To estimate Al^3+, 25.0 mL of the sample is treated with 40.0 mL of 0.0500 M EDTA (excess). The unreacted EDTA is back-titrated with 0.0400 M Zn^2+, requiring 18.5 mL. EDTA reacts 1:1 with both Al^3+ and Zn^2+. The molarity of Al^3+ in the sample is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3eq213

To estimate $Al X^{3 +}$ , $25.0 mL$ of the sample is treated with $40.0 mL$ of $0.0500 M$ EDTA (excess). The unreacted EDTA is back-titrated with $0.0400 M$ $Zn X^{2 +}$ , requiring $18.5 mL$ . EDTA reacts 1:1 with both $Al X^{3 +}$ and $Zn X^{2 +}$ . The molarity of $Al X^{3 +}$ in the sample is:

EDTA taken

= 40.0 \times 0.0500 = 2.000

mmol; excess (equal to

Zn X^{2 +}

)

= 18.5 \times 0.0400 = 0.740

mmol. EDTA bound to

Al X^{3 +}

= 1.260

mmol. Molarity

= 1.260/25.0 = 0.0504 M