`50 mL` of `1 M HCl, 100 mL` of `0.5 M HNO_(3)`, and `x mL` of `5 M H_(2)SO_(4)` are mixed together and the total volume is made upto `1.0 L` with water. `100 mL` of this solution exactly neutralises `10 mL` of `M//3 Al_(2) (CO_(3))_(3)`. Calculate the value of `x`.

Explanation

To solve the problem step by step, we will calculate the total milliequivalents of the acids mixed and then use the neutralization reaction to find the value of \( x \).

### Step 1: Calculate the milliequivalents of each acid

1. **HCl**:
- Volume = 50 mL
- Molarity = 1 M
- Milliequivalents = Volume (mL) × Molarity = \( 50 \, \text{mL} \times 1 \, \text{eq/L} = 50 \, \text{meq} \)

2. **HNO3**:
- Volume = 100 mL
- Molarity = 0.5 M
- Milliequivalents = Volume (mL) × Molarity = \( 100 \, \text{mL} \times 0.5 \, \text{eq/L} = 50 \, \text{meq} \)

3. **H2SO4**:
- Volume = \( x \, \text{mL} \)
- Molarity = 5 M
- Milliequivalents = Volume (mL) × Molarity × n-factor (2 for H2SO4) = \( x \, \text{mL} \times 5 \, \text{eq/L} \times 2 = 10x \, \text{meq} \)

### Step 2: Total milliequivalents of the acid mixture

Total milliequivalents of the acid mixture:
\[ \text{Total meq} = 50 + 50 + 10x = 100 + 10x \]

### Step 3: Calculate the normality of the acid mixture

The total volume of the mixture is made up to 1 L (1000 mL). Therefore, the normality (N) of the mixture is given by:
\[ N = \frac{\text{Total meq}}{\text{Total volume in mL}} = \frac{100 + 10x}{1000} \]

### Step 4: Use the neutralization reaction

We know that 100 mL of this acid mixture neutralizes 10 mL of \( \frac{1}{3} \, \text{M} \, \text{Al}_2(\text{CO}_3)_3 \).

1. **Calculate the normality of \( \text{Al}_2(\text{CO}_3)_3 \)**:
- Molarity = \( \frac{1}{3} \, \text{M} \)
- n-factor for \( \text{Al}_2(\text{CO}_3)_3 \) = 6 (as it produces 2 Al³⁺ and 3 CO₃²⁻)
- Normality \( N_2 = \text{Molarity} \times \text{n-factor} = \frac{1}{3} \times 6 = 2 \, \text{N} \)

2. **Calculate the equivalents from \( \text{Al}_2(\text{CO}_3)_3 \)**:
- Volume = 10 mL
- Equivalents = Normality × Volume = \( 2 \, \text{N} \times 10 \, \text{mL} = 20 \, \text{meq} \)

### Step 5: Set up the equation

Using the neutralization equation:
\[ \text{Equivalents of acid mixture} = \text{Equivalents of } \text{Al}_2(\text{CO}_3)_3 \]
\[ \frac{100 + 10x}{1000} \times 100 = 20 \]

### Step 6: Solve for \( x \)

1. Simplifying the equation:
\[ 100 + 10x = 20 \times 10 = 200 \]
\[ 10x = 200 - 100 \]
\[ 10x = 100 \]
\[ x = 10 \]

### Final Answer:
The value of \( x \) is **10 mL**.

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