Fully titrating a 0.1123 g sample of calcium carbonate required 40.75 ml of EDTA solution to reach the endpoint. What is the molar concentration of EDTA in the titrant solution?
Added by Robert H.
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The molar mass of CaCO₃ is approximately 100.09 g/mol. \[ \text{Moles of CaCO₃} = \frac{\text{mass of CaCO₃}}{\text{molar mass of CaCO₃}} = \frac{0.1123 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.00112 \, \text{mol} \] Show more…
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A 0.5118 g sample of CaCO3 is dissolved in 6 M HCl, and the resulting solution is diluted to 250.0 mL in a volumetric flask. The 25.00 mL of Ca2+ solution is titrated with EDTA to the Eriochrome Black T endpoint. It requires 30.16 mL of the EDTA to reach the endpoint. What is the molarity of the EDTA solution?
Dinesh S.
The concentration of a solution of EDTA was determined by standardizing against a solution of Ca2+ prepared from the primary standard CaCO3. A 0.4071-g sample of CaCO3 was transferred to a 500-mL volumetric flask, dissolved using a minimum of 6 M HCl, and diluted to volume. A 50.00-mL portion of this solution was transferred into a 250-mL Erlenmeyer flask and the pH adjusted by adding 5 mL of a pH 10 NH3–NH4Cl buffer containing a small amount of Mg2+–EDTA. After adding calmagite as a visual indicator, the solution was titrated with the EDTA, requiring 42.63 mL to reach the end point. Report the molar concentration of the titrant.
Sri K.
A 0.3205 g sample of CaCO3 was dissolved in HCl and the resulting solution was diluted to 250.00 mL in a volumetric flask. From this volumetric flask, 25.0 mL solution was obtained and titrated against EDTA. A 18.75 mL of EDTA was utilized to reach the end point (Eriochrome Black T used as the indicator). Calculate the moles of CaCO3 used. Hint: Moles = g / MW. Use 20Ca40, 6C12 and 8O16. A. 0.003205 moles B. 0.03205 moles C. 0.3205 moles D. 3.205 moles
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