Question

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?

Name: a 05118 g sample of caco3 is dissolved in 6 m hcl and the resulting solution is diluted to 2500 ml in a volumetric flask the 2500 ml of ca2 solution is titrated with edta to the eriochrome b 75387
Uploaded: 2022-02-19T08:23:56-08:00
Duration: 1 min 36 s
Channel: Dinesh Singh
Description: a 05118 g sample of caco3 is dissolved in 6 m hcl and the resulting solution is diluted to 2500 ml in a volumetric flask the 2500 ml of ca2 solution is titrated with edta to the eriochrome b 75387

          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?

Added by Jessica G.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Dinesh Singh

02/19/2022

Step 1

Given mass of CaCO3 = 0.5118 g Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + (3 * 16.00 g/mol) = 100.09 g/mol Moles of CaCO3 = 0.5118 g / 100.09 g/mol = 0.005115 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?