Question

A 1.000-mL sample of unknown containing Co²⁺ and Ni²⁺ was treated with 25.00 mL of 0.038 72 M EDTA. Back titration with 0.021 27 M Zn²⁺ at pH 5 required 23.54 mL to reach the xylenol orange end point. A 2.000-mL sample of unknown was passed through an ion-exchange column that retards Co²⁺ more than Ni²⁺. The Ni²⁺ that passed through the column was treated with 25.00 mL of 0.038 72 M EDTA and required 25.63 mL of 0.021 27 M Zn²⁺ for back titration. The Co²⁺ emerged from the column later. It, too, was treated with 25.00 mL of 0.038 72 M EDTA. How many milliliters of 0.021 27 M Zn²⁺ will be required for back titration?

Short answer

21.4 milliliters of 0.021 27 M Zn²⁺ will be required for back titration

Step-by-step solution

Given Information

Amount of unknown solution containing Ni²⁺ and Co²⁺ = 1 mL and 2 mL

Amount of EDTA added = 25.0 mL of 0.03872 M EDTA

Amount of Zn²⁺ required for back titration to obtain xylenol orange end point (1 mL of unknown sample) = 23.54mL of 0.02127 M Zn²⁺

Amount of Zn²⁺ required for back titration to obtain xylenol orange end point (2 mL of unknown sample) = 25.63 mL of 0.02127 M Zn²⁺

Determine the total amount of Ni2+ and Co2+ treated

For 1 mL of unknown Solution

Total amount of EDTA required

$$\begin{gathered} =(25.00\mathrm{mL})(0.03872\mathrm{MEDTA}) \\ =0.968\mathrm{mmol} \end{gathered}$$

Amount of Zn²⁺ required for back titration to reach xylenol orange end point

$$\begin{gathered} =(23.54\mathrm{mL})(0.02127{\mathrm{MZn}}^{2+} \\ =0.5\mathrm{mmol} \end{gathered}$$

Therefore, the total amount of Co²⁺ and Ni²⁺

$$\begin{gathered} =(0.968-0.5)\mathrm{mmol} \\ =0.468\mathrm{mmol} \end{gathered}$$

For 2 mL of unknown Solution

Total amount of EDTA required

$$\begin{gathered} =(25.00\mathrm{mL})(0.03872\mathrm{MEDTA}) \\ =0.968\mathrm{mmol} \end{gathered}$$

Amount of Zn²⁺ required for back titration to reach xylenol orange end point

$$\begin{gathered} =(25.63\mathrm{mL})(0.02127{\mathrm{MZn}}^{2+} \\ =0.545\mathrm{mmol} \end{gathered}$$

Therefore, the amount of Ni²⁺ in 2.0 mL

$$\begin{gathered} =(0.968-0.545)\mathrm{mmol} \\ =0.423\mathrm{mmol} \end{gathered}$$

The amount of Co²⁺ in 2mL unknown

$$\begin{gathered} =\left[2\right(0.468)-0.423]\mathrm{mmol} \\ =[0.936-0.423]\mathrm{mmol} \\ =0.513\mathrm{mmol} \end{gathered}$$

Determine the amount of Zn2+ required

Amount of Co²⁺ in 2mL unknown = Amount of EDTA reacted with Co²⁺

Therefore, amount of EDTA remained

$$\begin{gathered} =(0.968-0.513)\mathrm{mmol} \\ =0.455\mathrm{mmol} \end{gathered}$$

The excess EDTA will be back titrated using Zn²⁺ solution.

Therefore, the amount of Zn²⁺ needed

$$\begin{gathered} =\frac{0.455\mathrm{mmol}}{0.02127\mathrm{mmol}/\mathrm{mL}} \\ =21.4\mathrm{mL} \end{gathered}$$

Therefore, 21.4 mL of 0.021 27 M Zn²⁺ will be required for back titration