Problem 63 - Chemistry: The Central Science

Question

Complete the table by filling in the formula for the ionic compound formed by each pair of cations and anions, as shown for the first pair. $$ \begin{array}{l|l|l|l|l} \hline \text { Ion } & \mathrm{K}^{+} & \mathrm{NH}_{4}^{+} & \mathrm{Mg}^{2+} & \mathrm{Fe}^{3+} \\ \hline \mathrm{Cl}^{-} & \mathrm{KCl} & & & \\ \mathrm{OH}^{-} & & & & \\ \mathrm{CO}_{3}^{2-} & & & & \\ \mathrm{PO}_{4}^{3-} & & & & \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The completed table is: $$ \begin{array}{l|l|l|l|l} \hline \text{Ion} & \text{K}^{+} & \text{NH}_4^+ & \text{Mg}^{2+} & \text{Fe}^{3+} \\ \hline \text{Cl}^{-} & \text{KCl} & \text{NH}_4\text{Cl} & \text{MgCl}_2 & \text{FeCl}_3 \\ \text{OH}^{-} & \text{KOH} & \text{NH}_4\text{OH} & \text{Mg(OH)}_2 & \text{Fe(OH)}_3 \\ \text{CO}_{3}^{2-} & \text{K}_2\text{CO}_3 & \text{(NH}_4\text{)_2CO}_3 & \text{MgCO}_3 & \text{Fe}_2(\text{CO}_3)_3 \\ \text{PO}_{4}^{3-} & \text{K}_3\text{PO}_4 & \text{(NH}_4\text{)_3PO}_4 & \text{Mg}_3(\text{PO}_4)_2 & \text{FePO}_4 \\ \hline \end{array} $$

Step 1: Identify the charges in each pair

For each pair representing a cation and an anion, identify the charges on the ions. The charges can be seen in the table.

Step 2: Balance the charges to form an ionic compound

Determine the ratio of cations to anions in the compound to ensure that the charges are balanced. For example, if a cation has a charge of +2 and an anion has a charge of -1, their ratio should be 1:2 to form a neutral compound.

Step 3: Write the formula of the ionic compound

Write the formula using the balanced ratio of cations to anions. Use subscripts to indicate the numbers of each ion in the formula. Now to complete the table: \(NH_4^+\) with \(Cl^-\) will form \(\text{NH}_4\text{Cl}\). \(Mg^{2+}\) with \(Cl^-\) will form \(\text{MgCl}_2\). \(Fe^{3+}\) with \(Cl^-\) will form \(\text{FeCl}_3\). \(K^{+}\) with \(OH^-\) will form \(\text{KOH}\). \(NH_4^+\) with \(OH^-\) will form \(\text{NH}_4\text{OH}\). \(Mg^{2+}\) with \(OH^-\) will form \(\text{Mg(OH)}_2\). \(Fe^{3+}\) with \(OH^-\) will form \(\text{Fe(OH)}_3\). \(K^{+}\) with \(CO_3^{2-}\) will form \(\text{K}_2\text{CO}_3\). \(NH_4^+\) with \(CO_3^{2-}\) will form \(\text{(NH}_4\text{)_2CO}_3\). \(Mg^{2+}\) with \(CO_3^{2-}\) will form \(\text{MgCO}_3\). \(Fe^{3+}\) with \(CO_3^{2-}\) will form \(\text{Fe}_2(\text{CO}_3)_3\). \(K^{+}\) with \(PO_4^{3-}\) will form \(\text{K}_3\text{PO}_4\). \(NH_4^+\) with \(PO_4^{3-}\) will form \(\text{(NH}_4\text{)_3PO}_4\). \(Mg^{2+}\) with \(PO_4^{3-}\) will form \(\text{Mg}_3(\text{PO}_4)_2\). \(Fe^{3+}\) with \(PO_4^{3-}\) will form \(\text{FePO}_4\). So the completed table is: $$ \begin{array}{l|l|l|l|l} \hline \text{Ion} & \text{K}^{+} & \text{NH}_4^+ & \text{Mg}^{2+} & \text{Fe}^{3+} \\ \hline \text{Cl}^{-} & \text{KCl} & \text{NH}_4\text{Cl} & \text{MgCl}_2 & \text{FeCl}_3 \\ \text{OH}^{-} & \text{KOH} & \text{NH}_4\text{OH} & \text{Mg(OH)}_2 & \text{Fe(OH)}_3 \\ \text{CO}_{3}^{2-} & \text{K}_2\text{CO}_3 & \text{(NH}_4\text{)_2CO}_3 & \text{MgCO}_3 & \text{Fe}_2(\text{CO}_3)_3 \\ \text{PO}_{4}^{3-} & \text{K}_3\text{PO}_4 & \text{(NH}_4\text{)_3PO}_4 & \text{Mg}_3(\text{PO}_4)_2 & \text{FePO}_4 \\ \hline \end{array} $$