3. An EDTA solution is standardized against high purity CaCO3 by dissolving 0.3982 g CaCO3 in hydrochloric acid, adjusting the pH to 10.0 with buffer, and then titrating. If 38.26 mL of EDTA was required for the titration, what is the molarity of the EDTA?

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**Standardization of EDTA Solution**

**Problem Statement:**
An EDTA solution is standardized against high purity calcium carbonate (\( \text{CaCO}_3 \)) by dissolving \( 0.3982 \, \text{g} \, \text{CaCO}_3 \) in hydrochloric acid, adjusting the pH to 10.0 with a buffer, and then titrating. If \( 38.26 \, \text{mL} \) of EDTA was required for the titration, what is the molarity of the EDTA?

**Solution:**

To find the molarity of the EDTA solution, we need to determine the moles of \( \text{CaCO}_3 \) and use the volume of EDTA used in the titration.

1. **Calculate the Moles of \( \text{CaCO}_3 \):**
   - Molar mass of \( \text{CaCO}_3 = 40.08 \, (\text{Ca}) + 12.01 \, (\text{C}) + 3 \times 16.00 \, (\text{O}) = 100.09 \, \text{g/mol} \).
   - Moles of \( \text{CaCO}_3 = \frac{0.3982 \, \text{g}}{100.09 \, \text{g/mol}} \).

2. **Determine the Molarity of EDTA:**
   - Reacting ratio is 1:1 between \( \text{CaCO}_3 \) and EDTA in a balanced equation.
   - Moles of EDTA = Moles of \( \text{CaCO}_3 \).
   - Molarity of EDTA = \(\frac{\text{Moles of EDTA}}{\text{Volume of EDTA in Liters}}\).

Ensure the answer is expressed with appropriate significant figures based on the provided data.
Transcribed Image Text:**Standardization of EDTA Solution** **Problem Statement:** An EDTA solution is standardized against high purity calcium carbonate (\( \text{CaCO}_3 \)) by dissolving \( 0.3982 \, \text{g} \, \text{CaCO}_3 \) in hydrochloric acid, adjusting the pH to 10.0 with a buffer, and then titrating. If \( 38.26 \, \text{mL} \) of EDTA was required for the titration, what is the molarity of the EDTA? **Solution:** To find the molarity of the EDTA solution, we need to determine the moles of \( \text{CaCO}_3 \) and use the volume of EDTA used in the titration. 1. **Calculate the Moles of \( \text{CaCO}_3 \):** - Molar mass of \( \text{CaCO}_3 = 40.08 \, (\text{Ca}) + 12.01 \, (\text{C}) + 3 \times 16.00 \, (\text{O}) = 100.09 \, \text{g/mol} \). - Moles of \( \text{CaCO}_3 = \frac{0.3982 \, \text{g}}{100.09 \, \text{g/mol}} \). 2. **Determine the Molarity of EDTA:** - Reacting ratio is 1:1 between \( \text{CaCO}_3 \) and EDTA in a balanced equation. - Moles of EDTA = Moles of \( \text{CaCO}_3 \). - Molarity of EDTA = \(\frac{\text{Moles of EDTA}}{\text{Volume of EDTA in Liters}}\). Ensure the answer is expressed with appropriate significant figures based on the provided data.
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