Analytical Chem Lab report 2

Onukogu 1 Neutralization Titrations: Determination of Soda Ash Chioma Onukogu CHM3120L 10/02/2023

Onukogu 2 INTRODUCTION Titration is a method used to determine the molarity of an unknown solution with the use of a solution with known concentration. It is done by the slow addition of the solution with known concentration (titrant) to a known volume of the other solution with an unknown concentration until the endpoint (point of neutralization) is reached which is often indicated by a color change. 1 There are different types of titrations which include: acid-base titration, redox titration, precipitation titration, etc. One of the most common types is the acid-base titration where the substance being titrated if basic, the titrator (micropipette in the case of this experiment) will contain a strong acid and vice versa. 2 For this experiment hydrochloric acid (HCl: strong acid) and sodium hydroxide (NaOH: strong base) will serve as titrants. The titrants to be used must fulfill the requirements of either a primary or secondary standard. A primary standard is a reagent that can be used to dispense an accurate known amount of analyte because it is pure and has a known stoichiometry and reagents that do not meet this criteria are secondary standards whose concentrations are determined relative to a primary standard. 3 The purpose of this experiment is to determine the mass of sodium carbonate in a sample of soda ash (common name for Na 2 CO 3 ). This is done by using a HCl standard solution as the titrant and indicators such as phenolphthalein and methyl orange to detect the endpoint by a color change. The HCl solution titrant is standardized by dilution with deionized water and titration where a NaOH solution serves as the titrant. However, the NaOH solution also has to be standardized itself. NaOH is a secondary standard as it does not meet the requirements of a primary standard; therefore, its concentration is determined relative to the primary standard weak acid potassium hydrogen phthalate (KHC 8 H 4 O 4 ) in this experiment. 4,5

Onukogu 4

Onukogu 5

Onukogu 7 - Trial 1 calculations mols of KHP = 0.1063 (g) 204.22 (g/mol) ⁄ = 5.205 × 10 -4 mols KHP and NaOH have a 1:1 mol ratio from the reaction equation. Therefore, mol of NaOH is 5.205 × 10 -4 mols Volume of NaOH = 6450 μl ; [NaOH] = mols NaOH L ⁄ [NaOH] = 5.205 × 10 −4 (mols) 6450 × 10 −6 (L) ⁄ = 0.08069 M *These calculations were repeated for trial 2 and 3 Avg. [NaOH] = [0.08069 + 0.07569 + 0.07599] 3 ⁄ = 0.077 M ≅ 0.08 M Standard deviation = (equation that was used) = 2.798 × 10 -3 ≅ 0.003 M RSD = std dev avg [NaOH] ⁄ × 100 = 3.61 % ≅ 4 % Avg [NaOH] Standard deviation Relative Standard deviation 0.08 M 0.003 M 4 % 2) Molarity of HCl by titration with NaOH Avg. [NaOH] = 0.077 M HCl (aq) + NaOH (aq) → NaCl (s) + H 2 O (l) 1:1 ratio HCl and NaOH Vol. NaOH added ( μl ) Molarity of NaOH (M) Mols of NaOH (mols) Mols of HCl (mols) 8450 0.077 6.5 × 10 -4 6.5 × 10 -4 8710 0.077 6.7 × 10 -4 6.7 × 10 -4 8580 0.077 6.6 × 10 -4 6.6 × 10 -4 Trial 1 calculations Mols of NaOH= 0.077 × 8450 × 10 -6 = 6.5065 × 10 -4 ≅ 6.5 × 10 -4 mols Since it is a 1 to 1 mol ratio, then mol of HCl = 6.5 × 10 -4 mols. Volume of HCl used is 5 mL.

Onukogu 8 [HCl] = 6.5065 × 10 −4 (mols) 5.00 × 10 −3 (L) ⁄ = 0.13013 ≅ 0.13 M * These calculations were repeated for trial 2 and 3 Avg. [HCl] = 0.13013 + 0.13213 + 0.13413 3 ⁄ = 0.13213 ≅ 0.132 M Standard deviation = = 0.002 M RSD = 0.002 0.13213 ⁄ × 100 = 1.5 % ≅ 2% Average [HCl] Standard deviation RSD 0.13 M 0.002 M 2 % 3) Percent Weight of Na 2 CO 3 with HCl [HCl] = 0.132 M; Na 2 CO 3 = 106.00g/mol; Unknown Sample = # 57 Mass of Soda Ash Weighed (g) Total Volume of HCl used ( μl ) Mols of HCl (mols) Mols of Na 2 CO 3 (mols) Mass of Na 2 CO 3 (g) Weight percent of Na 2 CO 3 (%) 0.0798 4875 6.44 × 10 -4 3.22 × 10 -4 0.034 42.7 0.0797 4850 6.40 × 10 -4 3.20 × 10 -4 0.034 42.6 0.0824 5055 6.67 × 10 -4 3.34 × 10 -4 0.035 42.9 Trial 1 Calculations Mols of HCl = [HCl] × volume 0.132 (mols) 1 (L) ⁄ × 1 (𝐿) 10 6 (μl) ⁄ × 4875 ( μl ) = 6.435 × 10 -4 mols mol ratio of Na 2 CO 3 and HCl is 1:2 Therefore, mols of Na 2 CO 3 = 6.435 × 10 −4 (mols) 2 ⁄ = 3.2175 × 10 -4 ≅ 3.22 × 10 -4 mols

Onukogu 10 REFERENCES 1. LibreTexts Chemistry General Lab Techniques: Titration https://chem.libretexts.org/Ancillary_Materials/Demos_Techniques_and_Experiments/Ge neral_Lab_Techniques/Titration?readerView 2. Jacobs, A. Neutralization Titrations: Determination of Soda Ash; University of Florida/Department of Chemistry: Gainesville, FL, 2021 3. Harvey, David; LibreTexts Chemistry Analytical signals https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1 _(Harvey)/05%3A_Standardizing_Analytical_Methods/5.01%3A_Analytical_Signals 4. Harvey, David; LibreTexts Chemistry Analytical signals https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1 _(Harvey)/05%3A_Standardizing_Analytical_Methods/5.01%3A_Analytical_Signals 5. Jacobs, A. Neutralization Titrations: Determination of Soda Ash; University of Florida/Department of Chemistry: Gainesville, FL, 2021

Onukogu 11 Post Lab Questions 1) Give the required properties of a primary standard solid. Why does NaOH not meet these requirements? A primary standard: - must have a known stoichiometry - it should not decompose, absorb, or react with air i.e., stable for a long time - must have a known purity NaOH does not meet all these requirements as it contains impurities. It also reacts with carbon dioxide when exposed to air and forms sodium carbonate. 2) Define the term standard solution and describe the two general methods for preparation. Explain how the two standard solutions were prepared in this experiment. Standard solution is a solution where the concentration of a particular substance (element/compound) is known. The two general methods of preparation are the dilution method and the weighing method. The two standard solutions in this experiment were the 0.1 M HCl solution and the 0.1 M NaOH solution. They were made by the dilution method. The required amount of HCl and NaOH were obtained respectively using a micropipette. These samples were first added to a small amount of deionized water before being diluted and mixed with more DI water to make 300 mL individual standard solutions of the HCl and NaOH. 3) Sodium hydroxide solutions are usually prepared by diluting a concentrated (10% by weight) solution of NaOH. Explain why this method is preferred to dissolving solid NaOH directly. This method is preferred because the NaOH contains impurities and is hydroscopic so it cannot be weighed accurately and as a result must be standardized before use. 4) Write balanced net ionic equations for all titrations performed in this experiment. Give the indicators and color changes for each endpoint. H + (aq) + OH - (aq) → H 2 O (l) (for NaOH and HCl) H + (aq) + OH - (aq) → H 2 O (l) (for NaOH and KHC 8 H 4 O 4 ) CO 3 2- (aq) + 2H + (aq) → H 2 O (l) + CO 2 (g) (for HCl and Soda Ash) Phenolphthalein: pink → clear Methyl orange: green → bluish grey