# Answer to Question #145987 in Other Chemistry for Sunny Bae

Question #145987
As part of a geological team that studied a local cave, you brought with you a bunch of
1.00 g rock samples to be studied. Each rock was prepared and titrated against 0.050 M
EDTA

a. Calculate the percent calcite (CaCO3) content of rock A if it was titrated with
48.0 mL EDTA
b. Calculate the percent brucite (Mg(OH)2) content of rock B if it was titrated
with 76.5 mL EDTA
1
2020-11-24T05:53:02-0500

a. CaCO3 + EDTA4- → Ca(EDTA)-2 + CO32-

V(EDTA) = 48.0 mL

Proportion 1:

1000 mL – 0.05 mol

48.0 mL – n mol

n(EDTA)

n(CaCO3) = n(EDTA) = 0.0024 mol

M(CaCO3) = 100.08 g/mol

m(CaCO3)

Proportion 2:

1.00 g – 100 %

0.24 – x

x %

The percent calcite (CaCO3) content of rock A is 24 %.

b. Mg(OH)2 + EDTA4- → Mg(OH)2EDTA4-

V(EDTA) = 76.5 mL

Proportion 1:

1000 mL – 0.05 mol

76.5 mL – n mol

n(EDTA)

n(Mg(OH)2) = n(EDTA) = 0.003825 mol

M(Mg(OH)2) = 58.32 g/mol

m(Mg(OH)2)

Proportion 2:

1.00 g – 100 %

0.223 – x

x %

The percent brucite (Mg(OH)2) content of rock A is 22.3 %.

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