Answered

5. Next, similar to the table above, write the formula of the ionic compound formed from each pair of cations and anions shown below (see Aluminum sulfite as an example below):

\begin{tabular}{|l|l|l|l|l|l|}
\hline
& [tex]$Br ^{-}$[/tex] & [tex]$OH ^{-}$[/tex] & [tex]$HCO _3^{-}$[/tex] & [tex]$SO _3^{2-}$[/tex] & [tex]$PO _4^{3-}$[/tex] \\
\hline
[tex]$Na ^{+}$[/tex] & & & & & \\
\hline
[tex]$Ca ^{2+}$[/tex] & & & & & \\
\hline
[tex]$Al ^{3+}$[/tex] & & & & & \\
\hline
\end{tabular}

6. Using the table below (list of anions), write out the name of each of the following compounds:
a. [tex]$Na _2 CO _3$[/tex]



Answer :

GinnyAnswer
### Problem Statement
We need to write the formula for the ionic compounds formed from each given pair of cations and anions, and then name the given compound [tex]\( \text{Na}_2 \text{CO}_3 \)[/tex].

### Solution
To create the formulas for ionic compounds, we need to balance the charges. The charge of the cations and anions should balance each other so that the total charge of the compound is zero. Here is the step-by-step solution for each part:

#### Part 5: Ionic Compound Formulation

Given cations and anions:
- Cations (Positive Ions):
- [tex]\( \text{Na}^+ \)[/tex] (Sodium ion)
- [tex]\( \text{Ca}^{2+} \)[/tex] (Calcium ion)
- [tex]\( \text{Al}^{3+} \)[/tex] (Aluminum ion)

- Anions (Negative Ions):
- [tex]\( \text{Br}^- \)[/tex] (Bromide ion)
- [tex]\( \text{OH}^- \)[/tex] (Hydroxide ion)
- [tex]\( \text{HCO}_3^- \)[/tex] (Bicarbonate ion)
- [tex]\( \text{SO}_3^{2-} \)[/tex] (Sulfite ion)
- [tex]\( \text{PO}_4^{3-} \)[/tex] (Phosphate ion)

Let's fill in the table as requested:

\begin{tabular}{|l|l|l|l|l|l|}
\hline & [tex]$Br^-$[/tex] & [tex]$OH^-$[/tex] & [tex]$HCO_3^-$[/tex] & [tex]$SO_3^{2-}$[/tex] & [tex]$PO_4^{3-}$[/tex] \\
\hline [tex]$Na^+$[/tex] & NaBr & NaOH & NaHCO_3 & Na_2SO_3 & Na_3PO_4 \\
\hline [tex]$Ca^{2+}$[/tex] & CaBr_2 & Ca(OH)_2 & Ca(HCO_3)_2 & CaSO_3 & Ca_3(PO_4)_2 \\
\hline [tex]$Al^{3+}$[/tex] & AlBr_3 & Al(OH)_3 & Al(HCO_3)_3 & Al_2(SO_3)_3 & AlPO_4 \\
\hline
\end{tabular}

### Explanation of the Formulations:
- Sodium (Na) with charge [tex]\( +1 \)[/tex]:
- [tex]\( \ce{Na^+ + Br^- -> NaBr} \)[/tex] (Each ion is 1+, 1-)
- [tex]\( \ce{Na^+ + OH^- -> NaOH} \)[/tex] (Each ion is 1+, 1-)
- [tex]\( \ce{Na^+ + HCO3^- -> NaHCO3} \)[/tex] (Each ion is 1+, 1-)
- [tex]\( \ce{2 \ Na^+ + SO3^{2-} -> Na2SO3} \)[/tex] (2 atoms of Na+ needed to balance 2-)
- [tex]\( \ce{3 \ Na^+ + PO4^{3-} -> Na3PO4} \)[/tex] (3 atoms of Na+ needed to balance 3-)

- Calcium (Ca) with charge [tex]\( +2 \)[/tex]:
- [tex]\( \ce{Ca^{2+} + 2 \ Br^- -> CaBr2} \)[/tex] (2 ions of Br- needed to balance 2+)
- [tex]\( \ce{Ca^{2+} + 2 \ OH^- -> Ca(OH)2} \)[/tex] (2 ions of OH- needed to balance 2+)
- [tex]\( \ce{Ca^{2+} + 2 \ HCO3^- -> Ca(HCO3)2} \)[/tex] (2 ions of HCO3- needed to balance 2+)
- [tex]\( \ce{Ca^{2+} + SO3^{2-} -> CaSO3} \)[/tex] (Each ion is 2+, 2-)
- [tex]\( \ce{3 \ Ca^{2+} + 2 \ PO4^{3-} -> Ca3(PO4)2} \)[/tex] (3 atoms of Ca2+ needed to balance 6+, and 2 atoms of PO4 3- needed to balance 6-)

- Aluminum (Al) with charge [tex]\( +3 \)[/tex]:
- [tex]\( \ce{Al^{3+} + 3 \ Br^- -> AlBr3} \)[/tex] (3 ions of Br- needed to balance 3+)
- [tex]\( \ce{Al^{3+} + 3 \ OH^- -> Al(OH)3} \)[/tex] (3 ions of OH- needed to balance 3+)
- [tex]\( \ce{Al^{3+} + 3 \ HCO3^- -> Al(HCO3)3} \)[/tex] (3 ions of HCO3- needed to balance 3+)
- [tex]\( \ce{2 \ Al^{3+} + 3 \ SO3^{2-} -> Al2(SO3)3} \)[/tex] (2 atoms of Al3+ needed to balance 6+, and 3 atoms of SO3 2- needed to balance 6-)
- [tex]\( \ce{Al^{3+} + PO4^{3-} -> AlPO4} \)[/tex] (Each ion is 3+, 3-)

#### Part 6: Writing the Name of the Given Compound
Given compound:
- [tex]\( \text{Na}_2 \text{CO}_3 \)[/tex]

CO₃²⁻ is the carbonate ion. The name of the compound is "Sodium Carbonate". This is derived as follows:
- "Na" is the symbol for Sodium.
- "CO₃" is the symbol for Carbonate.

So the compound [tex]\( \text{Na}_2 \text{CO}_3 \)[/tex] is named Sodium Carbonate.