100 ml of `0.2"M H"_(2)SO_(4)` solution is mixed with 400ml of `0.05"M Ba"_(3)(PO_(4))_(2)`. Calculate the concentration of `[Ba^(+2)]` ion in resulting solution.

Options

A. `0.08` M
B. `0.04` M
C. `0.4` M
D. `0.8` M

Answer

To solve the problem, we need to calculate the concentration of the \( \text{Ba}^{2+} \) ions in the resulting solution after mixing the two solutions. Here are the steps to arrive at the solution:

### Step 1: Write the balanced chemical equation
When sulfuric acid (\( \text{H}_2\text{SO}_4 \)) reacts with barium phosphate (\( \text{Ba}_3(\text{PO}_4)_2 \)), the products are barium sulfate (\( \text{BaSO}_4 \)) and phosphoric acid (\( \text{H}_3\text{PO}_4 \)). The balanced equation is:

\[ 3 \text{Ba}_3(\text{PO}_4)_2 + 2 \text{H}_2\text{SO}_4 \rightarrow 3 \text{BaSO}_4 + 2 \text{H}_3\text{PO}_4 \]

### Step 2: Calculate the number of moles of \( \text{H}_2\text{SO}_4 \)
Using the formula:
\[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} \]
For \( \text{H}_2\text{SO}_4 \):
\[ \text{Volume} = 100 \, \text{ml} = 0.1 \, \text{L} \]
\[ \text{Moles of } \text{H}_2\text{SO}_4 = 0.2 \, \text{M} \times 0.1 \, \text{L} = 0.02 \, \text{mol} \]

### Step 3: Calculate the number of moles of \( \text{Ba}_3(\text{PO}_4)_2 \)
For \( \text{Ba}_3(\text{PO}_4)_2 \):
\[ \text{Volume} = 400 \, \text{ml} = 0.4 \, \text{L} \]
\[ \text{Moles of } \text{Ba}_3(\text{PO}_4)_2 = 0.05 \, \text{M} \times 0.4 \, \text{L} = 0.02 \, \text{mol} \]

### Step 4: Determine the limiting reagent
From the balanced equation, 2 moles of \( \text{H}_2\text{SO}_4 \) react with 1 mole of \( \text{Ba}_3(\text{PO}_4)_2 \). Thus, for 0.02 moles of \( \text{H}_2\text{SO}_4 \):
\[ \text{Required moles of } \text{Ba}_3(\text{PO}_4)_2 = \frac{0.02}{2} = 0.01 \, \text{mol} \]
Since we have 0.02 moles of \( \text{Ba}_3(\text{PO}_4)_2 \), \( \text{H}_2\text{SO}_4 \) is the limiting reagent.

### Step 5: Calculate the moles of \( \text{Ba}^{2+} \) produced
From the balanced equation, 2 moles of \( \text{H}_2\text{SO}_4 \) produce 3 moles of \( \text{Ba}^{2+} \). Therefore, from 0.02 moles of \( \text{H}_2\text{SO}_4 \):
\[ \text{Moles of } \text{Ba}^{2+} = \frac{3}{2} \times 0.02 = 0.03 \, \text{mol} \]

### Step 6: Calculate the total volume of the solution
The total volume after mixing is:
\[ 100 \, \text{ml} + 400 \, \text{ml} = 500 \, \text{ml} = 0.5 \, \text{L} \]

### Step 7: Calculate the concentration of \( \text{Ba}^{2+} \)
Using the formula:
\[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume (L)}} \]
\[ \text{Concentration of } \text{Ba}^{2+} = \frac{0.03 \, \text{mol}}{0.5 \, \text{L}} = 0.06 \, \text{M} \]

### Final Answer
The concentration of \( \text{Ba}^{2+} \) ions in the resulting solution is \( 0.06 \, \text{M} \).

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