The volume of 3 M `Ba(OH)^(2)` solution required to neutralize completely 120 mL of 1.5M `H_(3)PO_(4)` solution is :

Answer

To solve the problem of finding the volume of 3 M `Ba(OH)₂` solution required to neutralize 120 mL of 1.5 M `H₃PO₄`, we will follow these steps:

### Step 1: Determine the Normality of `H₃PO₄`
The normality (N) of an acid is calculated using the formula:
\[ N = \text{Basicity} \times \text{Molarity} \]

For `H₃PO₄`, the basicity is 3 (since it can donate three protons). Given that the molarity is 1.5 M, we calculate:
\[ N_{H₃PO₄} = 3 \times 1.5 = 4.5 \, \text{N} \]

### Step 2: Determine the Normality of `Ba(OH)₂`
The normality of a base is calculated using the formula:
\[ N = \text{Acidity} \times \text{Molarity} \]

For `Ba(OH)₂`, the acidity is 2 (since it can donate two hydroxide ions). Given that the molarity is 3 M, we calculate:
\[ N_{Ba(OH)₂} = 2 \times 3 = 6 \, \text{N} \]

### Step 3: Use the Neutralization Equation
The neutralization reaction can be represented as:
\[ N_1 V_1 = N_2 V_2 \]
Where:
- \( N_1 \) = Normality of `Ba(OH)₂` = 6 N
- \( V_1 \) = Volume of `Ba(OH)₂` (unknown)
- \( N_2 \) = Normality of `H₃PO₄` = 4.5 N
- \( V_2 \) = Volume of `H₃PO₄` = 120 mL

### Step 4: Rearranging the Equation to Solve for \( V_1 \)
We can rearrange the equation to find \( V_1 \):
\[ V_1 = \frac{N_2 V_2}{N_1} \]

### Step 5: Substitute the Values
Substituting the values we calculated:
\[ V_1 = \frac{4.5 \, \text{N} \times 120 \, \text{mL}}{6 \, \text{N}} \]

### Step 6: Calculate \( V_1 \)
Now, performing the calculation:
\[ V_1 = \frac{540}{6} = 90 \, \text{mL} \]

Thus, the volume of 3 M `Ba(OH)₂` solution required to neutralize completely 120 mL of 1.5 M `H₃PO₄` solution is **90 mL**.

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