Swirl the solution to ensure that the NaOH is properly dissolved in the deionised water. B) Prepare an Oxalic Acid Solution (approximately 0,05M) 1. Place a clean, dry glass beaker on the electronic scale. 2. Determine the mass of the glass beaker.
To have gotten a 0% error between the experimental and actual value for CH3COOH, the pH would have been measured at about 4.75, which is slightly more acidic than 4.80. The percent error was calculated to determine how accurate the Ka of acetic acid was: Since the calculations yielded a 20% error, this shows that experimental error occurred during the experiment. Factors that could have affected the results included improper reading of the meniscus for volume of NaOH, not allowing the NaOH to fully drip into the buret after removing the funnel, adding too much acetic acid after the indicator flashed pink to get an inaccurate equivalence point, and not allowing the solution in the beaker to mix thoroughly to get an accurate reading from the pH
The purpose of this experiment was to perform a Wittig reaction using two different methods: In method I, 250 mg aldehyde was mixed with 785 mg phosphonium salt in 5 M NaOH solvent. This mixture was stirred for thirty minutes and filter by vacuum filtration for the product. In method 2, 250 mg of aldehyde, 785 mg, benzyltriphenylphosphonium chloride, and 380 mg potassium phosphate tribasic were homogenize with a pestle and mortar. Vacuum filtration was also used in this method to attain the product. The products in both methods were used for recrystallization and TLC.
Experimental Methods: 1. SYNTHESIS OF 4-BENZOYL BUTYRIC ACID METHYL ESTER Materials required * 5 oxopentanoic acid : 2 gm (Aldrich) * Methanol : 50 ml * Acetic Acid (Rankem) Procedure: * 2 grams of 5 oxopentanoic acid was weighed and placed in a round bottom flask and then to it 50 ml of methanol was added. It was placed on a hot plate and the temperature was increased to 50 degrees under ambient air conditions. * To the RB, 2 ml of acetic acid was added and then by attaching a condenser the entire reaction was put on refluxing at 70 degrees Celsius in an oil bath. * For work up: * The reaction media was concentrated till about 10 ml and then dry silica gel was added.
There is one mole of OH- in the solution since NaOH goes to Na+ and OH-. Trial 1: 25.65mL NaOH x 0.100mol/1000mL = 2.57 x 10-3 mol NaOH = 2.57 x 10-3 mol HA = 2.57 x 10-3 mol H+. The equivalent mass is 0.356g Acid / 2.57 x 10-3 mol H+ = 139g/mol H+ Trial 2: 49.57mL NaOH x 0.100 mol / 1000 mL = 4.96 x 10-3 mol NaOH = 4.96 x 10-3 mol HA = 4.96 x 10-3 mol H+. The equivalent mass is 0.644g Acid / 4.96 x 10-3 mol H+ = 130.g/mol H+ Average = (139g Acid / 1 mol H+) + (130.g Acid / 1 mol H+) / 2 = 135g/mol H+. The average equivalent mass for the acid is 135g/mol H+.
Balanced Chemical Equation: Cu(s) + 4HNO3(aq) —> Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O (l) Reaction 2: when sodium hydroxide (NaOH) is added to copper (II) nitrate (Cu(NO3)2), a double displacement reaction will occur. Copper and sodium will displace each other to create copper (II) hydroxide and sodium nitrate. Balanced Chemical Equation: Cu(NO3)2 (aq) + 2NaOH (aq) —> CuOH2 (s) + 2NaNO3 (aq) Reaction 3: When copper (II) hydroxide is heated, a decomposition reaction will occur. The reaction will decompose forming two compounds, Copper (II) oxide, and water. Balanced Chemical Equation: Cu(OH)2 (s) + Heat —> CuO (s) + H2O (g) Reaction 4: when a sulphuric acid is added to the solution that contains copper (II) oxide, a double displacement reaction will occur.
In 10 g dried sediment sample added 7 ml 0.2 M NH4Cl solution. A mixture of 100 ml hexane: acetone (1:1) was used as a solvent to extract pesticides with overnight shaking for 12 h on reciprocal or wrist action shaker at 180 rpm. The extract was carefully decanted through activated florisil column (2-3 cm), giving twice wash with25 ml hexane: acetone (1:1) to the sediments. The elute was then washed with 200 ml water and then again aqueous layer was extracted with 50 ml hexane. Finally the hexane layer was washed with 100 ml water and then evaporated to dryness with a vacuum rotary evaporator.
An equivalent volume of saline solution was added (NaCl 0.16 M; pH = 7) and it was mixed completely. The resultant solution was centrifugated for 30 minutes at 9000 rpm at 4ºC in a rotor Beckman JA-20, eliminating the pellet. The supernatant was collected to obtain the lipoproteins by gradient density ultracentrifugation: 10ml of the yolk solution were taken and then 0.9 g of potassium bromide (KBr) was added. The bromide was dissolved with a smooth agitation, with care not to denature the proteins. Straight afterwards, saline solution was added (NaCl 0.16 M, pH=7) to the 10ml of the yolk solution with a Pasteur pipette, avoiding the sample diffusion, forming two phases and filling the tube completely.
The solution were tested by using calibrated pH meter to get the pH value of the solution. Results and Discussions pH ratio between acid and base: 7.3 = 6.82 + x x = 0.48 0.48 = log ([base])/([acid]) 100.48 =base/acid salt/acid = 3.02 There, 1 acid : 3 base calculate number of mole of acid and base to find the mass : molar = mol/L 50 mM = (mol )/(0.5 L) mol = 25 mol number of mole of NaH2PO4 25/4 = acid = 6.25 mol number of mole of Na2HPO4 25/4 × 3 = salt = 18.75 mol to calculate the mass of the acid and base : Mass of NaH2PO4 (6.25 mol)/(119.98 g/mol)=0.052g Mass of Na2HPO4 (18.75 mol)/(141.96
Grade) in 1 l water, standardize this solution with 0.0192(N) silver nitrate solution. The solution losses strength gradually and must be rechecked every week. 1 ml of solution = 1 mg CN (ix) Standard cyanide solution: Dilute 10 ml stock cyanide solution to 1 litre with distilled water, mix and make a second dilution of 10 ml to 100 ml. Prepare fresh solution daily. 1 ml = 1 µg CN (x) Chloramine –T: Dissolve 1 gm chloramine – T in 100 ml distilled water.