Answer to Question #145987 in Chemistry for Sunny Bae

Question

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

Answer 1

a. CaCO₃ + EDTA^4- → Ca(EDTA)^-2 + CO₃^2-

V(EDTA) = 48.0 mL

Proportion 1:

1000 mL – 0.05 mol

48.0 mL – n mol

n(EDTA) $= \frac{48.0 \times 0.05}{1000} = 0.0024\;mol$

n(CaCO₃) = n(EDTA) = 0.0024 mol

M(CaCO₃) = 100.08 g/mol

m(CaCO₃) $= 0.0024 \times 100.08 = 0.24 \;g$

Proportion 2:

1.00 g – 100 %

0.24 – x

x $= \frac{0.24 \times 100}{1} = 24$ %

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

b. Mg(OH)₂ + EDTA^4- → Mg(OH)₂EDTA^4-

V(EDTA) = 76.5 mL

Proportion 1:

1000 mL – 0.05 mol

76.5 mL – n mol

n(EDTA) $= \frac{76.5 \times 0.05}{1000} = 0.003825\;mol$

n(Mg(OH)₂) = n(EDTA) = 0.003825 mol

M(Mg(OH)₂) = 58.32 g/mol

m(Mg(OH)₂) $= 0.003825 \times 58.32 = 0.223 \;g$

Proportion 2:

1.00 g – 100 %

0.223 – x

x $= \frac{0.223 \times 100}{1} = 22.3$ %

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