Question 7 of 10 Available From: Not Due Date: Points Possibl 0 Grade Category:

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Question 7 of 10 Available From: Not Due Date: Points Possibl 0 Grade Category: Extra Map 12/8 Sapling Learning macmilan learning Suppose a solution contains 0.23 M Pb2 and 0.45 M A3t. Calculate the pH range that would allow Al(OH)3 to precipitate but not Pb(OH)2. The Ksp values for Al(OH)s and Pb(O) can be found here Policies: Hom minimum maxiumum You can check your ansv You can view solutions v give up on any question You can keep trying to a until you get it right or g You lose 5% of the point answer in your question attempt at that answer Number Number pH

Explanation / Answer

Ksp data

Pb(OH)2 1.43×10–20; Al(OH)3 4.6×10–33

[OH-] required for Al3+:Ksp = [Al+3][OH-]^3

[OH-] = (Ksp/[Al+3])^(1/3)

[OH-] = ((4.6*10^-33)/(0.45))^(1/3) = 2.170*10^-11

pOH = -log( 2.170*10^-11 = 10.66

pH = 14-10.66

pH = 3.34

for Pb:

[OH-] required for Pb+2 +:Ksp = [Pb+2][OH-]^2

[OH-] = (Ksp/[PB2+])^(1/2) = ((1.43*10^-20)/0.23))^0.5 = 2.4934*10^-10

pOH = -log(2.4934*10^-10 ) = 9.60

pH = 14-9.60

pH = 4.4

therefore, maintain pH:

3.4 < pH < 4.4

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