Question

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

Short answer

(e) For 55 mL the value ofpCu²⁺is 17.69.

Step-by-step solution

Introduction

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

Determine equilibrium constant

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

Equivalence point=50 mL

Determine the value of pCu2+

Past equivalence point (55mL) we can calculate

$\begin{array}{c}EDTA=\left(\frac{5}{105}\times 0.001M\right)\\ =4.76\times {10}^{-5}M\\ Cu{Y}^{2-}=\left(\frac{50}{105}\times 0.001M\right)\\ =4.76\times {10}^{-4}M\end{array}$

$\begin{array}{c}{K}_f^{\text{'}}=\frac{\left[Cu{Y}^{2-}\right]}{\left[C{u}^{2+}\right]\left[EDTA\right]}\\ 4.88\times {10}^{18}=\frac{\left[4.76\times {10}^{-4}\right]}{\left[C{u}^{2+}\right]\left[4.76\times {10}^{-5}\right]}\\ C{u}^{2+}=2.05\times {10}^{-18}M\end{array}$

Therefore, the value of pCu²⁺

^{$\begin{array}{c}pC{u}^{2+}=-\log \left(C{u}^{2+}\right)\\ =-\log \left(2.05\times {10}^{-18}\right)\\ =17.69\end{array}$}