25 ml of diluted unknown acid solution to 100ml beaker by using 25 volumetric pipet. 10ml of deionized water and 3 drops of phenlpthalin indicator the beaker labeled as 3. Potentiometric titration acid solutions 125 ml of NaOH was obtaining in a beaker and 50 ml of NaOH transfer to buret the tip and the meniscus is at below 0 ml. one magnetic stirring bar placed in a beaker contain one of the known solution on a stir. The pH recorded by using pH electrode before adding NaOH.
(35.7+36+36.6)/3=36.1 ml is the average volume of the Na2S2O3. (0.0361)(0.1) = 0.00361 0.00361/(2 ×2.5)=7.22 ×〖10〗^(-4) M NaClO 0.00481/(.005) = 0.96 M NaClO 0.358/5.4= 35.8 % |.358-0.060|= .298 |(0.358-0.06)/0.06|=4.9% percent error Conclusion In the experiment a redox titration was performed in order to determine the percent of sodium hypochlorite in commercial bleach. This done by reacting the bleach and the sodium thiosulfate in the presence of iodide ions and starch. Discussions of Theory Experimental Sources of Error Pre-Lab Questions Titration is the process, operation, or method of determining the concentration of a substance in solution by adding to it a standard reagent of known concentration in carefully measured amounts until a reaction of definite and known proportion is completed, as shown by a color change or by electrical measurement, and then calculating the unknown
Show the calculation to obtain the % Na2CO3 in your soda ash sample for trial 3. [(g Na2CO3) / (g soda ash)] ⋅ 100 = % Na2CO3 [(0.390 g Na2CO3) / (1.002 g soda ash)] ⋅ 100 = 38.9% Na2CO3 7. Show the calculation for determining the average % Na2CO3 in your soda ash sample. [(T1 % Na2CO3) + (T2 % Na2CO3) + (T3 % Na2CO3)] / 3 = average % Na2CO3 [39.1% + 39.1% + 38.9%] / 3 = 39.03 or 39.0% 8. Calculate the precision of your % Na2CO3 in your soda ash sample by determining the ppt (parts per thousand).
First, we had to calculate how many grams of copper (II) sulfate we needed to form 100 mL of a 0.200 M solution of copper (II) sulfate. We determined that we needed to use 4.994g of copper (II) sulfate to make the solution. We added distilled water to the 4.994g of copper (II) sulfate in a beaker until it reached 100 mL. Then we put the beaker on a hot plate and added a magnetic stirrer. We determined that the mass of zinc necessary to completely react with the copper (II) ions in the solution was 1.308g.
Place the solution on the burner again and wait until all of the H2O has evaporated leaving only NaCl residue. Weigh the beaker with the residue. Clean out the beaker and weigh the beaker alone. Subtract these from one another and this will give you the weight of the residue left. The one that weighs less is NaHCO3 and the one that weighs more is Na2CO3 Results (20 points) (1) Presents table(s) and calculations clearly and accurately.
CHEM 10050 Problem Set #3 Name_________________ 1. Calculate the molarity of normal saline (0.90% w/w NaCl) given the density of this solution to be 1.05 g/ml. Assume 100 g solution. Then: [pic] [pic] [pic] [pic] 2. Calculate the molarity of an aqueous solution resulting when 275 ml of 3.15 M glucose is diluted with 150 ml of water.
Part D: Density of Methanol 1) Find the mass of an empty 10 mL graduated cylinder, and then fill approximately 9 mL of methanol and record volume. 2) Find the mass of the cylinder and methanol, subtracting the initial mass from the new mass to determine the mass of methanol. 3) Determine the density and percent error. Data: Part A: Density of a Rectangular Solid/Cube Block # 9 (cube – brass) 1. Mass – 132.32g 2.
Comparing the rate of appearance of C and the rate of disappearance of A, we get[pic]. A) [pic] B) [pic] C) [pic] D) [pic] E) [pic] Answer: A Diff: 1 Page Ref: Sec. 14.2 A flask is charged with 0.124 mol of A and allowed to react to form B according to the reaction A(g) →B(g). The following data are obtained for [A] as the reaction proceeds: [pic] 5) The average rate of disappearance of A between 10 s and 20 s is __________ mol/s. A) [pic] B) [pic] C) [pic] D) 454 E) [pic] Answer: A Diff: 1 Page Ref: Sec.
Then prepare the unknown sample by pipetting 1 ml of the unknown solution into the cuvette and mix with 1 ml of dilute ferric nitrate. Then after that you calculate the concentration for each of the solutions. Cuvette # | 25 mg/dl Standard | H2O ml | Dilute Iron III Nitrate (ml) | .039 M HNO3 (ml) | Concentration | 1 | 0 | 1 | 0 | 1 | 0 | 2 | 0.1 | 0.9 | 1 | 0 | 0.09 | 3 | 0.3 | 0.7 | 1 | 0 | 0.287 | 4 | 0.5 | 0.5 | 1 | 0 | 0.407 | 5 | 0.7 | 0.3 | 1 | 0 | 0.706 | 6 | 1 | 0 | 1 | 0 | 1.316 | When then use the calibration curve to determine the concentration of the unknown and we get .747. IV. As you can see from our data as the amount of H2O decreases and the 25mg/dl standard increases as does our concentration for the salicylic acids.
Data See Graphs Calculations The calculations consisted of the deviations to create the pH curve graph. The Ka was calculated by taking the –log of the amount of acetic acid added and squaring it then dividing it by the amount of solution added. The moles for the NaOH was calculated by taking .1 moles and dividing it by 1,000 mL and then multiplying it by the amount of NaOH added. For the Molarity, the mole-mole ratio is used to convert it. Conclusion The pH curve came out zig zagged.