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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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JLeopard JLeopard
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10 years ago Edited: 10 years ago, JLeopard
Calculate the concentration of OH- in Ca(OH)2 solution
the concentration of Ca(OH)2  is 3.0*10^-7 M
the answer is 6.2*10^-7 M but I don't know the procedure please help me !
assume that the Ksp of  Ca(OH)2 is 5.5×10^–6.
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wrote...
#1
10 years ago
I think you start with Ca(OH)2 → Ca2+ + 2OH–

Then you use this relationship, where pH is 12.10

pH = 14 - pOH.
pOH = -log[OH-]

I don't know Face with Stuck-out Tongue
wrote...
#2
10 years ago
Here'a a similar question, but I'm not sure if it applies.

Similar question
Calculate the theoretical [OH–] in a saturated Ca(OH)2 solution. Use a literature value of Ksp = 6.5x 10-6.

Solution
Ca(OH)2 Leftwards Arrow---> Ca+2 + 2 OH-

We know that the Ksp = [Ca+2]([OH-]^2). We know that the concentration of the Ca will be half the concentration of the OH in the equilibrium. [Ca] = x, [OH-] = 2x

6.5*10^-6 = x * 4x^2
6.5*10^-6 = 4x^3
1.625*10^-6 = x^3
x = .012 M

[OH-] = 2x = .024 M

ANSWER: Theoretical [OH-] = .024 M
Source  http://www.chemteam.info/Equilibrium/calc-pH-from-Ksp.html
JLeopard Author
wrote...
#3
10 years ago
don't we need to consider the initial concentration of Ca(OH)2 ?
And 1.0*10^-7 M OH- dissociate from water
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