To make ethyl acetate, react ethanol with ethanoic acid in the presence of concentrated sulphuric acid. Another method is by preparing industrially using the Tishchenko reaction by combining two equivalents of acetaldehyde in the presence of an alkoxide catalyst.
When Sodium Acetate is mixed with NaOH, a double displacement reaction occurs, leading to the formation of water and sodium hydroxide, along with sodium acetate. The reaction can be represented as follows: CH3COONa + NaOH → CH3COONa + H2O The sodium acetate remains in the solution, while water and sodium hydroxide are formed as byproducts.
Add 50 mL of HCl 1 N in a 1 L volumetric flask, class A or B; add ca. 900 mL distilled water to the flask. Place the flask in a thermostat at 20 0C. After 30 min add slowly distilled water to the mark (1 L) and stir well the closed flask. Pour the solution in a bottle. Place a label with the date, concentration, name of the solution on the bottle.
This is an acid-base reaction Hydrochloric acid or HCl is a strong acid consisting of one hydrogen atom combined with one chlorine atom. Sodium acetate (NaCH3CO2) dissolves to produce sodium ions (Na+) and acetate ions (CH3CO2-). The acetate ion is a base and essentially "steals" a hydrogen ion (H+) from the HCl to form acetic acid (CH3CO2H) Chloride ions(Cl-) and Na+ ions are left behind to form sodium chloride (NaCl)
g HCl solution = 2500 mL of HCl * 1 liter/1000 mL * 1190 g/L = 2975 g 37% solution (37 g HCl/100 grams of solution) gives you the grams of HCl: g HCl = 0.37 * 2975 g = 1100.8 g HCl Moles HCl = 1100.8/(36.46 g/mole) = 30.2 moles Therefore the molarity, which equals the normality in this case = 30.2 moles/2.5 L = 12.07 M = 12.07 N If you want to make 100 mL of a 0.1 N solution, Volume of HCl solution needed = (0.1 N * 100 mL) /12.07 N = 0.83 mL Take 0.83 mL of the 37% HCl, and dilute it with water to 100 mL.
Ah, preparing a 0.02 M solution of HCl is a wonderful journey. Simply measure out the correct amount of hydrochloric acid and dilute it with water until you reach the desired concentration. Remember to handle chemicals with care and always wear appropriate safety gear. Happy experimenting, my friend!
HCl is not soluble in ethyl acetate. It typically forms a separate layer in the presence of ethyl acetate due to their immiscibility.
The ethyl acetate test is a qualitative test used to detect the presence of acetoacetic acid in urine. The equation involves the reaction of acetoacetic acid in urine with ethyl acetate in the presence of a mineral acid like hydrochloric acid. The equation can be represented as: CH3COCH2COOH (acetoacetic acid) + CH3COOCH2CH3 (ethyl acetate) + HCl → no visible change A positive result is indicated by the production of a pink or red color in the upper ethyl acetate layer due to the formation of sodium nitroprusside complex with acetoacetic acid.
One common method to test for acetate ion (CH₃COO⁻) is to add a few drops of dilute hydrochloric acid (HCl) to the unknown solution. Then, introduce a few drops of ferric chloride (FeCl₃) solution. If a red precipitate forms, it indicates the presence of acetate ion.
To make a 1% HCl solution from a 35% HCl solution, you would need to dilute the concentrated solution with water. The ratio of concentrated HCl to water would be approximately 1:34. So, to make 1% HCl, you would mix 1 part of the 35% HCl solution with 34 parts of water.
To make 0.25N HCl from 1.00N HCl, you would need to dilute the 1.00N HCl solution by adding three parts of water for every part of the original solution. For example, you can mix 1 mL of 1.00N HCl with 3 mL of water to obtain 0.25N HCl solution.
To prepare a 0.2N HCl solution from 1.0N HCl, you can dilute the 1.0N HCl solution by adding 4 parts of water to 1 part of the 1.0N HCl solution. This means combining 1 volume of 1.0N HCl with 4 volumes of water to obtain the desired 0.2N HCl solution.
To convert make a dillute solution from a concentrated one, take the amount of moles needed for the final solution as mL of concetrated solution, and dillute with water until the desired volume is reached.
To prepare a 10mM solution of Tris-HCl, you would weigh out the appropriate amount of Tris-HCl powder using a balance and dissolve it in water to make a final volume of solution. For example, to make 1L of 10mM Tris-HCl solution, you would need to dissolve 0.121g of Tris-HCl in 1L of water.
To prepare 1 M hydrochloric acid (HCl) solution, you usually need to dilute concentrated HCl solution (~35-37% HCl) with water in a specific ratio. The specific volume of concentrated HCl needed depends on its concentration and the final volume of the 1 M solution you want to make. In this case, 85 ml of concentrated HCl is likely to be the volume needed to make 1 L of 1 M HCl solution.
The question is in poorly worded. I will assume the question is "why adjust the pH of Tris buffer with HCl and not Sodium Acetate?" I would assume the answer is - because sodium acetate is the conjugate base of a weak acid, and HCl is a strong acid. Also the salts you would be putting into the solution as a result would be different. I think the question is actually, "The pH of Tris is adjusted with HCl, why isn't the pH of sodium acetate adjusted with HCl?". I'm not sure of the answer exactly, but I've always assumed its because if you adjust the pH with glacial acetic acid instead of HCl, you won't introduce chloride ions.
To make a 0.1M solution from a 1M HCL solution, you would dilute the 1M HCL with 10 parts of water (or whatever solvent you are using). For example, mix 1 mL of 1M HCL with 9 mL of water to obtain a 0.1M HCL solution.
No, a 38% HCl solution is not the same as a 12N HCl solution. The concentration of a solution is based on the amount of solute dissolved in a specific volume of solvent. A 38% HCl solution means there is 38 grams of HCl in 100 mL of solution, while a 12N HCl solution means there are 12 moles of HCl in one liter of solution.