91Ó°ÊÓ

Equations and data obtained in order to proceed for calculation are as follows

$\begin{array}{c}C{u}^{2+}+Y^{4−}⇌Cu{Y}^{2−}\\ K_f={10}^{18.78}=6.03×{10}^{18}\\ At\text{ â¶Ä‰â¶Ä‰}pH\text{ â¶Ä‰â¶Ä‰}11\text{ â¶Ä‰â¶Ä‰}{\mathrm{Î\pm }}_{Y^{4−}}=0.81\text{ }\left(Table\text{ }12−1\right)\\ \log {\mathrm{β}}_1=3.99\\ \log {\mathrm{β}}_2=7.33\\ \log {\mathrm{β}}_3=10.06\\ \log {\mathrm{β}}_4=12.03\end{array}$

The beta(β) values were obtained from appendix-1 for Cu²⁺ and NH₃

$\begin{array}{c}{\mathrm{Î\pm }}_{C{u}^{2+}}=\frac{1}{1+{\mathrm{β}}_1\left(1.00\right)+{\mathrm{β}}_2{\left(1.00\right)}^2+{\mathrm{β}}_3{\left(1.00\right)}^3+{\mathrm{β}}_4{\left(1.00\right)}^4}\\ =9.23×{10}^{−13}\\ K_f^{\text{'}}={\mathrm{Î\pm }}_{Y^{4−}}{K}_f\\ =0.81×6.03×{10}^{18}\\ =4.88×{10}^{18}\\ K_f^{"}={\mathrm{Î\pm }}_{C{u}^{2+}}×K_f^{\text{'}}\\ =9.23×{10}^{−13}×4.88×{10}^{18}\\ =4.51×{10}^6\end{array}$

Equivalence point=50 mL

At equivalence point (50mL) the following can be written

$\begin{array}{c}{K}_f^{"}=\frac{0.0005−x}{x^2}\\ \frac{0.0005−x}{x^2}=4.51×{10}^6\\ x=1.04×{10}^{−5}M\\ C_{C{u}^{2+}}=1.04×{10}^{−5}M\end{array}$

$\begin{array}{c}\left[C{u}^{2+}\right]={\mathrm{Î\pm }}_{C{u}^{2+}}×C_{C{u}^{2+}}\\ =\left(9.23×{10}^{−13}×1.04×{10}^{−5}\right)M\\ =9.62×{10}^{−18}M\end{array}$

Therefore, the value of pCu²⁺

^{$\begin{array}{c}pC{u}^{2+}=−\log \left(C{u}^{2+}\right)\\ =−\log \left(9.62×{10}^{−18}\right)\\ =17.02\end{array}$}