1 L sample of impure water containing sulphide ion is made ammoniacal and is titrated with 300 " mL of " 0.1 M `AgNO_(3)` solution. Which of the following statements is/are correct about the above reaction?

Options

A. The strength of `H_(2)S` in water is `0.51g L^(-1)`
B. The strength of `H_(2)S` in water is `5.1 gL^(-1)`
C. The concentration of `H_(2)S` in water in ppm is 510.
D. The concentration of `H_(2)S` in water in ppm is 51.

Explanation

To solve the problem, we need to analyze the titration of the sulfide ions in the impure water sample with silver nitrate (AgNO₃). Here’s a step-by-step solution:

### Step 1: Understand the Reaction
The sulfide ion (S²⁻) in the water sample reacts with silver nitrate (AgNO₃) to form silver sulfide (Ag₂S). The balanced chemical equation for the reaction is:
\[ 2 \text{AgNO}_3 + \text{H}_2\text{S} \rightarrow \text{Ag}_2\text{S} + 2 \text{HNO}_3 \]
This indicates that 2 moles of AgNO₃ react with 1 mole of H₂S.

### Step 2: Calculate Milliequivalents of AgNO₃
We are given that 300 mL of 0.1 M AgNO₃ is used. First, we calculate the milliequivalents of AgNO₃:
- Molarity (M) = 0.1 M
- Volume (V) = 300 mL = 0.3 L
- n-factor for AgNO₃ = 1 (since it provides 1 equivalent of Ag⁺)

Using the formula:
\[ \text{Milliequivalents} = \text{Molarity} \times \text{Volume (L)} \times \text{n-factor} \times 1000 \]
\[ \text{Milliequivalents of AgNO}_3 = 0.1 \times 0.3 \times 1 \times 1000 = 30 \text{ milliequivalents} \]

### Step 3: Relate Milliequivalents of H₂S
From the balanced equation, we know that 2 milliequivalents of AgNO₃ react with 1 milliequivalent of H₂S. Therefore, the milliequivalents of H₂S can be calculated as:
\[ \text{Milliequivalents of H}_2\text{S} = \frac{30}{2} = 15 \text{ milliequivalents} \]

### Step 4: Calculate the Weight of H₂S
To find the weight of H₂S in the 1 L sample, we need to use the equivalent weight of H₂S:
- Molecular weight of H₂S = 34 g/mol
- n-factor for H₂S = 2 (since it can donate 2 H⁺ ions)

The equivalent weight of H₂S is:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{n-factor}} = \frac{34}{2} = 17 \text{ g/equiv} \]

Now, we can calculate the weight of H₂S:
\[ \text{Weight of H}_2\text{S} = \text{Milliequivalents} \times \text{Equivalent weight} = 15 \times 17 = 255 \text{ mg} = 0.255 \text{ g} \]

### Step 5: Convert to g/L
Since we have 1 L of the sample, the concentration of H₂S in grams per liter is:
\[ \text{Concentration of H}_2\text{S} = 0.255 \text{ g/L} \]

### Step 6: Convert to PPM
To convert grams per liter to parts per million (PPM):
\[ \text{PPM} = \text{Concentration (g/L)} \times 1000 = 0.255 \times 1000 = 255 \text{ PPM} \]

### Final Results
1. The concentration of H₂S in water is **0.255 g/L**.
2. The concentration of H₂S in water is **255 PPM**.

### Conclusion
The correct statements about the reaction are:
- The strength of H₂S in water is 0.255 g/L.
- The strength of H₂S in water is 255 PPM.