If back titration required 13.00 mL Zn²⁺, what was the original concentration of Ni²⁺?

Analysis of Ni²⁺ can be done by a back titration using standard Zn²⁺ at pH 5.5 with xylenol orange indicator. A solution contains 25.00 mL of Ni²⁺ in dilute HCl. It was titrated with 25.00 mL of 0.05283 M Na₂EDTA. The solution turns yellow when a few drops of indicator are added. Titration with 0.02299 M Zn²⁺requires 13 mL to react. We need to calculate original concentration of Ni

The unknown was treated with 25.00 mL of 0.05283 M EDTA,

The number of moles of EDTA=(25.00 mL)(0.052 83 M) = 1.3208 mmol of EDTA.

Back titration required 13 mL Zn²⁺

Number of moles involved

$\left(13mL\right)\times (0.02299M)=0.2988mmolZ{n}^{2+}$

$$\begin{gathered} themolesofZ{n}^{2+}inthesecondreaction+themolesofN{i}^{2+}inthefirstreaction=thetotalmolesofs\tan dardEDTA \\ 0.2988+x=1.3208 \\ x=1.022mmolN{i}^{2+} \end{gathered}$$

The concentration of Ni²⁺

^{$$\begin{gathered} =\frac{1.022mmol}{25mL} \\ =0.041M \end{gathered}$$}

The concentration of Ni²⁺ is 0.041M