Concentration of Ions in Ba(OH)2 + NiSO4 Reaction

[KD]

While waiting for my other post to be moved, I'm putting this question up again.
25.00 mL of .1857 M Ba(OH)2 reacts with 30.00mL of .2228M NiSO4. Find the concentration of all ions after the reaction assuming that final volume is the sum of the initial volumes.
Ba(OH)2 + NiSO4 -> BaSO4 + Ni(OH)2
The answer is .03629M Ni2+ and .03629M SO42- [Ba2+] = [OH-] is about zero. How are these answers found? How would the concentrations change after the reaction? Please help. No specifics are needed just a general guide to figure this out. I may regret saying that.

 

[Borek]

What will precipitate, effectively removing ions from the solution?

 

[GCT]

Show your work

-First find the initial concentrations.

-Use the reaction equation to account for stoichiometry, but first you'll need to consider the solubility constant of ____

-Using the solubility constant equation, find the final concentrations.

 

[KD]

Okay, I get it now. Thanks for your help. To solve this I set up an ICE chart. Since Ba(OH)2 is the limiting reactant, all the moles of that will be used. So I would subtract those moles from the moles of the NiSO4 and divide that by .055 L (the total volume) to get the concentration. I don't have my book right in front of me, but I'm pretty sure that is what I did.