This solution was placed in a burette and 18.4 cm3 was required to neutralise 25 cm3 of 0.1 moldm-3 NaOH. Deduce the molecular formula of the acid and hence the value of n. 5. Sodium carbonate exists in hydrated form, Na2CO3.xH2O, in the solid state. 3.5 g of a sodium carbonate sample was dissolved in water and the volume made up to 250 cm3. 25.0 cm3 of this solution was titrated against 0.1 moldm-3 HCl and 24.5 cm3 of the acid were required.
An aqueous solution of ammonium sulfate is allowed to react with an aqueous solution of lead(II) nitrate. Identify the solid in the balanced equation. A) (NH4 )2 SO4 B) Pb(NO3 )2 C) PbSO4 D) NH4 NO3 E) There is no solid formed when the two solutions are mixed. ____ 11. An aqueous solution of sodium carbonate is reacted with an aqueous solution of calcium chloride.
CHE 111 Laboratory 3 Hydrates Introduction Hydrates Water molecules combine with the molecules of certain substances, forming loose chemical combinations called hydrates. An example of a hydrate is MgSO4•7H2O. This formula means 7 water molecules are loosely attached to a magnesium sulfate molecule. Other examples of hydrates are Na2SO4•10H2O and Ba(OH)2•8H2O. When the hydrate is heated, it easily loses water molecules attached and becomes an anhydrous salt.
The Ksp of Magnesium Oxalate Abstract The Ksp for the acid catalyzed titration of the saturated oxalate is 1.8 x 10-3. Introduction In this experiment, the solubility equilibrium for the salt magnesium oxalate must be found in order to determine a solubility product constant. Solubility equilibrium is a type of dynamic equilibrium which exists when a chemical compound in the solid state is in chemical equilibrium with a solution of that compound. At the point of equilibrium the solution becomes saturated. The chemical reaction used to find this constant is as follows: MgC2O4 (s) ↔Mg(aq)2++ C2O4 (aq)2- Kc= Mg2+[C2O42-][MgC2O4] Ksp=Mg2+[C2O42-] The solid salt magnesium oxalate is prepared through the following precipitation reaction: Mg(SO4)(aq)+NaC2O4 (aq) → MgC2O4 (s)+NaSO4 (aq) Next, the concentration of the Mg2+ and C2O42- ions is found through a redox titration.
3 x (C H5 N) = C3H15N3 Hydrated compounds Solving process: 1st- the difference between the initial mass and that of the dry sample is the mass of water that was driven off. Mass of hydrate minus mass of dry sample equals the mass of water 10.407 – 9.520 = 0.887 g 2nd- The mass of dry BaI2 and the mass of water are converted to MOLES. 9.520 g BaI2 x 1 mol BaI2 ∕ 391 g BaI2 = 0.0243 mol BaI2 anhydrate 0.887 g H2O x 1 mol H2O / 18.0 g H2O = o.o493 mol H2O 3rd: Dividing both results by the amt of 0.0243 mol, we get a ratio of 1 to 2.03, or 1 to 2, since the formula must have full numerical integers of water molecules, in other words no fractions of a water molecule. Thus, for every 1 mole of BaI2, there are two moles of water. The formula for the hydrate is written as BaI2 • 2H2O And it is named barium iodide dihydrate.
Aim Prepare standard NaOH and HCl to determine the concentration of acetic acid in vinegar as well as ammonia in window cleaner via titration using suitable indicators. Results and Calculations 1.1. Titration Set 1: Standardization of NaOH 1.2.1 Results of Titration of KHC8H4O4 against NaOH Titration | 1 | 2 | 3 | Final Volume of Titrant/mL | 8.60 | 13.55 | 18.50 | Initial Volume of Titrant/mL | 3.20 | 8.60 | 13.55 | Volume of Titrant used/mL | 5.40 | 4.95 | 4.95 | | | | | Indicator used | Phenolphthalein | Colour Change at Endpoint | Colourless to Pink | 1.2.2 Calculations Mass of KHC8H4O4 used = 1.0291 g Molar Mass of KHC8H4O4 = 204.22 g/mol Molarity of KHC8H4O4 = (1.0291 ÷ 204.22) ÷ 0.1000 = 0.05039 M No. of moles of KHC8H4O4 in 10.0 mL = (10.0 ÷ 1000) × 0.05039 = 5.039 ×10-4 mol Average volume of NaOH titrant used = 4.95 mL Chemical Equation of Reaction: KHC8H4O4 (aq) + NaOH (aq) KHC8H4O3-Na+ (aq) + H2O (l) KHC8H4O4 reacts with NaOH in a 1:1 ratio, thus No. of moles of NaOH in 4.95 mL = 5.039 ×10-4 mol Concentration of NaOH = (5.039 ×10-4) ÷ (4.95÷ 1000) = 0.1018 M 1.2.
The concentrations of OH- and Ca2+ were used to calculate the solubility product constant. For the solution containing Ca(OH)2, in pure water the concentration of the OH- ion was 0.06M and the concentration of the Ca2+ ion was 0.03M. For the solution containing Ca(OH)2, in NaOH the concentration of the OH- ion was 0.056M and the concentration of the Ca2+ ion was 0.028M. The solubility product constant was determined to be 1.8x10-4 for the solution of Ca(OH)2, in pure water and the solubility product constant for the solution of Ca(OH)2, in NaOH was determined to be 8.8x10-5. The solubility of the solution containing Ca(OH)2 in NaOH had a smaller solubility of 4.15 g/L than the solution containing Ca(OH)2 in water, which had a solubility of 4.45 g/L.
________________________________________ Chart 1.2: the difference between the readings in each of the trails. Chart 1.3: the difference between the sum and the average. CONCLUSION AND EVALUATION CONCLUSION: Equivalence point is the point at which the moles of H+ is equal to the moles of OH+,an indicator is used to show the equivalence point during a titration. in a titration the method is about totaling one reactant from the burette (regularly the acid),to a known volume of the other reactant in a conical flask(regularly the base) . In order to find the concentration of NaOH we need to tag along the following steps: a- note down the balanced chemical equation for the reaction C8H5O5K+NaOH→C8H4O4KNa+H2O b- pull out applicable information from the experiment: C8H5O5K V=0.025 dm3 C=0.2M NaOH V=0.021 dm3 C= ??
Lab 12 Experiment 12: Reactions of Amines Aim: 1) To investigate and to determine the chemical properties and chemical reactions of Amines 2) To observe the reactions of Amines Procedure: Refer to the Laboratory Manual of CHM152 page 72 to 74 Result: A. Solubility Compound | Butylamine | Diethylamine | Solubility | Very soluble | Not soluble | B. Salt Formation Solubility of dibutylamine in water: Not soluble Solubility of dibutylamine in acid: Soluble, white fume is produced Result of making the solution alkaline: Cloudy solution is produced C. Reaction with Nitrous Acid Compound | Observations | Butylamine | Bubble gas | Aniline | Yellow solid | N-Metylaniline | Yellow oil opaque green solution | Conclusion: Amines show various chemical properties and chemical reactions when tested with different reagents. Discussion: 1. Do not transfer any of the liquids from the reagent bottles to the beakers placed beside the bottles (the beakers are there to contain Pasteur pipettes). Transfer the liquid directly from the bottle to your test tube.
Acid Rain Components and pH Acid rain is formed by the combination of rain water with certain components such as carbon dioxide (CO2), sulfur dioxide (SO2), nitrous oxides, and other chemicals. Acid rain is measured using a scale called “pH.” The lower a substance's pH, the more acidic it is. Pure water has a pH of 7.0. However, normal rain is slightly acidic because carbon dioxide (CO2) dissolves into it forming weak carbonic acid, giving the resulting mixture a pH of approximately 5.6 at typical atmospheric concentrations of CO2. As of 2000, the most acidic rain falling in