91Ó°ÊÓ

Amount of cyanide solution treated= 12.73 mL

Amount of Ni²⁺ solution taken to convertcyanide into tetracyano nickelate (II)= 25.0 mL

Amount of EDTA required for titration of excess Ni²⁺= 10.15 mL of 0.01307 M EDTA

Amount of EDTA required to react with30.10 mLoriginalNi²⁺ solution= 39.35mL of 0.01307 M EDTA

Amount of EDTA required

$$\begin{gathered} =(39.35\mathrm{mL})(0.01307\mathrm{MEDTA}) \\ =0.514\mathrm{mmol} \end{gathered}$$

According to the question 30.10 mL of Ni²⁺ reacted with 0.514mmol of EDTA

Therefore, concentration of Ni²⁺

$$\begin{gathered} =\frac{0.514mmol}{30.10mL} \\ =0.0171M \end{gathered}$$

Therefore, the number of mols in 25.00 mL of Ni²⁺

$$\begin{gathered} =0.0171M×25ml \\ =0.4275mmol \end{gathered}$$

Number of mols in 10.15 mL of EDTA

$$\begin{gathered} =(10.15\mathrm{mL})(0.01307\mathrm{MEDTA}) \\ =0.1327\mathrm{mmol} \end{gathered}$$

Amount of Ni²⁺ reacted with CN^-

$$\begin{gathered} =(0.4275-0.1327) \\ =0.2948\mathrm{mmol} \end{gathered}$$

Therefore, the amount of cyanide reacting with Ni²⁺

$$\begin{gathered} =(4×0.2948)\mathrm{mmol} \\ =1.1792\mathrm{mmol} \end{gathered}$$

Original concentration of CN^-

$$\begin{gathered} =\frac{1.1792\mathrm{mmol}}{12.73\mathrm{mL}} \\ =0.093\mathrm{M} \end{gathered}$$

Therefore, the molarity of CN^- in the 12.73-mL sample is 0.093M