Question 5b2a9

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 ${\text{NH}}_{4}^{+}$ and the Group 1 elements are soluble.

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

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

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

5. $\text{CO"_3^"2-", "C"_2"O"_4^"2-", "OH"^"-", "O"^"2-", "SO"_3^"2-", "PO"_4^"3-", "CrO"_4^"2-", "S"^"2-}$ — 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.

${\text{NaCO"_3 + "Cl"^-) rightleftharpoons "NaCl" + "CO3}}^{2 -}$

So NaCl is soluble as we know

${K}_{s p} \text{of NaCl} = 39.4$

Doing a reaction with potassium ion.

${\text{NaCO"_3 + "Pb"^+ rightleftharpoons "PbCO"_3 + "Na}}^{+}$

Recall the first rule

All common compounds of the Group 1 elements are soluble.

Doing a reaction with $S {O}_{4}^{- 2}$

${\text{NaCO"_3 + "SO"_4^(-2) rightleftharpoons "NaSO"_4 +"CO3}}^{2 -}$

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

Thus $N a S {O}_{4}$ is soluble

Doing a reaction with magnesium

${\text{NaCO"_3 + "Mg"^+ rightleftharpoons "MgCO3" + "Na}}^{+}$

And that's your solution because ${\text{MgCO}}_{3}$ 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

${K}_{s p} {\text{of MgCO}}_{3} = 1 \times {10}^{-} 7.8$