A 100 mL sample of blakish water was made ammonical and sulphide ion in solution were titrated with `16.5` mL of `0.02 M AgNO_(3)`. The concentration of `H_(2)S` in water in ppm is

Options

A. `5.60 `
B. `560`
C. 56
D. `0.56`

Explanation

To find the concentration of \( H_2S \) in water in ppm, we will follow these steps:

### Step 1: Calculate the moles of \( AgNO_3 \) used in the titration.
Given:
- Volume of \( AgNO_3 \) solution = \( 16.5 \, \text{mL} = 0.0165 \, \text{L} \)
- Concentration of \( AgNO_3 \) = \( 0.02 \, \text{M} \)

Using the formula:
\[ \text{Moles of } AgNO_3 = \text{Concentration} \times \text{Volume} = 0.02 \, \text{mol/L} \times 0.0165 \, \text{L} = 0.00033 \, \text{mol} \]

### Step 2: Determine the moles of \( H_2S \) that reacted.
The reaction between \( H_2S \) and \( AgNO_3 \) is:
\[ H_2S + 2AgNO_3 \rightarrow Ag_2S + 2HNO_3 \]
From the balanced equation, 1 mole of \( H_2S \) reacts with 2 moles of \( AgNO_3 \). Therefore, the moles of \( H_2S \) can be calculated as:
\[ \text{Moles of } H_2S = \frac{\text{Moles of } AgNO_3}{2} = \frac{0.00033}{2} = 0.000165 \, \text{mol} \]

### Step 3: Calculate the mass of \( H_2S \).
The molar mass of \( H_2S \) is:
\[ \text{Molar mass of } H_2S = 2 \times 1.01 \, \text{g/mol (H)} + 32.07 \, \text{g/mol (S)} = 34.09 \, \text{g/mol} \]
Now, we can find the mass:
\[ \text{Mass of } H_2S = \text{Moles} \times \text{Molar mass} = 0.000165 \, \text{mol} \times 34.09 \, \text{g/mol} = 0.00562 \, \text{g} \]

### Step 4: Calculate the concentration in ppm.
To find the concentration in ppm (parts per million), we use the formula:
\[ \text{ppm} = \left( \frac{\text{mass of solute (g)}}{\text{volume of solution (L)}} \right) \times 10^6 \]
Given that the volume of the sample is \( 100 \, \text{mL} = 0.1 \, \text{L} \):
\[ \text{ppm} = \left( \frac{0.00562 \, \text{g}}{0.1 \, \text{L}} \right) \times 10^6 = 56200 \, \text{ppm} \]

### Final Answer:
The concentration of \( H_2S \) in water is **56200 ppm**.

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