The chromate ion present in water sample is reduced to insoluble chromium hydroxide , `Cr(OH)_(3)` by dithionation, in basic solution .
`S_(2)O_(4)^(2-) + CrO_(4)^(2-) + 2H_(2)O to 2SO_(3)^(2-) + Cr(OH)_(3)+ OH^(-)`
100 L of water requires 387 g of `Na_(2)S_(2)O_(4)` . The molarity of `CrO_(4)^(2-)` in waste water is

Options

A. 0.0448
B. 4.448
C. 0.0148
D. 0.0224

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Write down the balanced chemical equation
The reaction given is:
\[ S_2O_4^{2-} + CrO_4^{2-} + 2H_2O \rightarrow 2SO_3^{2-} + Cr(OH)_3 + OH^{-} \]

### Step 2: Determine the oxidation states
- For \( S_2O_4^{2-} \):
- Let the oxidation state of sulfur be \( x \).
- The equation for the charge balance is \( 2x + 4(-2) = -2 \).
- This simplifies to \( 2x - 8 = -2 \) → \( 2x = 6 \) → \( x = +3 \).

- For \( CrO_4^{2-} \):
- Let the oxidation state of chromium be \( y \).
- The equation for the charge balance is \( y + 4(-2) = -2 \).
- This simplifies to \( y - 8 = -2 \) → \( y = +6 \).

- For \( SO_3^{2-} \):
- Let the oxidation state of sulfur be \( z \).
- The equation for the charge balance is \( z + 3(-2) = -2 \).
- This simplifies to \( z - 6 = -2 \) → \( z = +4 \).

### Step 3: Calculate the change in oxidation states
- Sulfur changes from +3 to +4 (a change of +1).
- Chromium changes from +6 to +3 (a change of -3).

### Step 4: Determine the n-factor
- The n-factor for \( S_2O_4^{2-} \) is 2 (2 sulfur atoms changing by +1 each).
- The n-factor for \( CrO_4^{2-} \) is 3 (1 chromium atom changing by -3).

### Step 5: Calculate the equivalent weight of \( Na_2S_2O_4 \)
- Given mass of \( Na_2S_2O_4 = 387 \, g \).
- Molecular weight of \( Na_2S_2O_4 = 174 \, g/mol \).
- Calculate moles of \( Na_2S_2O_4 \):
\[ \text{Moles} = \frac{\text{mass}}{\text{molecular weight}} = \frac{387 \, g}{174 \, g/mol} \approx 2.22 \, mol \]

### Step 6: Calculate the total equivalents of \( Na_2S_2O_4 \)
- Total equivalents = moles × n-factor = \( 2.22 \, mol \times 2 = 4.44 \, equivalents \).

### Step 7: Set up the equivalence relation
According to the balanced equation:
\[ \text{Equivalents of } Na_2S_2O_4 = \text{Equivalents of } CrO_4^{2-} \]
Let \( n \) be the moles of \( CrO_4^{2-} \):
\[ n \times 3 = 4.44 \implies n = \frac{4.44}{3} \approx 1.48 \, mol \]

### Step 8: Calculate the molarity of \( CrO_4^{2-} \)
- Volume of solution = 100 L.
- Molarity \( M \) is given by:
\[ M = \frac{\text{moles}}{\text{volume in L}} = \frac{1.48 \, mol}{100 \, L} = 0.0148 \, M \]

### Final Answer
The molarity of \( CrO_4^{2-} \) in the wastewater is \( 0.0148 \, M \).

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