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Chapter 10: Problem 14

A solution is prepared by diluting \(0.7850 \mathrm{~L}\) of \(1.262 \mathrm{M}\) potassium sulfide solution with water to a final volume of \(2.000 \mathrm{~L}\). (a) How many grams of potassium sulfide were dissolved to give the original solution? (b) What are the molarities of the potassium sulfide, potassium ions, and sulfide ions in the diluted solution?

Short Answer

Expert verified
Answer: The mass of potassium sulfide in the original solution is 109.277 g. In the diluted solution, the molarities are as follows: K2S has a molarity of 0.497125 M, K+ ions have a molarity of 0.99425 M, and S^2- ions have a molarity of 0.497125 M.

Step by step solution

01

Calculate the mass of K2S in the original solution

To find the mass of K2S in the original solution, we'll use the formula mass = molarity × volume × molar mass. Molarity (M1) = 1.262 M Volume (V1) = 0.7850 L Molar mass of K2S = 39.098 + 39.098 + 32.07 = 110.266 g/mol Mass = (1.262 mol/L) × (0.7850 L) × (110.266 g/mol) = 109.277 g There are 109.277 grams of potassium sulfide dissolved in the original solution.
02

Calculate the molarity of K2S in the diluted solution

To find the molarity of K2S in the diluted solution, we'll use the dilution formula M1V1 = M2V2. Final volume (V2) = 2.000 L M1 = 1.262 M V1 = 0.7850 L M2 = (M1 × V1) / V2 = (1.262 mol/L × 0.7850 L) / 2.000 L= 0.497125 M The molarity of K2S in the diluted solution is 0.497125 M.
03

Calculate the molarities of K+ and S^2- ions

As each molecule of K2S dissociates into two potassium ions (K+) and one sulfide ion (S^2-): Molarity of K+ ions = 2 × Molarity of K2S = 2 × 0.497125 M = 0.99425 M Molarity of S^2- ions = Molarity of K2S = 0.497125 M In the diluted solution: Molarity of potassium sulfide (K2S) = 0.497125 M Molarity of potassium ions (K+) = 0.99425 M Molarity of sulfide ions (S^2-) = 0.497125 M

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Most popular questions from this chapter

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