50mL of standard hard water (contains 10g of CaCO3 per liter) required 27mL of EDTA solution for end point. 100 mL of sample water required 12 mL of EDTA solution while 100mL of water after boiling required 7mL of EDTA solution. Calculate total, permanent and temporary hardness of water.
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- A standard CaCO3 solution is prepared by dissolving 0.4193g in enough dilute HCl to effect solution and then diluted to 500ml solution. A 25.00ml aliquot requires 23.62ml EDTA solution for titration. A 25.00ml water sample determined for total hardness, requires 8.45ml of the EDTA solution using Eirochrome Black T indicator. Calculate the total hardness in the water sample expressed in ppm CaCO3.The distribution coefficient for X between chloroform and water is 9.6. What volume of CHCl3 is required to decrease the concentration of X to 1.0 × 10-4 if 25.0 ml of 0.0500 M X is extracted witha) 25.0 ml portions of CHCl3b) 10.0 ml portions of CHCl3c) 2.0 ml portions of CHCl360 ml of standard hardness containing 1 mg of pure CaCO3 per ml consumed 22 ml of EDTA. 40 ml of water sample consumed 20 ml of EDTA solution using EBT indicator. 40 ml of water sample after boiling, filtering consumed 15 ml of EDTA. Calculate the temporary and permanent hardness of water sample.
- Create a flowchart for the procedures in the determination of water hardness.A 2.054 g of macrogol monostearate (average molecular weight 706. 5) was added to a 200 ml flask and 25 ml of an ethanolic solution of potassium hydroxide (molecular weight 56.1, ca 0.5 M) was added. The sample was heated under a reflux condenser for 1 hour. The excess of alkali was then titrated with 0.5016 M hydrochloric acid using phenolphthalein solution as an indicator. The operation was repeated without the macrogol monostearate. Results Volume of HCI required to titrate the excess alkali = 18.35 ml Volume of HCI required to titrate the blank = 24.03 ml Calculate the saponification value for the macrogol stearate. Answer: Blank mg/gIf 30.0 mL of water that is 0.0500 M in Q is to be extracted with four 10.0-mL portions of an immiscible organic solvent, what is the minimum distribution coefficient that allows transfer of all but the following percentages of the solute to the organic layer: (a) 1.00 X10-24 (b) 1.00 X10-22
- 20 aspirin tablets labeled 80mg were dissolved in 100mL of 90% ethanol. A 10mL aliquot was taken and was used for assay. The analyte followed usual process and was treated with 50mL of 0.1000N NaH and was titrated with 35mL 0.1050N H2O4 until the solution achieved completion. Calculate the % content of the total aspirin capsules and the actual label claimFor Indirect Iodometric Analysis of Copper... ~0.0896g KIO3 necessary to consume 350mL of 0.1 M Na2S2O3, Na2S2O3 is stored in an amber glass bottle until ready for use. Primary Standard KIO3 has 2g of KI, 50mL of DI water, and 10 mL of 1.0M HCl is added then immediately titrated with Na2S2O3 until medium yellow or straw... then 5mL of starch indicator is added and titrated again until blue black color turns clear. Unknown CuO use 1.2G of Unknown, 20mL of HNO3 heated until sample dissolved, 25 mL of DI water added and boiled until clear light blue color, after cooling 1:1 NH3 added (~34.47 mL of NH4OH reagent) until permanent deep blue color amine complex, 2g of NH4HF2 added and swirled until dissolved, 3 g of KI is added then titrated immediately with Na2S2O3 until brown color of iodide is nearly gone (brown milk color), 2 g of KSCN and 3 mL of starch indicator is then added with titration continuing until disappearance of new blue black color. 1. Na2CO3 is often added to thiosulfate…For Indirect Iodometric Analysis of Copper... ~0.0896g KIO3 necessary to consume 350mL of 0.1 M Na2S2O3, Na2S2O3 is stored in an amber glass bottle until ready for use. Primary Standard KIO3 has 2g of KI, 50mL of DI water, and 10 mL of 1.0M HCl is added then immediately titrated with Na2S2O3 until medium yellow or straw... then 5mL of starch indicator is added and titrated again until blue black color turns clear. Unknown CuO use 1.2G of Unknown, 20mL of HNO3 heated until sample dissolved, 25 mL of DI water added and boiled until clear light blue color, after cooling 1:1 NH3 added (~34.47 mL of NH4OH reagent) until permanent deep blue color amine complex, 2g of NH4HF2 added and swirled until dissolved, 3 g of KI is added then titrated immediately with Na2S2O3 until brown color of iodide is nearly gone (brown milk color), 2 g of KSCN and 3 mL of starch indicator is then added with titration continuing until disappearance of new blue black color. 4. Why is the starch indicator solution…
- 1. A 1.2-gram sample of lanolin was treated with Wij’s solution and excess potassium iodide solution. The liberated iodine reacted with 30 ml of 0.1 N sodium thiosulfate solution. If the iodine value was determined as 12.69, what is the volume used in blank titration? 2. A fat sample with combination of acids contain standard hydrochloric acid for blank and sample with 8mL and 5mL respectively. The normality of the standard hydrochloric acid is 0.93N and the weight of the sample is 3 grams. Calculate the saponification value. 3. A 3.50-gram sample of Streptomycin powder was tested for its water content. If the water equivalence factor of the KF reagent was 4.6, what is the percentage water content of the sample if 9.2 ml of the KF reagent was used? 4. A 500mg oil sample is taken from a conical flask and is dissolved in 50mL distilled alcohol. An indicator is added and is then titrated against 0.112N KOH until a slight pink color appears. It took 17.6mL of the titrant to reach the…You are required to prepare working standard solutions of 1.00 × 10−5, 2.00 × 10−5, 5.00 × 10−5, and 1.00 × 10−4 M glucose from a 0.100 M stock solution. You have available 100-mL volumetric flasks and pipets of 1.00-, 2.00-, 5.00-, and 10.00-mL volume. Outline a procedure for preparing the working standards.A 5.000-g soil sample was analyzed for potassium content by extracting the potassium using 10.00 mL aqueous ammonium acetate solution. Following the extraction, the soil was filtered and rinsed. The filtrate with rinsings was diluted to exactly 50.00 mL. Then, 1.00 mL of this solution was diluted to 25.00 mL, and this dilution was tested with an instrument. The concentration in this 25.00 mL was found to be3.18 ppm. What is the concentration of the potassium in the soil in ppm?
