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5.00 mL of a water sample was titrated for hardness with 4.15 mL of 0.0200 M EDTA. How many moles of Ca2+ were in the 5.00 mL of water?
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- A 5.00 mL tap water sample was measured out with a volumetric pipette, and added to a 25 mL Erlenmeyer flask. It was then titrated with a 0.0100 M Na2EDTA.2H2O solution, and found to take 0.635 mL of the EDTA solution to reach the blue endpoint.A deepwell water sample of 100 mL was titrated with 0.010 M EDTA at pH 12 and consumed 31 mL. Calculate the Calcium content of the water sample in mg/L.A weak acid, HA, has a pKa of 6.5. If you are utilizing ion-exchange chromatography with an anion exchange resin, what pH range is beneficial to the binding to the stationary phase? And Why?
- Five white, 500-mg uncoated ascorbic acid (AA) tablets with an average weight of 0.6100-g were pulverized in a mortar. A sample of the powdered ascorbic acid weighing 0.4610-g was placed in an iodine flask and was dissolved in 50-mL H2SO4 then 5-g of KBr was added to the resulting solution. The solution was titrated with 46.73-mL of 0.0152 M STD. KBrO3 to reach a faint yellow endpoint then 3-g KI and 5-mL Starch TS. The blue color solution is then titrated with 2.78-mL of 0.1047 M STD. Na2S2O3 to reach the disappearance of the blue iodostarch complex. MW: KBrO3 = 167.0 ; KIO3 = 214.0 ; Na2S2O3 = 158.11 ; C6H8O6 = 176.12 Compute the milligrams of pure AA per tablet from the assay. None of the choices 349.7 mg 264.3 mg 462.7 mga. A deepwell water sample of 100 mL was titrated with 0.010 M EDTA at pH 10 ( with Mg-EDTA pH 10 buffer reagent) and consumed 31 mL. Calculate the Total Hardness of the water sample as mg CaCO3. b. A deepwell water sample of 100 mL was titrated with 0.010 M EDTA at pH 12 and consumed 31 mL. Calculate the Calcium content of the water sample in mg/L.A 5.00 mL tap water sample was measured out with a volumetric pipette, and added to a 25 mL Erlenmeyer flask. It was then titrated with a 0.0100 M Na2EDTA.2H2O solution, and found to take 0.635 mL of the EDTA solution to reach the blue endpoint. a.) How many moles of the 0.0100 M Na2EDTA.2H2O were added in the titration reaction? Show calculation. b.) How many moles of calcium ions were reacted?
- What is the function of masking agent in EDTA titration?A 28.93 mL aliquot of trimethylamine that has a concentration of 0.661 M will be titrated with 0.752 M HCl. Calculate the pH of the solution upon the addition of 32.6 mL of HCl. The Kb at the temperature of the experiment is equal to 6.11 x 10-5. Kw = 1.00 x 10-14Calculate the pCa for the titration of 50 ml of 0.02 M Ca2+ at pH = 8 with 0.4 M EDTA after addition of 40 ml EDTA? Ca2+ + Y4- ⇄ CaY2- Kf = 5x1010 a. 9.6 b. 4.9 c. 7.7 d. 10.2 e. 7.6
- DETERMINATION OF WATER HARDNESS BY EDTA TITRATION 1. Why is it needed to add a pH 10 buffer to the sample during EDTA titration? 2. What is EDTA and why can it be used to determine the hardness of water?Which of the following experiments can be described as a displacement titration? A. Calcium in powdered milk is determined by dry ashing a 1.50 g sample and then titrating the calcium with 12.1 mL of 0.008949 M EDTA. B. The Tl in a 9.57-g sample of rodenticide was oxidized to the trivalent state and treated with an unmeasured excess of Mg/EDTA solution. The reaction is Tl3+ + MgY2- -> TlY- + Mg2+ Titration of the liberated Mg2+ required 12.77 mL of 0.03610 M EDTA. C. A 3.650-g sample containing bromide was dissolved in sufficient water to give 250.0 mL. After acidification, silver nitrate was introduced to a 25.00-mL aliquot to precipitate AgBr, which was filtered, washed, and then redissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) -> 2Ag(CN)2- + Ni2+ + 2Br- The liberated nickel ion required 26.73 mL of 0.02089 M EDTA. D. A solution contains 1.569 mg of CoSO4 (155.0 g/ mol) per milliliter. 50.00 mL of 0.007840 M EDTA…Calculate the PCa vs. EDTA titration curve for 50.0 mL of 0.0060 M Ca2+ being titrated with 0.0150 M EDTA in a solution buffered to a constant pH of 10.00. After the addition of 0.00 mL After the addition of 5.00 mL After the addition of 26.00 mL