To solve the problem step by step, we will follow the outlined procedure based on the information given in the question.
### Step 1: Understand the Reaction of Bromine with NaOH
When bromine (Br₂) reacts with sodium hydroxide (NaOH), it undergoes disproportionation to form bromide ions (Br⁻) and bromate ions (BrO₃⁻). The balanced chemical equation for this reaction is:
\[
3 \text{Br}_2 + 6 \text{NaOH} \rightarrow 5 \text{NaBr} + \text{NaBrO}_3 + 3 \text{H}_2\text{O}
\]
### Step 2: Calculate the Moles of Br₂
Given that we have a 10 mL solution of 1 M Br₂:
\[
\text{Moles of Br}_2 = \text{Volume (L)} \times \text{Concentration (M)} = 0.010 \, \text{L} \times 1 \, \text{mol/L} = 0.010 \, \text{mol}
\]
### Step 3: Determine the Moles of BrO₃⁻ Produced
From the balanced equation, we see that 3 moles of Br₂ produce 1 mole of BrO₃⁻. Thus, the moles of BrO₃⁻ produced from 0.010 moles of Br₂ is:
\[
\text{Moles of BrO}_3^- = \frac{1}{3} \times \text{Moles of Br}_2 = \frac{1}{3} \times 0.010 = 0.00333 \, \text{mol}
\]
### Step 4: Convert Moles of BrO₃⁻ to Milliequivalents
Milliequivalents (mEq) can be calculated as:
\[
\text{mEq of BrO}_3^- = \text{Moles} \times 1000 = 0.00333 \, \text{mol} \times 1000 = 3.33 \, \text{mEq}
\]
### Step 5: Reaction of BrO₃⁻ with Calcium Oxalate
The reaction between BrO₃⁻ and calcium oxalate (C₂O₄²⁻) can be represented as:
\[
\text{BrO}_3^- + 3 \text{C}_2\text{O}_4^{2-} \rightarrow \text{Br}^- + 3 \text{CO}_2 + 3 \text{H}_2\text{O}
\]
From the stoichiometry, 1 mole of BrO₃⁻ reacts with 3 moles of C₂O₄²⁻.
### Step 6: Calculate Moles of C₂O₄²⁻ Reacted
Using the relationship from the reaction, we can find the moles of C₂O₄²⁻:
\[
\text{Moles of C}_2\text{O}_4^{2-} = \frac{1}{3} \times \text{Moles of BrO}_3^- = \frac{1}{3} \times 0.00333 = 0.00111 \, \text{mol}
\]
### Step 7: Calculate the Mass of C₂O₄²⁻
Using the molar mass of calcium oxalate (128 g/mol):
\[
\text{Mass of C}_2\text{O}_4^{2-} = \text{Moles} \times \text{Molar Mass} = 0.00111 \, \text{mol} \times 128 \, \text{g/mol} = 0.14208 \, \text{g}
\]
### Step 8: Calculate the % Purity of the Sample
Given that the total mass of the impure sample is 2 g, the % purity can be calculated as:
\[
\text{% Purity} = \left(\frac{\text{Mass of pure C}_2\text{O}_4^{2-}}{\text{Total mass of sample}}\right) \times 100 = \left(\frac{0.14208 \, \text{g}}{2 \, \text{g}}\right) \times 100 \approx 7.04\%
\]
### Final Answer
The % purity of the oxalate sample is approximately **7.04%**.
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