Problem 44 - Chemistry and Chemical Reactivity

Question

Equal molar quantities of ammonia and sodium dihydrogen phosphate \(\left(\mathrm{NaH}_{2} \mathrm{PO}_{4}\right)\) are mixed. (a) Write a balanced, net ionic equation for the acid-base reaction that can, in principle, occur. (b) Does the equilibrium lie to the right or left?

Step-by-Step Solution

Answer

Net ionic equation: \\(\text{NH}_3+\text{H}_2\text{PO}_4^- \rightarrow \text{NH}_4^+ + \text{HPO}_4^{2-}\\); equilibrium lies to the left.

Step 1: Identify the Species Present in Solution

When equal molar quantities of ammonia \(\text{(NH}_3\) and sodium dihydrogen phosphate \(\text{(NaH}_2\text{PO}_4)\) are mixed, they exist as ions in solution. Ammonia contributes \(\text{NH}_3\) and sodium dihydrogen phosphate dissociates to give \(\text{H}_2\text{PO}_4^-\). Thus, the reaction considered will be between \(\text{NH}_3\) and \(\text{H}_2\text{PO}_4^-\).

Step 2: Write the Initial Equation for the Reaction

Ammonia, \(\text{NH}_3\), can act as a base to accept a proton and form \(\text{NH}_4^+\). \(\text{H}_2\text{PO}_4^-\) can donate a proton to become \(\text{HPO}_4^{2-}\). Thus, the initial reaction equation is: \[\text{NH}_3+\text{H}_2\text{PO}_4^- \rightarrow \text{NH}_4^+ + \text{HPO}_4^{2-}.\]

Step 3: Write the Balanced Net Ionic Equation

To form the net ionic equation, identify the actual chemical change, which excludes spectator ions. Here, there are no spectator ions, so the net ionic equation remains: \[\text{NH}_3+\text{H}_2\text{PO}_4^- \rightarrow \text{NH}_4^+ + \text{HPO}_4^{2-}.\] This equation is already balanced.

Step 4: Determine the Direction of Equilibrium

To determine the direction of the equilibrium, compare the strength of the acids and bases on each side. The \(\text{H}_2\text{PO}_4^-\) is a stronger acid compared to \(\text{NH}_4^+\), and \(\text{NH}_3\) is a weaker base compared to \(\text{HPO}_4^{2-}\). Therefore, the equilibrium will favor the weaker acid and base, lying to the left.

Key Concepts

Understanding the Net Ionic Equation

When dealing with an acid-base reaction, it is crucial to focus on the net ionic equation. This equation shows only the particles that participate in the reaction and excludes spectator ions.
In the reaction between ammonia \(\text{(NH}_3\) and sodium dihydrogen phosphate \(\text{(NaH}_2\text{PO}_4)\), both compounds exist as ions in solution. Ammonia acts as a base, picking up a proton from \(\text{H}_2\text{PO}_4^-\) to form \(\text{NH}_4^+\). At the same time, \(\text{H}_2\text{PO}_4^-\) donates this proton to become \(\text{HPO}_4^{2-}\).
The net ionic equation for this process is:
\[\text{NH}_3 + \text{H}_2\text{PO}_4^- \rightarrow \text{NH}_4^+ + \text{HPO}_4^{2-}.\]
This equation captures the essence of the interaction, illustrating how the acid-base process occurs without any unnecessary details.

The Role of Chemical Equilibrium

Chemical equilibrium is a concept where the forward and reverse reactions occur at the same rate, leading to a stable mixture of reactants and products.
In our ammonia and sodium dihydrogen phosphate reaction, determining the equilibrium position helps us understand which side of the reaction is favored.

• If the products are favored, the equilibrium lies to the right.
• If the reactants are favored, it lies to the left.

In this case, we compare \(\text{NH}_3\) with \(\text{HPO}_4^{2-}\) and \(\text{H}_2\text{PO}_4^-\) with \(\text{NH}_4^+\). The stronger acid \(\text{H}_2\text{PO}_4^-\) compared to \(\text{NH}_4^+\), and weaker base \(\text{NH}_3\) over \(\text{HPO}_4^{2-}\), suggest an equilibrium favoring reactants.
Therefore, the equilibrium position is to the left, meaning more reactants than products are present at equilibrium.

Exploring Acid and Base Strength

The strength of acids and bases influences the direction of an equilibrium reaction. In acid-base reactions like the one involving \(\text{NH}_3\) and \(\text{H}_2\text{PO}_4^-\), the stronger acid and weaker base can sway the equilibrium.

• Acid Strength: \(\text{H}_2\text{PO}_4^-\) is a stronger acid than \(\text{NH}_4^+\). It can donate a proton more readily, pushing the reaction backward to form more of the reactants.
• Base Strength: \(\text{NH}_3\) is a weaker base compared to \(\text{HPO}_4^{2-}\), meaning it’s less likely to accept protons.

The combination of a stronger acid on the reactants' side and a weaker base on this side means that the original reactants are more stable. As a result, equilibrium shifts to retain more reactants than products, hence the equilibrium mostly lies to the left in this scenario.