# Titrations Lecture 2, Oct 11. Homework Ch 5 Problems 3,4,6,9,12,13,14 Due Wed Oct 16.

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Titrations Lecture 2, Oct 11

Homework Ch 5 Problems 3,4,6,9,12,13,14 Due Wed Oct 16

Calculations for Titrations In a back titration, the calculation consists of 1)Calculation of the total number of moles of the excess reactant. 2)From the titration with the second reactant, calculate the number of moles of analyte that has reacted with the first reactant.

Although most of the metallic ions of the periodic table can be determined with edta, several require a back titration because of slow kinetics or an unsuitable indicator.

Problem 1 - A 25.00 mL aliquot of a solution of Hg +2 was adjusted to a pH of 10.0 with ammonia and treated with 10.00 mL of 0.04882 M edta. The excess edta required 24.66 mL of 0.01137 M Mg +2 to reach the Eriochrome black T end point. What is the conc of Hg +2 in the original solution?

Edta reacts 1:1 with all metals through the first 2 transition series. # mmol of edta = 10.00mL x 0.04882 M = 0.4882 mmol

Problem 1 - A 25.00 mL aliquot of a solution of Hg +2 was adjusted to a pH of 10.0 with ammonia and treated with 10.00 mL of 0.04882 M edta. The excess edta required 24.66 mL of 0.01137 M Mg +2 to reach the Eriochrome black T end point. What is the conc of Hg +2 in the original solution? Edta reacts 1:1 with all metals through the first 2 transition series. # mmol of edta = 10.00mL x 0.04882 M = 0.4882 mmol # mmol Mg +2 = 24.66mL x 0.01137 = 0.2804 mmol

Problem 1 - A 25.00 mL aliquot of a solution of Hg +2 was adjusted to a pH of 10.0 with ammonia and treated with 10.00 mL of 0.04882 M edta. The excess edta required 24.66 mL of 0.01137 M Mg +2 to reach the Eriochrome black T end point. What is the conc of Hg +2 in the original solution? Edta reacts 1:1 with all metals through the first 2 transition series. # mmol of edta = 10.00mL x 0.04882 M = 0.4882 mmol # mmol Mg +2 = 24.66mL x 0.01137 = 0.2804 mmol # mmol difference = # mmol Hg +2 = 0.2078 mmol

Problem 1 - A 25.00 mL aliquot of a solution of Hg +2 was adjusted to a pH of 10.0 with ammonia and treated with 10.00 mL of 0.04882 M edta. The excess edta required 24.66 mL of 0.01137 M Mg +2 to reach the Eriochrome black T end point. What is the conc of Hg +2 in the original solution? Edta reacts 1:1 with all metals through the first 2 transition series. # mmol of edta = 10.00mL x 0.04882 M = 0.4882 mmol mmol Mg +2 = 24.66mL x 0.01137 = 0.2804 mmol # mmol difference = # mmol Hg +2 = 0.2078 mmol Conc Hg +2 = 0.2078mmol/25.00 mL = 8.312 x 10 -3 M

Problem 1 - A 25.00 mL aliquot of a solution of Hg +2 was adjusted to a pH of 10.0 with ammonia and treated with 10.00 mL of 0.04882 M edta. The excess edta required 24.66 mL of 0.01137 M Mg +2 to reach the Eriochrome black T end point. What is the conc of Hg +2 in the original solution? Edta reacts 1:1 with all metals through the first 2 transition series. # mmol of edta = 10.00mL x 0.04882 M = 0.4882 mmol mmol Mg +2 = 24.66mL x 0.01137 = 0.2804 mmol # mmol difference = # mmol Hg +2 = 0.2078 mmol Conc Hg +2 = 0.2078mmol/25.00 mL = 8.312 x 10 -3 M Mass of Hg +2 per L = 8.312 x 10 -3 x 200.59 g/mol = 1.667 g/L

Problem 2 - The purity of ethyl acetate was determined by treatment of a 3.278 g sample with 75.00 mL of 0.3861 M KOH and heating to saponify the ester. (products are ethyl alcohol and acetic acid). The excess KOH required 32.53 mL of 0.2066 M HCl to reach the equivalence point. What is the purity of the ester?

# mmol KOH = 75.00mL x 0.3861M = 28.957mmol

Problem 2 - The purity of ethyl acetate was determined by treatment of a 3.278 g sample with 75.00 mL of 0.3861 M KOH and heating to saponify the ester. (products are ethyl alcohol and acetic acid). The excess KOH required 32.53 mL of 0.2066 M HCl to reach the equivalence point. What is the purity of the ester? # mmol KOH = 75.00mL x 0.3861M = 28.957mmol #mmol HCl = 32.53mL x 0.2066M = 6.721 mmol

Problem 2 - The purity of ethyl acetate was determined by treatment of a 3.278 g sample with 75.00 mL of 0.3861 M KOH and heating to saponify the ester. (products are ethyl alcohol and acetic acid). The excess KOH required 32.53 mL of 0.2066 M HCl to reach the equivalence point. What is the purity of the ester? # mmol KOH = 75.00mL x 0.3861M = 28.957mmol #mmol HCl = 32.53mL x 0.2066M = 6.721 mmol diff = # mmol HAc = 22.236 mmol

Problem 2 - The purity of ethyl acetate was determined by treatment of a 3.278 g sample with 75.00 mL of 0.3861 M KOH and heating to saponify the ester. (products are ethyl alcohol and acetic acid). The excess KOH required 32.53 mL of 0.2066 M HCl to reach the equivalence point. What is the purity of the ester? # mmol KOH = 75.00mL x 0.3861M = 28.957mmol #mmol HCl = 32.53mL x 0.2066M = 6.721 mmol diff = # mmol HAc = 22.236 mmol MM ethylacetate = MM 88.11 g/mol

Problem 2 - The purity of ethyl acetate was determined by treatment of a 3.278 g sample with 75.00 mL of 0.3861 M KOH and heating to saponify the ester. (products are ethyl alcohol and acetic acid). The excess KOH required 32.53 mL of 0.2066 M HCl to reach the equivalence point. What is the purity of the ester? # mmol KOH = 75.00mL x 0.3861M = 28.957mmol #mmol HCl = 32.53mL x 0.2066M = 6.721 mmol diff = # mmol HAc = 22.236 mmol MM ethylacetate = MM 88.11 g/mol Mass of ethylacetate = 22.236 x 10 -3 mol x 88.11 g/mol = 1.959 g

Problem 2 - The purity of ethyl acetate was determined by treatment of a 3.278 g sample with 75.00 mL of 0.3861 M KOH and heating to saponify the ester. (products are ethyl alcohol and acetic acid). The excess KOH required 32.53 mL of 0.2066 M HCl to reach the equivalence point. What is the purity of the ester? # mmol KOH = 75.00mL x 0.3861M = 28.957mmol #mmol HCl = 32.53mL x 0.2066M = 6.721 mmol diff = # mmol HAc = 22.236 mmol MM ethylacetate = MM 88.11 g/mol Mass of ethylacetate = 22.236 x 10 -3 mol x 88.11 g/mol = 1.959 g % purity = 1.959 x 100 / 3.278 = 59.76 % pure

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