Question #5b2a9

1 Answer
May 5, 2017

Answer:

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#