Posted by dude on Monday, October 24, 2011 at 8:25pm.
What is the ion in the resulting solution?
100.0 mL of 0.015 M K2CO3 and 100.0 mL of 0.0075 M BaBr2?
25 mL of 0.57 M Ni(NO3)2 and 100.0 mL of 0.150 M NaOH
- chemistry - DrBob222, Monday, October 24, 2011 at 9:20pm
100 mL x 0.015M = 1.5 millimoles K2CO3.
100 mL x 0.0075M = 0.75 mmoles BaBr2.
.........BaBr2 + K2CO3 ==>BaCO3 + 2KBr
BaCO3 is a ppt. KBr is soluble; you will have K^+ and Br^- from the KBr. Also you will have some of the K2CO3 that is unreacted and that is soluble to provide K^+ and CO3^2-.
Concn K^+ = (1.5 + 1.5)mmoles/200 mL
Concn CO3^2- = 0.75 mmoles/200 mL
Concn Br^- = 1.5 mmoles/200 mL
- chemistry - dude, Monday, October 24, 2011 at 9:31pm
what about Ba?
- chemistry - DrBob222, Monday, October 24, 2011 at 9:52pm
All of the Ba is ppted as the carbonate. You can calculate the concn of Ba^2+ from the Ksp of BaCO3 (it will be VERY small) but I don't think that is the intent of the problem.
- chemistry - dude, Monday, October 24, 2011 at 10:15pm
actually, that's what it's asking me to do?
- chemistry - dude, Monday, October 24, 2011 at 10:16pm
- chemistry - dude, Monday, October 24, 2011 at 10:17pm
- chemistry - DrBob222, Monday, October 24, 2011 at 11:13pm
Not quite. The carbonate ion acts as a common ion so
(Ba^2+) = Ksp/(CO3^2-). I think I gave the (CO3^2-) in my first response but you may want to check it to make sure it is correct. If it's concn you want the above will get it for you; if it's grams you want watch it since the volume is not 1L but 200 mL.
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