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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: hannah2329 on May 13, 2012, 06:00:08 PM

Title: molar solubility
Post by: hannah2329 on May 13, 2012, 06:00:08 PM

Calculate the molar solubility of Cr(OH)3 in 0.54 M NaOH. Kf for Cr(OH)4- is 8x10^29 please help with this problem! ive tried everything!

Title: Re: molar solubility
Post by: JustinCh3m on May 14, 2012, 12:00:05 AM

Kf = [Cr3+][3(OH)]^3

Kf = (x) (3x)^3

this should get you started...

Title: Re: molar solubility
Post by: hannah2329 on May 14, 2012, 09:39:11 AM

so do I plug the .54 in there and make it 8x10^29=(x)(.54-3x)^3??

Title: Re: molar solubility
Post by: AWK on May 14, 2012, 02:21:04 PM

http://answers.yahoo.com/question/index?qid=20080811093541AAqlWyO