# For the reaction (NH_4)_2SO_4(aq) + H_2SO_4(aq) \to ?, write: (1) a balanced chemical equation...

## Question:

For the reaction {eq}\displaystyle (NH_4)_2SO_4(aq) + H_2SO_4(aq) \to \ ? {/eq}, write:

(1) a balanced chemical equation (molecular equation),

(2) a total ionic equation, and

(3) a net ionic equation. Include the appropriate phase for each species.

## Ionic Equations:

An ionic equation is a tool to express a chemical reaction in terms of how the compounds in an aqueous solution will dissociate into individual ions. A total ionic equation will include all of the ions present in solution, while a net ionic equation only includes the ions that were involved in the chemical reaction and experienced a change. The ions that were unchanged in a chemical reaction are referred to as spectator ions and are omitted from the net ionic equation.

### (1) Write a balanced chemical equation

When ammonium sulfate ({eq}(NH_4)_2SO_4 {/eq}) reacts with sulfuric acid ({eq}H_2SO_4 {/eq}), they undergo a synthesis reaction to form ammonium bisulfate ({eq}NH_4HSO_4 {/eq}), which is highly soluble in water.

The unbalanced chemical equation is:

{eq}(NH_4)_2SO_4 (aq) + H_2SO_4 (aq) \rightarrow NH_4HSO_4 (aq) {/eq}

We can balance this equation by counting the atoms of each element on each side of the chemical reaction.

 Reactants Products N = 1x2 = 2 N = 1 H = 4x2 + 2 = 10 H = 4 + 1 = 5 S = 1 + 1 = 2 S = 1 O = 4 + 4 = 8 O = 4

We can see that each element has twice as many atoms present in the reactants compared to the products. Since we only have one product, adding a coefficient of 2 in front of it will balance the entire equation.

The balanced chemical equation is {eq}(NH_4)_2SO_4 (aq) + H_2SO_4 (aq) \rightarrow 2 \ NH_4HSO_4 (aq) {/eq}

### (2) Write a total ionic equation

To write a total ionic equation, we need to break all aqueous reactants and products into the individual ions they would dissolve into in water. Be careful to keep the total ionic equation balanced by including the subscripts and coefficients.

For the reactants, {eq}(NH_4)_2SO_4 (aq) {/eq} breaks down into {eq}2 \ NH_4^{+} (aq) + SO_4^{2-} (aq) {/eq} and {eq}H_2SO_4 (aq) {/eq} dissolves into {eq}2 \ H^{+} (aq) + SO_4^{2-} (aq) {/eq}.

For the products, {eq}2 \ NH_4HSO_4 (aq) {/eq} becomes {eq}2 \ NH_4^{+} (aq) + 2 \ HSO_4^{-} (aq) {/eq}.

The total ionic equation is {eq}2 \ NH_4^{+} (aq) + 2 \ SO_4^{2-} (aq) + 2 \ H^{+} (aq) \rightarrow 2 \ NH_4^{+} (aq) + 2 \ HSO_4^{-} (aq) {/eq}

### (3) Write a net ionic equation

To transition from a total ionic equation to a net ionic equation, we can cancel out any ions that are present in both the reactants and the products. These ions are called spectator ions since they are not changed during the course of the chemical reactant. They are present during the reaction, but not actively involved in the chemical reaction. In this case, we can eliminate {eq}2 \ NH_4^{+} (aq) {/eq} from both sides of the equation.

The net ionic equation is {eq}2 \ SO_4^{2-} (aq) + 2 \ H^{+} (aq) \rightarrow 2 \ HSO_4^{-} (aq) {/eq}