The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that [tex]$NH _3$[/tex] is a weak base.
| 0.9 mol of HBr is added to 1.0 L of a [tex]$0.9 M NH _3$[/tex] solution. | acids: $\square$ bases: $\square$ other: $\square$ |
| 0.41 mol of KOH is added to 1.0 L of a solution that is 1.5 M in both [tex]$NH _3$[/tex] and [tex]$NH _4 Br$[/tex]. | acids: $\square$ bases: $\square$ other: $\square$ |
Answer (1)
In the first scenario, HBr reacts completely with N H 3 to form N H 4 + and B r − , where N H 4 + is an acid and B r − is neither.
In the second scenario, KOH reacts with N H 4 + to produce N H 3 , resulting in N H 3 , N H 4 + , K + , and B r − as major species, where N H 4 + is an acid, N H 3 is a base, and K + and B r − are neither.
Therefore, the final answer is: Acids: N H 4 + Bases: N H 3 Other: B r − , K +
In the first case, there is no base, so we leave it blank.
Explanation
Problem Analysis We are given two scenarios involving aqueous solutions and asked to identify the major species present at equilibrium, classifying them as acids, bases, or neither. We will analyze each scenario separately, considering the chemical reactions that occur.
First Scenario Analysis In the first scenario, 0.9 mol of HBr is added to 1.0 L of a 0.9 M N H 3 solution. HBr is a strong acid, and N H 3 is a weak base. They will react to form N H 4 + and B r − . The reaction is: H B r ( a q ) + N H 3 ( a q ) → N H 4 + ( a q ) + B r − ( a q ) Since we have 0.9 mol of HBr and 0.9 mol of N H 3 , the reaction goes to completion. Thus, the major species present at equilibrium are N H 4 + and B r − . N H 4 + is an acid, and B r − is neither an acid nor a base.
Second Scenario Analysis In the second scenario, 0.41 mol of KOH is added to 1.0 L of a solution that is 1.5 M in both N H 3 and N H 4 B r . KOH is a strong base. It will react with the acid N H 4 + to produce N H 3 and H 2 O . The reaction is: K O H ( a q ) + N H 4 + ( a q ) → N H 3 ( a q ) + K + ( a q ) + H 2 O ( l ) Initially, we have 1.5 mol of N H 3 and 1.5 mol of N H 4 + . After the reaction with 0.41 mol of KOH, the amount of N H 4 + will be 1.5 - 0.41 = 1.09 mol, and the amount of N H 3 will be 1.5 + 0.41 = 1.91 mol. The major species present will be N H 3 , N H 4 + , K + , and B r − . N H 4 + is an acid, N H 3 is a base, and K + and B r − are neither acids nor bases.
Final Answer Based on the analysis above, we can now fill in the table:
acids: □ bases: □ other: □
0.9 mol of HBr is added to 1.0 L of a 0.9 MN H 3 solution.
acids: N H 4 + bases: other: B r −
0.41 mol of KOH is added to 1.0 L of a solution that is 1.5 M in both N H 3 and N H 4 B r .
acids: N H 4 + bases: N H 3 other: K + , B r −
Examples
Understanding acid-base chemistry is crucial in many real-world applications. For example, in environmental science, it helps in monitoring and controlling the pH levels of water bodies to protect aquatic life. In medicine, it's essential for understanding how medications interact with the body's chemical environment. Also, in agriculture, managing soil pH is vital for optimal plant growth, ensuring efficient nutrient uptake.
