A student was given an unknown sample containing the chloride ion, Cl and was to

Question

A student was given an unknown sample containing the chloride ion, Cl and was to determine the percentage of Cl in the sample. The student dissolved 0.250 grams of the unknown in 50 mL of distilled water, added 3 drops of K2CrO4 and then titrated the solution with 0.05005 M AgNO3. The presence ofthe red Ag CrO precipitate or buff mixture indicated that the reaction had reached completion after 26.45 mL of the 0.05005 M AgNO, solution was added. Using the data given, complete the following table to determine the percentage of Cl in the unknown sample. Show your work. Measurement Titration 1 1. Mass of sample 2. Volume of AgNOs added 3. Moles of AgNO added 4. Moles of Ag 5. Moles of Cl 6.Mass of CI % Clin unknown

Explanation / Answer

Potassium chromate react with silver nitrate and red precipitate of Ag2CrO4 (s) is formed. Chromate ion is yellow in color. When end point came, chloride solution begins to turn red.

K2CrO4 + 2AgNO3 ----> Ag2CrO4 (s)+ 2KNO3

Mass of sample = 0.250 g

Volume of AgNO3 added = 26.45 ml

Molarity of AgNO3 added = 0.05005M

3. Moles of AgNO3 added = Molarity of AgNO3 added x Volume of AgNO3 added in L

= 0.05005 moles/L * (26.45 /1000) L = 0.0013238 moles

4. Moles of Ag+ in sample = 0.00132 moles

5. Moles of Cl- in sample = 0.00132 moles

6. Mass of Cl- in sample

Molar mass of Cl- = 35.45 g /mol

To calculate mass of Cl- in sample, multiply moles of Cl- with molar mass of Cl-

Mass of Cl- in sample = 0.00132 moles* 35.45 g /mol = 0.047 g

% Cl- in unknown sample = (Mass of Cl- in sample/ Mass of sample) x 100

% Cl- in unknown sample = (0.047 g/0.250 g) x 100

% Cl- in unknown sample = 18.8%

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