Question

A 50.0-mL sample containing Ni²⁺ was treated with 25.0 mL of 0.050 0 M EDTA to complex all the Ni²⁺ and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring 5.00 mL of 0.050 0 M Zn²⁺. What was the concentration of Ni²⁺ in the original solution?

Short answer

The concentration of Ni²⁺ in the original solution is 0.02 M

Step-by-step solution

Given Information

Amount of sample taken= 50.0 mL

Amount of EDTA required for titration =25 mL of 0.05M EDTA

Amount of Zn²⁺ required for back titration= 5 mL of 0.050 M Zn²⁺

Determine the number of moles of EDTA and Zn2+ required

Number of moles of EDTA required

$$\begin{gathered} =\left(25\mathrm{mL}\right)\left(0.050\mathrm{M}\mathrm{EDTA}\right) \\ =1.25\mathrm{mmol} \end{gathered}$$

Number of moles of Zn²⁺ required

$\begin{array}{l}=\left(5\mathrm{mL}\right)(0.050\mathrm{EDTA})\\ =0.25\mathrm{mmol}\end{array}$

Determine the concentration of Ni2+

Let the number of mols of Ni²⁺ be x

$$\begin{gathered} \mathrm{mmol}\mathrm{EDTA}={\mathrm{mmolNi}}^{2+}+{\mathrm{mmolZn}}^{2+} \\ 1.250\mathrm{mmol}\mathrm{EDTA}=\mathrm{x}\mathrm{mmol}{\mathrm{Ni}}^{2+}+0.250{\mathrm{mmolZn}}^{2+} \\ \mathrm{x}=1\mathrm{mmol}{\mathrm{Ni}}^{2+} \end{gathered}$$

50.0-mL of sample containing Ni²⁺ was taken

Therefore, concentration of Ni²⁺in the solution

$$\begin{gathered} =\frac{1\mathrm{mmol}}{50\mathrm{mL}} \\ =0.02\mathrm{M} \end{gathered}$$