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The volume of water needed to dissolve 0.0694 grams of lead fluoride is ______ L. Assume no volume change upon addition of the solid. (Ksp = 3.7 x 10-8)

The volume of water needed to dissolve 0.0694 grams of lead fluoride is ______ L. Assume no volume change upon addition of the solid. (Ksp = 3.7 x 10-8)

Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter8: Thermochemistry
Section: Chapter Questions
Problem 24QAP
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  1. The volume of water needed to dissolve 0.0694 grams of lead fluoride is ______ L. Assume no volume change upon addition of the solid. (Ksp = 3.7 x 10-8)
  2. The attached photos is an example of how to solve the problem 
xL. The volume of water needed to dissolve 0.0700 grams of lead sulfate is 183x10^-8 Assume no volume change upon addition of the solid. Incorrect STEP 1 Determine the number of moles of PBSO4(s) that can dissolve in ONE liter: Let 's' = number of moles per liter that dissolve. %3D PBSO4 (s) = Pb2+ (aq) + SO,²- (aq) Initial some Change + S Equilibrium some + s + s STEP 2 Set up the Ksp expression and solve for s: Ksp =[Pb2+] [So,2] = (s)(s) = s? = 1.8×108 (from the table) 1.34x104 moles PbSO4 Previous. Next 1/2
Transcribed Image Text:xL. The volume of water needed to dissolve 0.0700 grams of lead sulfate is 183x10^-8 Assume no volume change upon addition of the solid. Incorrect STEP 1 Determine the number of moles of PBSO4(s) that can dissolve in ONE liter: Let 's' = number of moles per liter that dissolve. %3D PBSO4 (s) = Pb2+ (aq) + SO,²- (aq) Initial some Change + S Equilibrium some + s + s STEP 2 Set up the Ksp expression and solve for s: Ksp =[Pb2+] [So,2] = (s)(s) = s? = 1.8×108 (from the table) 1.34x104 moles PbSO4 Previous. Next 1/2
1.34x10 4 moles PbSO4 Ksp 1/2 1.34x10 4 M = S = %3D L solution We'll carry 3 significant figures and round to 2 at the end. STEP 3 Determine the moles of PbSO, to dissolve: 1 mol PbS04 0.0700 g PbSO4x 2.31×101 moles PbSO4 303.3 g PbS04 STEP 4 Determine the volume needed: 1 L solution V = 2.31x104 moles PbSO4 1.7 L solution needed 1.34×10 4 mol PbSO4
Transcribed Image Text:1.34x10 4 moles PbSO4 Ksp 1/2 1.34x10 4 M = S = %3D L solution We'll carry 3 significant figures and round to 2 at the end. STEP 3 Determine the moles of PbSO, to dissolve: 1 mol PbS04 0.0700 g PbSO4x 2.31×101 moles PbSO4 303.3 g PbS04 STEP 4 Determine the volume needed: 1 L solution V = 2.31x104 moles PbSO4 1.7 L solution needed 1.34×10 4 mol PbSO4
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