The amount of ascorbic acid, C6H8O6, in orange juice was determined by oxidizing the ascorbic acid to dehydroascorbic acid, C6H6O6, with a known excess of I3–, and back titrating the excess I3– with Na2S2O3. A 5.00-mL sample of filtered orange juice was treated with 50.00 mL of excess 0.01003 M I3–. After the oxidation was complete, 12.2722 mL of 0.06729 M Na2S2O3 was needed to reach the starch indicator end point. Report the concentration of ascorbic acid in milligrams per 100 mL.

100.0 cm3 water is titrated with 0.0049 M ascorbic acid solution. What is the analytical concentrationof the weak acid at 12.73 cm3 added titrant volume?

A 27.56 mL portion of a 0.337 M solution of hydrochloric acid, a monoprotic acid, is to be titrated with a standarized 0.411 M solution of KOH. What is the expected volume in milliliters of the KOH solution needed to reach the phenolphthalein end point? Report your answer with a precision of two decimal places.

5 b) Benzoic acid, HC7H5O2, is a weak monoprotic acid. A stock HC,H5O2 solution with unknown concentration was prepared. A 25.00 mL sample of the stock HCH5O2 solution was titrated with 0.1234 M sodium hydroxide to the equivalence point, using phenolphthalein as an indicator. The titration took 20.26 mL of titrant (sodium hydroxide). (b) Write a balanced complete ionic equation for the chemical reaction that occurs in this titration.

Malic acid (C4H6O5) is a dicarboxylic acid that naturally occurs in the production of bignay wine. Maria was tasked to find out the acid content in a 200.0 mL wine sample by titrating it with a standardized solution of KOH. He was able to establish the working concentration of the KOH solution by using 39.12 mL of it to titrate a primary standard of KHP that weighed 3.021 g. The grams and %w/w of Malic acid present in the 200.0 mL wine sample were then determined by titrating a 20.00 mL aliquot using 48.95 mL of the standardized KOH solution.   C4H6O5 (aq) + 2KOH (aq) ↔ K2C4H4O5 (aq) + 2H2O (l)   Assume the density of the wine sample is 1.000 g/mL MM of KHP = 204.22 g/mol MM of C4H6O5 = 134.0874 g/mol                                                                                                                                                     What is the molar concentration of the prepared KOH solution?…

A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 51 mg of oxalic acid (H2C2O4), a diprotic acid that can be purchased inexpensively in high purity, and dissolves it in 250 mL of distilled water. The student then titrates the oxalic acid solution with her sodium hydroxide solution. When the titration reaches the equivalence point, the student finds she has used 19.2 mL of sodium hydroxide solution. Calculate the molarity of the student's sodium hydroxide solution.

The protein in a 0.2976 -g sample of cheese is determined by the Kjeldahl method. The sample is digested with H2SO4, the resulting solution made basic with NaOH, and the ammonia distilled into 50.00 mL of 0.09756 M HCl. The excess HCl is then back titrated using 39.87 mL of 0.05345 M NaOH. Calculate the % N and % protein of the sample.

Chaya the chemistry student carries out a titration by adding an aqueous solution of NaOH to an aqueous solution of the weak acid HY. The images below represent the solutions formed at various points during this titration. (Na+ and H₂O are omitted for clarity.) HY OY &H30+ О ОН A B C Select the image corresponding to each of the following points in Chaya's titration. Before adding any NaOH : -- + Partway to the equivalence point: -- At the equivalence point : -- After the equivalence point: [ → D E y selecting h menu.

A pure organic compound has the formula C4H8SOx. A sample of a compound wasdecomposed; the sulfur was converted to sulfate and titrated in a modified version of theadsorption indicator procedure. From the following data, determine the correct formula forthe compound.Given: weight of sample = 12.64 mgBa(ClO4)2 needed for titration = 10.60 mL of 0.0100 M Ba(ClO4)2

A 0.5504-g sample of KHP (potassium hydrogen phthalate, KHC8H4O4; molar mass = 204.22 g/mol) is completely dissolved in enough deionized water to make 50.00 mL of solution. The titration of this solution with a KOH solution requires 20.50 mL of the base (KOH solution) to reach the “end-point”. What is the molar concentration of KOH solution? The reaction occurs as follows:                       KOH(aq)  +  KHC8H4O4(aq)  -->  K2C8H4O4(aq)  +  H2O(l)           (A) 0.2457 M           (B) 0.1315 M           (C) 0.05390 M           (D) 0.03823 M