3.75 g sample of calcium carbonate was dissolved using 6 m hcl and diluted with water to 500 ml. a 50.00 ml portion of this solution is transferred to a flask and the ph was adjusted by adding 5ml of NH3-NH4Cl buffer. after adding EBT indicator, the solution required 47 ml EDTA to reach the end point. given: CaCO3 = 3.75g HCl Conc = 6M Water = 500 mL or 0.5 L EDTA = 47 mL
3.75 g sample of calcium carbonate was dissolved using 6 m hcl and diluted with water to 500 ml. a 50.00 ml portion of this solution is transferred to a flask and the ph was adjusted by adding 5ml of NH3-NH4Cl buffer. after adding EBT indicator, the solution required 47 ml EDTA to reach the end point. given: CaCO3 = 3.75g HCl Conc = 6M Water = 500 mL or 0.5 L EDTA = 47 mL
Principles of Instrumental Analysis
7th Edition
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Chapter24: Coulometry
Section: Chapter Questions
Problem 24.10QAP: A 0.0712-g sample of a purified organic acid was dissolved in an alcohol-water mixture and titrated...
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A 3.75 g sample of calcium carbonate was dissolved using 6 m hcl and diluted with water to 500 ml. a 50.00 ml portion of this solution is transferred to a flask and the ph was adjusted by adding 5ml of NH3-NH4Cl buffer. after adding EBT indicator, the solution required 47 ml EDTA to reach the end point.
given:
CaCO3 = 3.75g
HCl Conc = 6M
Water = 500 mL or 0.5 L
EDTA = 47 mL
QUESTION:
What is the computed molar concentration of EDTA?
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