What is the mass of pure dry CaCO3 (MW: 100.0869 g/mol) standard used in the standardization of EDTA solution? Assume that 1.758 M EDTA solution was computed based on the titration of CaCO3 standard. The volume of titrant needed to reach the endpoint was 35.00 mL. Ca2+ + Y4- → CaY2- 0.1500 g 5.630 g 6.158 g 0.6158 g

What is the mass of pure dry CaCO3 (MW: 100.0869 g/mol) standard used in the standardization of EDTA solution? Assume that 1.758 M EDTA solution was computed based on the titration of CaCO3 standard. The volume of titrant needed to reach the endpoint was 35.00 mL. Ca2+ + Y4- → CaY2- 0.1500 g 5.630 g 6.158 g 0.6158 g

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter5: Errors In Chemical Analyses

Section: Chapter Questions

Problem 5.10QAP

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What is the mass of pure dry CaCO3 (MW: 100.0869 g/mol) standard used in the standardization of EDTA solution? Assume that 1.875 M EDTA solution was computed based on the titration of CaCO3 standard. The volume of titrant needed to reach the endpoint was 30.00 mL. Ca2+ + Y4- → CaY2- 5.630 g 0.6158 g 0.1500 g 6.158 g
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. Ans in 3 sig figures. no need to write the unit.
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.
C1. A carefully weighed 280mg Calcium carbonate was used in the standardization of an EDTA solution. Initially, 25mL of the titrant was consumed but only after the addition of another 10mL of the titrant did an endpoint was visible. What is the MW of calcium carbonate? (use C1 as reference)* 102 g/mol 98 g/mL 100 g/mL 100 g/mol What is the molar concentration of the standardized EDTA solution? (use C1 as reference)* 0.08m 0.8M 0.08N 0.08M How many grams of EDTA (MW: 292 g/mole) is required to prepare 250mL of a 0.025M solution?* 0.1825 0.1852 1.825 1.1852
Mr. Clean recently bought a laboratory-grade sodium carbonate from a chemical company known as Brand X. He was supposed to use it in the production of detergents. Unfortunately, he was scammed by the company. He suspected that he purchased a crude sodium carbonate so he tasked the Quality Assurance Department to determine the components of the purchased chemical. The chemist assigned to analyze the sample used double indicator method. For the standardization of HCl titrant, 0.1025 g Na2CO3 of 99.5% purity (FW: 106.00) required 8.20 mL of the titrant to reach the phenolphthalein endpoint. FW: NaOH (40.00), NaHCO3 (84.01), Na2CO3 (106.00) a. What is the molarity of the titrant? The chemist obtained a 3.150 g sample and dissolved it in distilled water to produce a 50.0 mL solution. An aliquot of 10.00 mL was obtained and diluted in a 100.0 mL volumetric flask. A 50.00-mL aliquot of the diluted sample was taken and it required 25.70 mL of titrant for the methyl orange endpoint, while…
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.
Bay Water Titration The concentration of Cl- in ocean water is about 500-600 mM. The baywater is diluted by a factor of 12.5 using a 20.00 mL volumetric pipet and a 250 mL volumetric flask. Using a 15.00 mL volumetric pipet, 15.00 mL is transferred into three clean Erlenmeyer flasks. 10 mL of 1% dextrin solution, 20 mL of de-ionized water, and 3-4 drops of indicator are added and titrated each with the AgNO3 solution. From the procedure, find: Dilution factor Volume of diluted bay water Then calculate: The [Cl-] of diluted bay water The [Cl-] of bay water Consider this information: [AgNO3] = 0.04043177 trial # Veq 1 13.91 2 13.73 3 13.9 4 13.86 5 13.87 6 13.84 average 13.85167 [Cl-]Bay = ([AgNO3](Veq(ave))/Vdil bay water pipeted) (Vflask/Vbay water)
1) 100 mL tap water is titrated with 20 mL of 0.02 M EDTA solution to determine the hardness of the water. The second 100 mL tap water is precipitated as CaC2O4 and removed by filtration and the filtrate is titrated with 8.0 mL of the same EDTA solution. What are the concentrations (as ppm) of Ca2+ (40.078 g/mol) and Mg2+ (24.305 g/mol) ions in the water? 2) 10 g of a solution, containing one or more of NaOH, Na3PO4, Na2HPO4, NaH2PO4, H3PO4 and HCl substances, is titrated with 1.0 M NaOH using the phenolphthalein indicator and 34.5 mL NaOH is used. Another 10 g mixture was titrated to reach the bromocresol green end point and 20.50 mL NaOH is used. What is the composition of the solution? What are the percentages of the species(H:1.0078, O:15.99, Na: 22.98, P:30.974, Cl:35.453)?
When standardizing two liters of a potassium permanganate solution with a primary standard solution of sodium oxalate, approximately 0.0100 M, it was planned to use between 30.00 mL and 45.00 mL of the titrated reagent. In which mass range (in grams) the primary standard should be weighed? Data: MM KMnO4 = 158.00 g/mol; MM Na2C2O4 = 134.00 g/mol 2CO2 + 2e- ↔ C2O4 2- Eo = -0,432 V MnO4 - + 8 H+ + 5e- ↔ Mn2+ + 4 H2O Eo = 1,52 V
If you titrate 50 mL of a sample with 0.06 N EDTA and the titration takes 12.4 mL of EDTA to reach the endpoint, what is the hardness of the sample in ppm as CaCO3?Identify the nature of the water sample from the hardness value.
In a titration, the following masses were recorded from an unknown substance that contain some amount of KHP in three experimental solutions 0.44, 0.437, and 0.437, if the volume of NaOH IN THE exp are 5, 10, and 8 respectively in order to the masses of unknown substances.1. Find the mass of KHP of each experiment 2.Find % KHP 3. Find average % KHP 4.find standard deviation.( Note; in a previous three 100% KHP experiments the average molarity of NaOH in each experiment are shown below average molarity=5.30x 10^-4+ 4.137x10^-4 + 5.0x10^-4 divided by 3 = 4.81x10^-4 ).
In a Complexometric Titration Experiment, the preparation of the EDTA solution is as follows: In a 250 mL beaker, weigh 0.10 g MgCl2*6H2O and 1.0 g NaOH. Add approximately 200 mL distilled water and stir to dissolve. Add about 2.0 g of EDTA and transfer the solution quantitatively to a 1-L volumetric flask. Dilute to the mark with distilled water. What is the purpose of adding MgCl2*6H2O? Please show COMPLETE solution.