September 20, 2020, 04:04:39 PM
Forum Rules: Read This Before Posting

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

0 Members and 1 Guest are viewing this topic.

#### Edher

• Guest
##### [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

#### Borek

• Mr. pH
• Deity Member
• Posts: 25957
• Mole Snacks: +1697/-402
• Gender:
• I am known to be occasionally wrong.
##### 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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### AWK

• Retired Staff
• Sr. Member
• Posts: 7522
• Mole Snacks: +525/-88
• Gender:
##### 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 %.