A 200.00 mL sample of a citrus fruit drink is acidified and 10.00 ml of 0.05000 M I2 solution is added. After the reaction is complete the excess I2 is titrated with 3.86 mL of 0.1200 M Na,S,03 solution. Calculate the concentration of ascorbic acid, C,H3O, (M.M. = 176.14) in the juice drink sample in mg/100 mL sample. CeHsOs CaH.Os + 2H + 2e A 32.41 mg CH3O6/100 mL sample 23.64 mg CgHgO6/100 mL sample C 47.28 mg CoHsO6/ 100 mL sample D 400 mg CsHsO/ 100 mL sample

A 0.8159 g sample of HCl was placed into a 50 mL volumetric flask and the sample was thoroughly dissolved in water to make 50 mL of solution. It required 22.07 mL of NaOH to reach the endpoint in the titration. What is the molarity of the NaOH solution? I mg IHE/ 2.022 4 mol 0.81599 Hcl 36.4bg.

A solution is prepared by pipetting 25.0 mL of 0.350 M CuCl2 into a 100.0 mL volumetric flask, diluting to the calibration mark on the flask with water and quantitatively mixing the contents of the flask. What is the molar concentration of Cl- in the solution?

Mark and John, two vinegar enthusiasts, are each tasked to determine the acetic acid content of their respective vinegar concoctions by titration. First, a 1 M-labeled KOH solution was standardized against the KHP (MW = 204.22 g/mol) standard that is 99.4% pure. In the process, 0.540 g KHP was found to require 2.80 mL of the KOH solution to completely react up to the phenolphthalein endpoint. Then, Mark and John both prepared their samples by taking 10.0-mL aliquots of each vinegar and diluting them to 25.0 mL. Using the same titrant and indicator, Mark’s vinegar required 18.60 mL of the standardized titrant to reach the endpoint, while John’s vinegar required 16.50 mL of the same titrant to reach the same endpoint. Question: What is the acetic acid concentration of Mark’s and John’s vinegar in molarity given that the Concentration of KOH is 9386 M?