Question

For the titration of a $10 \, mL$ (aq) solution of $CaCO_3$, $2 \,mL$ of $0.001 \,M \,Na_2\,EDTA$ is required to reach the end point. The concentration of $CaCO_3$ is

• A. $5 \times 10^{-4}\, g/mL$
• B. $2 \times 10^{-4}\, g/mL$
• C. $5 \times 10^{-5}\, g/mL$
• D. $2 \times 10^{-5}\, g/mL$

Answer 1

Answer: (D)

$Na_2EDTA + CaCO_3 \to Ca-EDTA + 2Na^+$
$N_1 V_1 (CaCO_3) = N_2V_2(Na_2EDTA)$
$N_1 \times 10 = 2 \times 0.002 $ (For $Na_2EDTA, 1 \,M = 2N)$
$N_1 = \frac{4\times 10^{-3}}{10} = 4 \times 10^{-4}$
For $CaCO_3$, Molarity $ = \frac{\text{Normality}}{2}$
$ = \frac{4\times 10^{-4}}{2} = 2\times 10^{-4} \,mol/L$
$ = \frac{2 \times 10^{-4} \times 100}{1000} = 2\times 10^{-5} \,g/mL$
(M.W. of $CaCO_3 = 100)$