acid with know Volume Phenolphthalein indicator->

  4. Procedure: Titration • Prepare 200 ml of 0.10 M KOH • Grind up one 325 mg Aspirin tablet with a pistol and mortar. • Measure out 100 ml of distilled water and dissolve the Aspirin in the water in an Erlenmeyer flask. • Add 1.0 ml of Phenolphthalein (10 dps) • Pour slightly more than 50.0 ml of the 0.10 M KOH in the burette. • Turn on the stopcock and allow the solution to drip into an empty flask until the level at the top is at the “zero” mark.

  5. Continue • Place the Erlenmeyer flask with the aspirin and indicator under the tip of the burette and cautiously drip the KOH into the acid solution. • Place a magnetic stirring bar into the flask and put the apparatus on a magnetic stirring device. • When pink areas occur when the drop hits the acid solution, slow down the process because the end point is near. • Continue to drop the KOH into the Acid until the whole solution is pink.

  6. Titration Lab • Burette 50.0 ml 0.10 M KOH Scale reading = 0.0 ml Ma = Mb Vb Va = 0.10 x 20.0 ml 100.0 ml = 0.020 M 20.0 ml 0.10 M KOH Scale reading = 15.0 ml Volume of 0.10 M KOH = 20.0 ml Erlenmeyer flask Unknown Concentration of acid Volume = 100.0 ml Phenolphthalein indicator->

  7. Find the mass. Calculations • Aspirin ( Acetylsalicylic Acid C9H8O4 ) 0.020 mol/L X 0.100 L x 180.2 g/mol = 0.360 g of Aspirin O OH Size of Aspirin = 325 mg = 0.325 g O %Error = H – L x 100% H = 0.360 -0.325 x 100 0.360 = 9.7 % O Molar Mass = 180.2 g/mol

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