Problem: What is the molar solubility of Pb(OH)2 in a 0.138-M solution of CH3NH2?

🤓 Based on our data, we think this question is relevant for Professor Dixon's class at UCF.

FREE Expert Solution

CH3NH2 → weak base

Step 1: Calculate OH in the solution


Kb of CH3NH2 = 4.4×10‒4


Kb=[CH3NH3+][OH-][CH3NH2]Kb=xx0.138-xKb=x20.138-x

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4.4×10-4=x20.138-x4.4×10-4(0.138-x)=x26.072×10-5-4.4×10-4x=x20=x2+4.4×10-4x-6.072×10-5


x=-b±b2-4ac2a

x=-(4.4×10-4)±(4.4×10-4)2-4(1)(-6.072×10-5)2(1)x=-4.4×10-4±0.015592

x=-4.4×10-4+0.015592

x = 7.5754 × 10–3

x=-4.4×10-4-0.015592

x = –8.0154 × 10–3


x = [OH-]

[OH-] = 7.5754 × 10–3 M


Step 2:

Pb(OH)2(s)  Pb2+(aq) + 2 OH(aq)

Ksp = 1.43×10‒20

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Problem Details

What is the molar solubility of Pb(OH)2 in a 0.138-M solution of CH3NH2?

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Based on our data, we think this problem is relevant for Professor Dixon's class at UCF.

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Our data indicates that this problem or a close variation was asked in Chemistry - OpenStax 2015th Edition. You can also practice Chemistry - OpenStax 2015th Edition practice problems.