Calculate pCu2+ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu2+ with 0.00100 M EDTA at pH 11.00 in a solution with [NH3] fixed at 1.00 M:

(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL

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

$\begin{array}{c}C{u}^{2+}+Y^{4-}\rightleftharpoons Cu{Y}^{2-}\\ K_f={10}^{18.78}=6.03\times {10}^{18}\\ AtpH11{\alpha }_{Y^{4-}}=0.81\left(Table12-1\right)\\ \log {\beta }_1=3.99\\ \log {\beta }_2=7.33\\ \log {\beta }_3=10.06\\ \log {\beta }_4=12.03\end{array}$

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

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

Equivalence point=50 mL

The concentration of the remaining productcan be calculated using the following equation

=Fraction remaining × Initial concentration × Dilution factor

If 1 mL solution is added then copper concentration will be

$$\begin{gathered} \begin{array}{c}\left[Cc{u}^{2+}\right]=\left(\frac{50-1}{50}\right)\left(0.001M\right)\left(\frac{50}{50+1}\right)\\ =9.61\times {10}^{-4}M\end{array} \\ \begin{array}{c}\left[C{u}^{2+}\right]={\alpha }_{C{u}^{2+}}\times C_{C{u}^{2+}}\\ =\left(9.23\times {10}^{-13}\times 9.61\times {10}^{-4}\right)M\\ =8.87\times {10}^{-16}M\end{array} \end{gathered}$$

Therefore, the value of pCu²⁺

^{$\begin{array}{c}pC{u}^{2+}=-\log \left(C{u}^{2+}\right)\\ =-\log \left(8.87\times {10}^{-16}\right)\\ =15.05\end{array}$}