Concept Explanation: Redox Titration and Percentage Purity Calculation

This problem involves determining the percentage purity of iron in a wire sample using redox titration with potassium dichromate. The key concepts are:

1. Redox Reaction: Iron(II) is oxidized to iron(III) by dichromate in acidic medium
2. Equivalent Concept: Number of equivalents of oxidant = Number of equivalents of reductant
3. Dilution Factor: Only 20 mL of 100 mL solution is titrated

Step-by-Step Solution:

Step 1: Write the balanced redox reaction

${\mathrm{Cr}}_2{O}_7^{2-}+6{\mathrm{Fe}}^{2+}+14H^{+}\to 2{\mathrm{Cr}}^{3+}+6{\mathrm{Fe}}^{3+}+7H_{2}O$

Step 2: Calculate equivalents of K₂Cr₂O₇ used

Normality = N/30 = 0.0333 N

Volume = 30 mL = 0.03 L

Equivalents of K₂Cr₂O₇ = Normality × Volume(L) = $\frac{1}{30}\times \frac{30}{1000}=\frac{1}{1000}$ equivalents

Step 3: These equivalents correspond to Fe in 20 mL aliquot

Equivalents of Fe in 20 mL = $\frac{1}{1000}$

Step 4: Calculate equivalents of Fe in 100 mL solution

Equivalents of Fe in 100 mL = $\frac{1}{1000}\times \frac{100}{20}=\frac{1}{200}$ equivalents

Step 5: Calculate mass of pure Fe

Equivalent weight of Fe = Atomic weight/1 = 56 g/equiv

Mass of pure Fe = Equivalents × Equivalent weight = $\frac{1}{200}\times 56=0.28$ g

Step 6: Calculate percentage purity

Percentage purity = $\frac{\mathrm{Mass}\mathrm{of}\mathrm{pure}\mathrm{Fe}}{\mathrm{Mass}\mathrm{of}\mathrm{sample}}\times 100=\frac{0.28}{0.35}\times 100=80\%$

Final Answer: 80%

Related Topics:

• Redox Titration Principles
• Equivalent Weight Calculations
• Dilution and Aliquot Concepts
• Percentage Purity Determination

Important Formulae:

Number of equivalents = Normality × Volume(L)

Equivalent weight of element = Atomic weight/Change in oxidation number

Percentage purity = (Mass of pure substance/Mass of sample) × 100

Dilution factor = Final volume/Aliquot volume