Question

Question #5b2a9

Answer 1

B. Magnesium

Explanation:

The trick here is the to find the ion which reacts to form a insoluble compounds.

SOLUBILITY RULES

1. All common compounds of `"NH"_4^+` and the Group 1 elements are soluble.

2. `"NO"_3^"-", "ClO"_3^"-", "ClO"_4^"-", "C"_2"H"_3"O"_2^"-"` — all common nitrates, chlorates, perchlorates, and acetates are soluble.

3. `"F"^"-", "Cl"^"-", "Br"^"-", "I"^"-"` — all halides are soluble except those of `"Ag"^+, "Hg"_2^"2+", "Pb"^"2+"` and the fluorides of `"Mg"^"2+", "Ca"^"2+", "Sr"^"2+"`, and`"Ba"^"2+"`.

4. `"SO"_4^"2-"` — most sulfates are soluble EXCEPT those of `"Sr"^"2+", "Ba"^"2+", "Ca"^"2+", "Pb"^"2+"`, `"Hg"_2^"2+"`, and `"Hg"^"2+"`.

5. `"CO"_3^"2-", "C"_2"O"_4^"2-", "OH"^"-", "O"^"2-", "SO"_3^"2-", "PO"_4^"3-", "CrO"_4^"2-", "S"^"2-"` — all carbonates, oxalates, hydroxides, oxides, phosphates, chromates, and sulfides are insoluble. Remember the first rule. all the Group 1 elements are soluble independent of their ion attached to them.

So if you do a reaction with `Cl^-)`ion.

`"NaCO"_3 + "Cl"^-) rightleftharpoons "NaCl" + "CO3"^(2-)`

So NaCl is soluble as we know

`K_(sp) "of NaCl" = 39.4`

Doing a reaction with potassium ion.

`"NaCO"_3 + "Pb"^+ rightleftharpoons "PbCO"_3 + "Na"^+`

Recall the first rule

All common compounds of the Group 1 elements are soluble.

Doing a reaction with `SO_4^(-2)`

`"NaCO"_3 + "SO"_4^(-2) rightleftharpoons "NaSO"_4 +"CO3"^(2-)`

Recall the 4th rule.
`"SO"_4^"2-"` — most sulfates are soluble EXCEPT those of `"Sr"^"2+", "Ba"^"2+", "Ca"^"2+", "Pb"^"2+"`, `"Hg"_2^"2+"`, and `"Hg"^"2+"`.

Thus `NaSO_4` is soluble

Doing a reaction with magnesium

`"NaCO"_3 + "Mg"^+ rightleftharpoons "MgCO3" + "Na"^+`

And that's your solution because `"MgCO"_3` is insoluble in water

Most carbonates are insoluble except the elements in group 1.

Magnesium is not in group 1` and thus it is insoluble

`K_(sp) "of MgCO"_3 = 1 xx 10^-7.8`