Question #5b2a9

1 Answer
May 5, 2017

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