# chemistry

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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 -

100 mL x 0.015M = 1.5 millimoles K2CO3.
100 mL x 0.0075M = 0.75 mmoles BaBr2.
.........BaBr2 + K2CO3 ==>BaCO3 + 2KBr
initial..0.75.....1.5.......0......0
change..-0.75....-0.75....+0.75..+1.5
equil.....0.......0.75.....0.75...1.5

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 -

• chemistry -

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 -

actually, that's what it's asking me to do?

• chemistry -

5.6e-9=x^2?

• chemistry -

**5.1

• chemistry -

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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