University:
Campbell University
Course:
CHEM 111 | General Chemistry I
Academic year:

2021
Views:

318

Pages:

2
Author:

rawnwinen9wn

Applications of Solubility 1. When 25.0 mL of NaCl solution having an unknown concentration is titrated with 0.100 M AgNO₃, using chromate ion as an indicator, 36.8 mL of the AgNO₃ solution are required to reach the equivalence point. What is the [Cl⁻]? AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) **Answers:** mol Ag⁺ = 0.100 mol/L * 0.0368 L = 0.00368 mol Ag⁺ mol Cl⁻ = 0.00368 mol Cl⁻ [Cl⁻] = 0.00368 mol / 0.0250 L = 0.147 M 2. What volume of 0.0988 M Cl⁻ solution is required to titrate 25.0 mL of 0.0750 M AgNO₃, using chromate indicator? AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) 1:1 ratio for Ag⁺:Cl⁻ **Answers:** mol Ag⁺ = 0.0750 mol/L * 0.0250 L = 0.001875 mol Ag⁺ mol Cl⁻ = 0.001875 mol Cl⁻ volume of Cl⁻ = 0.001875 mol / 0.0988 M = 0.018977 L = 19.0 mL 3. A solution of potassium chloride is made by dissolving 3.25 g KCl in water and diluting to 500.0 mL. If 9.48 mL of AgNO₃ solution is required to titrate 25.00 mL of the KCl solution, what is the molar concentration of the AgNO₃? AgCl ⇌ Ag⁺ + Cl⁻ 1:1 ratio for Ag⁺:Cl⁻ **Answers:** mol Cl⁻ = 3.25 g KCl * (1 mol KCl / 74.55 g KCl) * (1 mol Cl⁻ / 1 mol KCl) * (1 / 0.500 L) = 0.0871 mol Cl⁻ mol Ag⁺ = 0.002179 mol Ag⁺ [Ag⁺] = 0.002179 mol / 0.00948 L = 0.22993 M 4. A student is assigned the task of finding the Ksp value for silver acetate, AgCH₃COO. Several grams of AgCH₃COO are added to distilled water and stirred overnight. The next day a 50.0 mL sample of the saturated AgCH₃COO solution is titrated with 30.6 mL of 0.100 M NaCl. What is the value of Ksp for AgCH₃COO? AgCH₃COO ⇌ Ag⁺ + CH₃COO⁻ 1:1 ratio for Ag⁺:CH₃COO⁻ **Answers:** mol Cl⁻ = 0.100 mol/L * 0.0306 L = 0.00306 mol Cl⁻ mol Ag⁺ = 0.00306 mol Ag⁺ [Ag⁺] = 0.00306 mol / 0.0500 L = 0.0612 M [CH₃COO⁻] = 0.0612 M Ksp = (0.0612)(0.0612) = 3.74544 × 10⁻³ 5.) A 4.75 g silver coin was dissolved in nitric acid and the resulting solution diluted to 250 mL. When a 25.0 mL sample of 0.200 M NaCl was titrated with the silver solution, using chromate indicator, 28.8 mL of silver solution was required. What was the percentage purity of the silver in the coin, assuming any impurities present were unreactive? AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) 1:1 ratio for Ag⁺:Cl⁻ **Answers:** mol Cl⁻ = 0.200 mol/L * 0.0250 L = 0.00500 mol Cl⁻ mol Ag⁺ = 0.00500 mol Ag⁺ [Ag⁺] = 0.00500 mol / 0.0288 L = 0.173611 M 0.173611 mol Ag⁺/L * 0.250 L * (107.87 g Ag / 1 mol Ag) = 4.681857 g Ag %Ag = (4.681857 g / 4.75 g) * 100 = 98.6% 6.) A 95.6 g sample of chicken from a restaurant was checked for Cl⁻ (in the form of NaCl). The chicken was blended with water and suction filtered. The solution obtained was then diluted to 1.00 L. A 25.0 mL sample of the solution was titrated with 0.0200 M AgNO₃ solution, using chromate indicator, and 15.3 mL was found to be needed. a.) What was the [Cl⁻] in the solution? AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq) 1:1 ratio for Ag⁺:Cl⁻ **Answers:** mol Ag⁺ = 0.0200 mol/L * 0.0153 L = 3.06 × 10⁻⁴ mol Ag⁺ mol Cl⁻ = 3.06 × 10⁻⁴ mol Cl⁻ [Cl⁻] = 3.06 × 10⁻⁴ mol / 0.0250 L = 0.01224 M b.) How many grams of the NaCl were extracted from the meat? **Answers:** 0.01224 mol Cl⁻ * (1 mol NaCl / 1 mol Cl⁻) * (58.44 g NaCl / 1 mol NaCl) = 0.715305 g NaCl NaCl = 0.715 g c.) What was the percentage of NaCl in the original chicken sample? **Answers:** % NaCl = (0.715 g / 95.6 g) * 100 % NaCl = 0.748 %