After equal volume of `0.10 M` solutions of `(NH_(4))_(2)SO_(4)` and `Ba(OH)_(2)` have been mixed, which of the following species is present in greatest concentration in solution?

Options

A. `NH_(4)^(+) (aq)`
B. `Ba^(+2) (aq)`
C. `NH_(3) (aq)`
D. `BaSO_(4) (aq)`

Answer

To solve the problem, we need to analyze the reaction that occurs when equal volumes of `0.10 M` solutions of `(NH₄)₂SO₄` and `Ba(OH)₂` are mixed.

### Step 1: Determine the initial concentrations and volumes
Given that both solutions are `0.10 M` and we are mixing equal volumes, we can assume we are mixing `1 L` of each solution for simplicity.

- Moles of `(NH₄)₂SO₄` = Molarity × Volume = `0.10 mol/L × 1 L = 0.10 moles`
- Moles of `Ba(OH)₂` = Molarity × Volume = `0.10 mol/L × 1 L = 0.10 moles`

### Step 2: Write the balanced chemical reaction
The reaction between ammonium sulfate and barium hydroxide can be written as:

\[ (NH₄)₂SO₄ + Ba(OH)₂ \rightarrow BaSO₄ + 2NH₄OH \]

### Step 3: Determine the limiting reactant
From the balanced equation, we see that `1 mole` of `(NH₄)₂SO₄` reacts with `1 mole` of `Ba(OH)₂` to produce `1 mole` of `BaSO₄` and `2 moles` of `NH₄OH`.

Since we have `0.10 moles` of each reactant, both will completely react with each other, and we will have:

- `0.10 moles` of `(NH₄)₂SO₄` will produce `0.10 moles` of `BaSO₄` and `0.20 moles` of `NH₄OH`.

### Step 4: Calculate the concentrations after the reaction
After the reaction, the total volume of the solution will be `2 L` (1 L from each solution).

- Concentration of `BaSO₄` = Moles / Total Volume = `0.10 moles / 2 L = 0.05 M`
- Concentration of `NH₄OH` = Moles / Total Volume = `0.20 moles / 2 L = 0.10 M`

### Step 5: Identify the species present in greatest concentration
Among the species present after the reaction, we have:
- `BaSO₄` at `0.05 M`
- `NH₄OH` at `0.10 M`

Thus, the species present in the greatest concentration is `NH₄OH`.

### Final Answer
The species present in greatest concentration in the solution is `NH₄OH`.

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