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Topic: [OH-] after precipitation  (Read 3504 times)

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Edher

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[OH-] after precipitation
« on: February 28, 2005, 06:31:39 PM »
What [OH-] should be maintained in a solution if, after precipitation of Mg+2 as
Mg(OH)2, the remaining Mg+2 is to be at a level of 1 microgram Mg2+ / L?
Ksp = 1.8x10^-11.

I found the initial concentration of OH- to be
After I did stoicheometry I foud that 1 microgram Mg2+ / L is tantamount to 4.11x10^-8 moles.

What exactly do I do with this result? Am I supposed to add it to the initial [OH-]
that's what I did and I didn't get the right answer. According to the book, the answer should be 0.02M.

Edher

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Re:[OH-] after precipitation
« Reply #1 on: February 28, 2005, 08:25:05 PM »
You have the definition of Qsp:

Qsp = [Mg2+ ][OH- ]2

You know what final concentration (M) of Mg2+ is to be.

Simply rearrange above definition and you will find what concentration of OH- is necessary to keep Mg2+ concentration below given treshold.
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Offline AWK

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Re:[OH-] after precipitation
« Reply #2 on: March 01, 2005, 03:12:20 AM »
Quote
After I did stoicheometry I foud that 1 microgram Mg2+ / L is tantamount to 4.11x10^-8 moles.

Should be 4.17x10^-8

Correction added later - this number is for rounded mass of Mg to 24 - this causes error of 1.3 %.
 Your result is correct!
« Last Edit: March 01, 2005, 11:00:21 AM by AWK »
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