For the titration of a 10 mL (aq) solution of CaCO3, 2 mL of 0.001 M Na2 EDTA is required to reach the end point. The concentration of CaCO3 is

Question

For the titration of a 10 mL (aq) solution of CaCO3, 2 mL of 0.001 M Na2 EDTA is required to reach the end point. The concentration of CaCO3 is (A) 5 × 10-4 g/mL (B) 2 × 10-4 g/mL (C) 5 × 10-5 g/mL (D) 2 × 10-5 g/mL

Tardigrade Admin · Accepted Answer

Na2EDTA + CaCO3 → Ca-EDTA + 2Na+ N1 V1 (CaCO3) = N2V2(Na2EDTA) N1 × 10 = 2 × 0.002 (For Na2EDTA, 1 M = 2N) N1 = (4× 10-3/10) = 4 × 10-4 For CaCO3, Molarity = ( textNormality/2) = (4× 10-4/2) = 2× 10-4 mol/L = (2 × 10-4 × 100/1000) = 2× 10-5 g/mL (M.W. of CaCO3 = 100)

Q. For the titration of a $10 m L$ (aq) solution of $C a C O_{3}$ , $2 m L$ of $0.001 M N a_{2} E D T A$ is required to reach the end point. The concentration of $C a C O_{3}$ is

$5 \times 1 0^{- 4} g / m L$

$2 \times 1 0^{- 4} g / m L$

$5 \times 1 0^{- 5} g / m L$

$2 \times 1 0^{- 5} g / m L$

Q. For the titration of a 10mL (aq) solution of CaCO3​, 2mL of 0.001MNa2​EDTA is required to reach the end point. The concentration of CaCO3​ is

5×10−4g/mL

2×10−4g/mL

5×10−5g/mL

2×10−5g/mL

Na2​EDTA+CaCO3​→Ca−EDTA+2Na+ N1​V1​(CaCO3​)=N2​V2​(Na2​EDTA) N1​×10=2×0.002 (For Na2​EDTA,1M=2N) N1​=104×10−3​=4×10−4 For CaCO3​, Molarity =2Normality​ =24×10−4​=2×10−4mol/L =10002×10−4×100​=2×10−5g/mL (M.W. of CaCO3​=100)

Q. For the titration of a $10 m L$ (aq) solution of $C a C O_{3}$ , $2 m L$ of $0.001 M N a_{2} E D T A$ is required to reach the end point. The concentration of $C a C O_{3}$ is

$5 \times 1 0^{- 4} g / m L$

$2 \times 1 0^{- 4} g / m L$

$5 \times 1 0^{- 5} g / m L$

$2 \times 1 0^{- 5} g / m L$