CHEm help please

Question

A solution contains one or more of the following ions: Ag+, Ca2+, and Co2+. Lithium bromide is added to the solution and no precipitate forms. An excess of lithium sulfate is then added to the solution and a precipitate forms. The precipitate is filtered off and lithium phosphate is added to the remaining solution, producing a precipitate. Which ions are present in the original solution? Ca2+ Ag+ Co2+

Write a net ionic equation for the formation of the precipitate observed after the addition of lithium sulfate. Include physical states.

net ionic equation:

Write a net ionic equation for the formation of the precipitate observed after the addition of lithium phosphate. Include physical states.

net ionic equation:

Answer 1

Ca2+ and Co2+ are present. Ag+ is absent.

Here is how each test pins one ion down. The key idea is that a test which produces no precipitate is just as informative as one that does - it rules something out.

Test 1: lithium bromide added, no precipitate

Bromides are soluble except those of silver, lead and mercury(I). If Ag+ were in solution, AgBr would have dropped out immediately as a pale yellow solid. Nothing formed, so Ag+ is absent. CaBr2 and CoBr2 are both soluble, so this test says nothing about the other two ions.

Test 2: excess lithium sulfate added, precipitate forms

Sulfates are soluble except those of barium, lead, and calcium and strontium (which are only slightly soluble). CoSO4 is soluble, and Ag+ is already eliminated, so the only candidate left is calcium. Ca2+ is present.

Notice the word excess. That is deliberate - it drives the calcium out as completely as possible so it will not interfere with the next test.

Test 3: filter, then add lithium phosphate, precipitate forms

Phosphates are insoluble except those of the group 1 metals and ammonium. The calcium has been filtered away as CaSO4, and Ag+ was never there, so the ion producing this precipitate has to be cobalt. Co2+ is present.

Net ionic equations

Strip out the spectator ions - here that is Li+ every time, along with the bromide that never reacted.

After lithium sulfate:

Ca2+(aq) + SO4^2-(aq) -> CaSO4(s)

After lithium phosphate:

3 Co2+(aq) + 2 PO4^3-(aq) -> Co3(PO4)2(s)

Check that second one carefully, because it is where most of the marks are lost. Charge must balance as well as atoms: 3(+2) = +6 on the left from cobalt, 2(-3) = -6 from phosphate, net zero, matching the neutral solid. Cobalt(II) and phosphate need a 3:2 ratio for exactly that reason.

The habit to take from this

Qualitative analysis schemes are built so that each reagent removes one ion and clears the field for the next test, which is why the order matters and why the filtration step is not optional. When you work one of these, keep a running list of the three ions and mark each as confirmed, eliminated, or still unknown after every step. Doing that turns what looks like a puzzle into simple bookkeeping against a solubility table.