A 100 mL sample of water was treated to convert any iron present to `Fe^(2+)`. Addition of 25 mL of 0.002 M `K_(2)Cr_(2)O_(7)` resulted in the reaction :
`6Fe^(2+)+Cr_(2)O_(7)^(2-)+14H^(+)rarr6Fe^(3+)+2Cr^(3+) + 7H_(2)O`
The excess `K_(2)Cr_(2)O_(7)` was back-titrated with 7.5 mL of 0.01 M `Fe^(2+)` solution. Calcution the parts per million (ppm) of iron in the water sample.

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of K₂Cr₂O₇ used in the reaction.
Given:
- Volume of K₂Cr₂O₇ = 25 mL = 0.025 L
- Molarity of K₂Cr₂O₇ = 0.002 M

Using the formula:
\[ \text{Moles of K}_2\text{Cr}_2\text{O}_7 = \text{Molarity} \times \text{Volume (L)} \]
\[ \text{Moles of K}_2\text{Cr}_2\text{O}_7 = 0.002 \, \text{mol/L} \times 0.025 \, \text{L} = 0.00005 \, \text{mol} \]

### Step 2: Determine the moles of Cr₂O₇²⁻ used in the reaction.
From the stoichiometry of the reaction:
\[ 6 \, \text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} + 14 \, \text{H}^+ \rightarrow 6 \, \text{Fe}^{3+} + 2 \, \text{Cr}^{3+} + 7 \, \text{H}_2\text{O} \]
1 mole of Cr₂O₇²⁻ reacts with 6 moles of Fe²⁺. Therefore, the moles of Cr₂O₇²⁻ used is equal to the moles of K₂Cr₂O₇ calculated above:
\[ \text{Moles of Cr}_2\text{O}_7^{2-} = 0.00005 \, \text{mol} \]

### Step 3: Calculate the moles of Fe²⁺ that reacted with Cr₂O₇²⁻.
Using the stoichiometry:
\[ \text{Moles of Fe}^{2+} = 6 \times \text{Moles of Cr}_2\text{O}_7^{2-} \]
\[ \text{Moles of Fe}^{2+} = 6 \times 0.00005 = 0.0003 \, \text{mol} \]

### Step 4: Calculate the moles of Fe²⁺ that were back-titrated.
Given:
- Volume of back-titrated Fe²⁺ = 7.5 mL = 0.0075 L
- Molarity of Fe²⁺ = 0.01 M

Using the formula:
\[ \text{Moles of Fe}^{2+} = \text{Molarity} \times \text{Volume (L)} \]
\[ \text{Moles of Fe}^{2+} = 0.01 \, \text{mol/L} \times 0.0075 \, \text{L} = 0.000075 \, \text{mol} \]

### Step 5: Calculate the moles of Fe²⁺ that reacted with the original K₂Cr₂O₇.
The moles of Fe²⁺ that reacted with K₂Cr₂O₇ is the initial moles minus the moles back-titrated:
\[ \text{Moles of Fe}^{2+} \text{ reacted} = 0.0003 - 0.000075 = 0.000225 \, \text{mol} \]

### Step 6: Calculate the mass of Fe in grams.
Using the molar mass of Fe (approximately 56 g/mol):
\[ \text{Mass of Fe} = \text{Moles of Fe} \times \text{Molar mass of Fe} \]
\[ \text{Mass of Fe} = 0.000225 \, \text{mol} \times 56 \, \text{g/mol} = 0.0126 \, \text{g} \]

### Step 7: Calculate the concentration of Fe in ppm.
Since the sample volume is 100 mL, we can convert grams to ppm:
\[ \text{ppm} = \left( \frac{\text{Mass of Fe (g)}}{\text{Volume of sample (L)}} \right) \times 10^6 \]
\[ \text{ppm} = \left( \frac{0.0126 \, \text{g}}{0.1 \, \text{L}} \right) \times 10^6 = 126 \, \text{ppm} \]

### Final Answer:
The concentration of iron in the water sample is **126 ppm**.

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