Can someone please explain to me how to do this problem?

Question

You dissolve 22g of calcium hydroxide in 220 mL of 1.23 M nickel (II) chloride. A nickel precipitate is formed. At equilibrium, [Ni2+] = 0.795 M and [OH-] = 1.83 M.

Find the mass of product formed, the concentrations of the ions at equilibrium, and the final pH.

Answer 1

Need a reaction:
Ca(OH)2 + NiCl2 -> Ni(OH)2 + CaCl2. To start, your Ca(OH)2 molarity = (22g/74g/mole)/0.22 L = 1.3 M. If there was no nickel, the OH- would be 2.6 M, but it is 1.83M, so the difference is in the ppt. Ni(OH)2. We know the Ni+2 and OH- at equilibrium, and that Ca+2 and Cl- would stay dissolved (hopefully). We also know that the Ni(OH)2 mass pptd = (1.23-.795)x95(appx) x 0.22L=
9 grams. The final pOH =log ( 10-1.83)=
-2+Log 1.7 = -1.75 appx. So pH= 15.75.

Note: I'd be surprised if you could really dissolve that much Ca(OH)2 in 220 mL water. But that's what the problem says.

Answer 2

The Ni in solution starts at 1.23M and ends up at .795M.
Therefore, .455moles went to the precipitate.
Figure out the molarity of the Ca(OH)2, remembering to double it as the OH is in solution as a single ion. See how many moles were used up since you still have 1.83M in solution.
See if all the chlorine that was originally there balances with the calcium in solution. If it does, your precipitate will be strictly Ni and OH. If it doesn't, you may have a complex salt.
This problem is strictly accounting. What you started with and where it ends up.