Question

A 50.0-mL solution containing Ni²⁺ and Zn²⁺ was treatedwith 25.0 mL of 0.045 2 M EDTA to bind all the metal. The excess unreacted EDTA required 12.4 mL of 0.012 3 M Mg²⁺ for complete reaction. An excess of the reagent 2,3-dimercapto-1-propanol was then added to displace the EDTA from zinc. Another 29.2 mL of Mg²⁺ were required for reaction with the liberated EDTA. Calculate the molarity of Ni²⁺ and Zn²⁺ in the original solution

Short answer

The molarity ofNi²⁺ and Zn²⁺ in the original solution are 0.0124M and 0.0072M respectively.

Step-by-step solution

Given Information

Amount of solution containing Ni²⁺ and Zn²⁺ = 25 mL

Amount of EDTA added = 25.0 mL of 0.0452 M EDTA

Amount of Mg²⁺ required to complete reaction with excess unreacted EDTA= 12.4 mL of 0.0123 M Mg²⁺

Amount of Mg²⁺ required for reaction with liberated EDTA=29.2 mL

Determine the total amount of Ni2+ and Zn2+ treated

Total amount of EDTA

$$\begin{gathered} =(25.00\mathrm{mL})(0.0452\mathrm{MEDTA}) \\ =1.13\mathrm{mmol} \end{gathered}$$

Amount of Mg²⁺ required

$$\begin{gathered} =(12.4\mathrm{mL})(0.0123\mathrm{M}) \\ =0.153\mathrm{mmol} \end{gathered}$$

Therefore, the total amount of Ni²⁺ and Zn²⁺

$$\begin{gathered} =(1.13-0.153)\mathrm{mmol} \\ =0.977\mathrm{mmol} \end{gathered}$$

Determine the concentration of Zn2+ in the solution

Amount of Zn²⁺ = EDTA displaced by 2,3-dimercapto-1-propanol

Amount of EDTA displaced

$$\begin{gathered} =(29.2\mathrm{mL})(0.0123\mathrm{M}) \\ =0.36\mathrm{mmol} \end{gathered}$$

Concentration of Zn²⁺ in the original solution will be

$$\begin{gathered} =\frac{0.36\mathrm{mmol}}{50\mathrm{mL}} \\ =0.0072\mathrm{M} \end{gathered}$$

Determine the concentration of Ni2+ in the solution

$\frac{\begin{array}{c}\mathrm{Ni}(2+)+{\mathrm{Zn}}^{2+}=0.977\mathrm{mmol}\\ {\mathrm{Zn}}^{2+}=0.36\mathrm{mmol}\end{array}}{{\mathrm{Ni}}^{2+}=0.617\mathrm{mmol}}$

50.0-mL solution containing Ni²⁺ and Zn²⁺ was taken

Therefore, concentration of Ni²⁺ in the original solution will be

$$\begin{gathered} =\frac{0.617\mathrm{mmol}}{50\mathrm{mL}} \\ =0.0124\mathrm{M} \end{gathered}$$