Discussion
For the titration of a 10 mL (aq) solution of CaCO3, 2 mL of 0.001 M Na2EDTA is required to reach the end point. The concentration of CaCO3 (assume molecular weight of CaCO3 = 100) is
Answer: D
The concentration of CaCO3 is 2 × 10-5 g/mL.
EDTA binds Ca2+ in a 1:1 ratio, so the moles of CaCO3 equal the moles of EDTA used:
\[ \begin{aligned} n &= 2\ \text{mL} \times 0.001\ \text{mol/L} \\ &= 2 \times 10^{-3}\ \text{L} \times 10^{-3}\ \text{mol/L} = 2 \times 10^{-6}\ \text{mol} \\ m &= 2 \times 10^{-6} \times 100 = 2 \times 10^{-4}\ \text{g} \end{aligned} \]
That mass is in 10 mL of solution, so the concentration is 2 × 10-4 g / 10 mL = 2 × 10-5 g/mL.
2 × 10-4 g/mL forgets to divide by the 10 mL volume. The 5 × values do not follow from the 1:1 Ca2+-EDTA stoichiometry with these volumes.
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