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To determine the concentration of acetic acid in an unknown sample of vinegar by titrating a known volume ofthe acid solution against a standardized basic solution.

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S.6A Karen Kwok (8) VA2 22th, September, 2002. Chemistry Laboratory Report Title: Acid-base Titration Date: 20th, September, 2002. Objective: (1) To determine the concentration of acetic acid in an unknown sample of vinegar by titrating a known volume of the acid solution against a standardized basic solution. Procedures: Part A: Standardization of 0.1M sodium hydroxide solution using potassium hydrogen phthalate as the primary standard. 1. 5.010g of analytical grade potassium hydrogen phthalate was weighted and dissolved in deionized water in a beaker. 2. After all the KHC8H4O4 was dissolved, the solution was then transfer to a 250 cm3 volumetric flask. 3. Then the volumetric flask was filled to the graduation mark with deionized water. 4. The volumetric flask was stopped. The flask was turned upside down in order to mix the solution well. 5. 25 cm3 of KHC8H4O4 solution from the volumetric flask was transfered in a conical flask using a pipette and pipette filler. 6. ...read more.


Therefore the average volume of 0.0500 M sodium carbonate required for neutralization: = 25.55 + 25.30 +25.50 3 cm3 =25.45 cm3 No. of moles of 250 cm3 KHC8H4O4 in volumetric flask = Mass Molar Mass = 5.010 204.23 mol =0.025 mol No of mole of 25 cm3 of KHC8H4O4 =0.025 � 25 250 =0.0025mol According to the equation, 1 mole of NaOH required 1 mole of KHC8H4O4 for complete neutralization. ?No of moles of NaOH= 0.0025mol Molarity of NaOH solution = No. of moles of NaOH Volume of solution = 0.0025mol 0.02545dm3 =0.098M The volumes of 0.098M NaOH solution required for neutralization are: 24.50 cm3, 22.65 cm3. Therefore the average volume of 0.098M sodium hydroxide required for neutralization: = 24.50 + 22.65 2 cm3 = 23.58 cm3 NaOH (aq) + CH3COOH(aq)?CH3COONa (aq) + H2O(l) 0.098M ?M 23.58 cm3 25 cm3 No. of moles of NaOH in 23.58 cm3 = Molarity of solution �Volume of solution = 0.098 mol dm3 �0.02358 dm3 = 0.0023mol According to the equation, 1 mole of NaOH required 1 mole of CH3COOH for complete neutralization. ...read more.


Potassium hydrogen phthalate can be act as a primary standard to standardize NaOH. It is because it is a monoprotic acid whose formula is KHC8H4O4 and has a high molecular weight (204.22 g/mol). Also, the exact mass can be easily determi ned by weighing the dried acid on an analytical balance. It is stable and not hydroscopic, and it fits all the propeties of a primary standard. 3. Is methyl orange a suitable indicator in this experiment? Why? Ans: Methyl orange is not a suitable indicator in this experiment. Because for strong base (NaOH) with weak acid (KHC8H4O4), the pH at the equivalence point is always greater than 7 because of the conjugate base is basic. And from the titration curve for strong acid with weak base, we know that pH at the equivalence point is approximately 8.8, and the dramatic change in pH cannot cover the pH range of methyl orange which is (3.1-4.4). But it can cover the pH range of phenolpthalein (8.3-10.0), therefore it is more suitable for us to use phenolpthalein in this experiment. Titration for weak acid with strong base ...read more.

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