Titration Lab

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Los Angeles Valley College *

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051

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Chemistry

Date

Feb 20, 2024

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pdf

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Uploaded by PresidentHippopotamusMaster230

AP Chemistry Lab Titration Labs Pre-Lab Questions 1) What is meant by the term standardization? Standardization refers to the process of determining the exact concentration (molarity) of a solution. 2) What is a primary standard? A primary standard refers to a chemical reagent used to make standard solutions for titrations and when preparing secondary standards. It is typically used to measure an unknown concentration of another known chemical. 3) What is oxalic acid’s formula and molar mass? Oxalic acid’s formula is C2H2O4 and it’s molar mass is 90.03 g/mol. 4) What is meant by the term “ accurately weigh out approximately 2 g of NaOH”? As sodium hydroxide pellets absorb water from the surrounding air in a quick manner, it causes them to melt quickly if left in the open. Therefore the jar should be kept almost always capped, hence, getting an accurate measurement would refer to being quick and eﬃcient with the process, as it may be hard to take one’s time weighing it out. The scale measures an extra 2 decimal places. 5) Write the net ionic equation for the reaction between NaOH and Oxalic Acid. The reaction is H2C2O4 + 2OH^(-) -> C2O4^(2-) + 2H2O 6) Write the net ionic equation for the reaction between NaOH and acetic. OH^(-) + CH3COOH -> C3COO^(-) + H2O 7) What do all the stoichiometries have in common in this lab? All the stoichiometries in this lab help us to determine the concentration of the unknown solutions in the titration. Sample Data Table Sample 1 Sample 2 Sample 3

Mass of Oxalic Acid Used 0.22g 0.28g Initial Buret Reading 48.4 mL 49.5 mL Final Buret Reading 35.7 mL 35.2mL Volume of NaOH used 12.7 mL 14.3 mL Moles oxalic acid Present 0.22g C2H2O4 x C2H2O4/90.03 g/mol = 2.44 × 10 ^(-3) Moles C2H2O4 0.28g C2H2O4 x C2H2O4/90.03 g/mol = 3.11 × 10 ^(-3) Moles C2H2O4 Molarity of NaOH 2.44 × 10 ^(-3) moles NaOH / 0.0127 L = 0.19 M 3.11 × 10 ^(-3) moles NaOH/ 0.0143 mL = 0.22 M Average Molarity of NaOH 0.205 M Post Lab Questions These questions can be answered in the data tables. 1) Calculate the concentration of the standardized NaOH solution. Report this as an average. 0.1 M

2) Calculate the percentage of oxalic acid in the vinegar used in this lab. Report this as an average. 2.44 × 10 ^(-3) Moles C2H2O4 / 0.1 L = 0.0244 M C2H2O4 3.11 × 10 ^(-3) Moles C2H2O4 / 0.1 L = 0.0311 M C2H2O4 These questions should be answered as essays at the end of the lab. Imagine the following procedural errors were made. Explain what effect, if any, they would have on the outcome of the lab. 3) After rinsing the buret with distilled water, the buret is filled with the standard NaOH solution; the Oxalic acid is titrated to its equivalence point. The molar mass of the oxalic acid would be lower. This is because it would cause the molarity of NaOH to be a lower concentration due to the dilution caused by the extra distilled water in the buret, hence, causing more of the solution to be used for the titration.

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4) Extra water is added to the 0.4 - 0.5 gram sample of Oxalic acid. This would not affect the results because the additional water would not affect the moles of acid that are measured initially or the amount of base needed to reach the endpoint. 5) An indicator that changes color at pH 5 is used to signal the equivalence point of the vinegar titration. The molar mass of oxalic acid would be higher, as the equivalence point would be reached with a lesser volume of the base. 6) An air bubble passes unnoticed through the tip of the buret during the titration This would cause the volume of NaOH to be larger than expected, thus, causing a lower molar mass for oxalic acid. Create a table to numerically show how the following would affect the concentration of the vinegar determined in the experiment. Make sure you show the correct calculation in one column and the incorrect one in another column. 7) The student thought the formula of KHP contained one potassium, one hydrogen, and one phosphorus atom. The molar mass of the perceived KHP will be lower bc the actual KHP, the oxalic acid. the balanced equation would be different and the actual molar mass would be different 8) The student recorded a value of 0.6001 g of KHP used but it was actually 0.6100 g.