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The mass of lead chromate that is dissolved in 250 mL of a saturated solution is _____ grams. (Ksp = 1.8 x 10-14)

The mass of lead chromate that is dissolved in 250 mL of a saturated solution is _____ grams. (Ksp = 1.8 x 10-14)

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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  1. The mass of lead chromate that is dissolved in 250 mL of a saturated solution is _____ grams. (Ksp = 1.8 x 10-14)
  2. The attached photo is an example of how to solve the problem
The mass of iron(III) hydroxide that is dissolved in 225 mL of a saturated solution is 2.36x10^-9 grams. Incorrect STEP 1 Determine the number of moles of Fe(OH)3(s) that can dissolve: Let 's' = number of moles per liter that dissolve. %3D Fe(OH)3 (s) Fe3+ (aq) + 30H¯ (aq) Initial some Change Equilibrium + 3s + 3s + s some
Transcribed Image Text:The mass of iron(III) hydroxide that is dissolved in 225 mL of a saturated solution is 2.36x10^-9 grams. Incorrect STEP 1 Determine the number of moles of Fe(OH)3(s) that can dissolve: Let 's' = number of moles per liter that dissolve. %3D Fe(OH)3 (s) Fe3+ (aq) + 30H¯ (aq) Initial some Change Equilibrium + 3s + 3s + s some
STEP 2 Set up the Ksp expression and solve for s: Ksp =[Fe3+] [OH13 = (s)(3s)³ = 27s1 = 38 (from the table) 6.3x10 %3D %3D %3D 2.20x10 10 moles Fe(OH)3 S = (Ksp /27)1/4 2.20x10 10M = %3D L solution We'll carry 3 significant figures and round to 2 at the end. STEP 3 In 225 mL of solution: 2.20×10¯10 mol Fe(OH)3 1 L # moles Fe(0H)3 4.95x10 11 mol Fe %3D 225 mL L 1000 mL 106.9 g Fe(0H)3 # grams Fe(OH)3 4.95×10 11 mol Fe(OH)3 5.3x109 g Fe(0H)3 di %3D %3D mol Fe(OH)3
Transcribed Image Text:STEP 2 Set up the Ksp expression and solve for s: Ksp =[Fe3+] [OH13 = (s)(3s)³ = 27s1 = 38 (from the table) 6.3x10 %3D %3D %3D 2.20x10 10 moles Fe(OH)3 S = (Ksp /27)1/4 2.20x10 10M = %3D L solution We'll carry 3 significant figures and round to 2 at the end. STEP 3 In 225 mL of solution: 2.20×10¯10 mol Fe(OH)3 1 L # moles Fe(0H)3 4.95x10 11 mol Fe %3D 225 mL L 1000 mL 106.9 g Fe(0H)3 # grams Fe(OH)3 4.95×10 11 mol Fe(OH)3 5.3x109 g Fe(0H)3 di %3D %3D mol Fe(OH)3
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