Chem 218 Exp. 8

# Plot the titration curves for the titration of khp

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Plot the titration curve(s) for the titration of KHP, determine the equivalence point and molarity of NaOH. Compare this molarity with that form part (1); discuss any discrepancies. The equivalence point is roughly 21 mL of NaOH. This makes the molarity .109 which is only a difference of .005. There were no discrepancies. Titration Curve for the titration of KHP 0 2 4 6 8 10 12 0 5 10 15 20 25 30 mL of NaOH Solution pH

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3. Calculate average volume of NaOH used in titrations of acetic acid and corresponding concentration of vinegar sample (before dilution) both in terms of molarity and % (V/V). Average volume of NaOH used in titrations of acetic acid & vinegar: Average mL NaOH = (26 + 25.8 + 26 + 25.7)/4 = 25.875 mL Molarity: × 0.025875 L = 0.002691 mol NaOH 0.002691 mol/0.02 L = 0.13455 M vinegar in 100 mL of water (stock solution) = (0.13455 M)(100 mL) = ()(20 mL) 13.455 = ()(20 mL) = 0.67275 M of vinegar solution B % (V/V): 0.67 mol acetic acid × = 40.2 g acetic acid 0.42 g acetic acid × = 38.29 mL acetic acid × 100 = 3.829 % (V/V) 4. Plot the titration curve(s) for the titration of acetic acid, determine the equivalence point and concentration of vinegar. Compare the concentration with that from part (2) ad with 5% (V/V) advertised on the label; discuss any discrepancies. The Equivalency point is roughly 25.6mL. This changes the molarity to about .2876 which changes the V/V % to 3.997%. This is extremely close to our original percent. There were no discrepancies. Titration Curve for the titration of Acetic Acid 0 2 4 6 8 10 12 0 5 10 15 20 25 30 mL of NaOH solution pH pH
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• Fall '12
• professoridon'tknow
• pH, KHP

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