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32 ml naoh 2 4 6 8 10 12 ph 464 476 488 506 519 531

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Initial pH (of pH meter in buffer solution) = 6.32 mL NaOH 2 4 6 8 10 12 pH 4.64 4.76 4.88 5.06 5.19 5.31 mL NaOH 14 16 18 20 22 23 pH 5.56 5.71 5.91 6.30 7.15 8.80 mL NaOH 23.5 + 1 drop + 2 drops + 3 drops 24 26 pH 9.52 9.52 9.60 9.58 9.66 10.82 Molarity of titrant for 1 st sample: 0.133 M Molarity of titrant for 2 nd sample: 0.103 M Molarity of titrant for 3 rd sample: 0.105 M Molarity of titrant for 4 th sample: 0.104 M Average Value of NaOH molarities: 0.104 M Part II 1 st titration: 3 drops phenolphthalein 26 mL NaOH
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2 nd titration: 3 drops phenolphthalein 25.8 mL NaOH 3 rd titration: 3 drops phenolphthalein 26 mL NaOH 4 th titration: 3 drops phenolphthalein 25.7 mL NaOH pH values for titration with 4 th sample of acetic acid mL NaOH 2 4 6 8 10 12 pH 3.76 4.02 4.27 4.49 4.63 4.75 mL NaOH 14 16 18 20 22 24 pH 4.89 5.08 5.24 5.49 5.95 6.82 mL NaOH 25 26 pH 7.40 10.26 Molarity of original sample before dilution: 0.67275 M Molarity converted into % (V/V): 3.829 % Sample Calculations Molarity of titrant for 1 st sample (of KHP): 0.49 g KHP × = 0.0024 mol KHP (Moles KHP = Moles NaOH → 1:1 ratio) 0.0024 mol NaOH/0.018 L NaOH = 0.133 M NaOH Average Value of NaOH molarities: (0.103 + 0.105 + 0.104)/3 = 0.104 M Molarity of original sample (before dilution): Average mL NaOH = (26 + 25.8 + 26 + 25.7)/4 = 25.875 mL × 0.025875 L = 0.002691 mol NaOH in 20 mL of water (NaOH : Vinegar → 1:1 ratio) 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 Molarity converted into % (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)
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Results and Discussion By observing the molarities of NaOH in titrating the KHP we can conclude that one value does not correspond with the others. For the first titration, we got a molarity of 0.133 M, while all the other titrations’ molarities were around 0.104 M. This may be due to using a titration flask that was not completely cleaned. Also, this flask may have contained remnants of some base solution from a previous experiment. However, our other three values are clearly within 1% of one another. This signifies that our titrations of KHP with NaOH were successful. Our titration values for the acetic acid solutions also agreed to within 1% of one another. From these titrations, we can see that the equivalence point of
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