Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started 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!
-
Kf = [Cr3+][3(OH)]^3
Kf = (x) (3x)^3
this should get you started...
-
so do I plug the .54 in there and make it 8x10^29=(x)(.54-3x)^3??
-
http://answers.yahoo.com/question/index?qid=20080811093541AAqlWyO