Chapter 23, Problem 23.70QP

Interpretation Introduction

Interpretation: The concentration of lead ion ${\text{[Pb]}}^{\text{2+}}$ in the given equilibrium to be calculated.

Concept Introduction:

Spectrochemical series: The list of ligands arranged in an ascending order of $\text{(Δ)}$(the splitting of d-orbitals in presence of various ligands).

$\begin{array}{l}\stackrel{{}^{{\text{I}}^{\text{-}}\text{<}{\text{Br}}^{\text{-}}\text{<}{\text{SCN}}^{\text{-}}\text{<}{\text{Cl}}^{\text{-}}\text{<}{\text{S}}^{\text{2-}}\text{<}{\text{F}}^{\text{-}}\text{<}{\text{OH}}^{\text{-}}\text{<}{\text{O}}^{\text{2-}}\text{<}{\text{H}}_{\text{2}}\text{O}\text{<}{\text{NCS}}^{\text{-}}\text{<}{\text{edta}}^{\text{4-}}\text{<}{\text{NH}}_{\text{3}}\text{< en}\text{<}{\text{NO}}^{\text{2-}}\text{<}{\text{CN}}^{\text{-}}\text{<}\text{CO}}}{\to }\\ {}_{{}^{{}^{\begin{array}{l}\text{weak-field}\text{increasing(Δ)}\text{strong-field}\\ \text{ligands}\text{ligands}\end{array}}}}\end{array}$

Crystal field splitting: The energy gap between the splitting of d-orbitals of the metal ion in presence of ligands is known as the crystal field splitting $\text{(Δ)}$. The magnitude of $\text{(Δ)}$ is depends on the nature of metal ions and the ligands.

Equilibrium constant: The ratio of equilibrium concentration of products by the equilibrium concentration of the reactants.

${\text{K}}_{\text{f}}\text{=}\frac{{\text{[Products]}}_{\text{eq}}}{{\text{[reactants]}}_{\text{eq}}}$

To Identify: The concentration of lead ion ${\text{[Pb(EDTA)}}^{\text{2-}}\text{]}$ in the given equilibrium to be calculated.

Answer to Problem 23.70QP

The calculated equilibrium concentration of ${\text{[Pb(EDTA)}}^{\text{2-}}\text{]}$ is $\text{0}\text{.0 M}$

Explanation of Solution

Find the concentration of lead ion ${\text{[Pb(EDTA)}}^{\text{2-}}\text{]}$.

$\begin{array}{l}\underset{\_}{\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{[Pb]}}^{\text{2+}}\text{+}{\text{EDTA}}^{\text{4-}}\rightleftarrows {\text{[Pb(EDTA)]}}^{\text{2-}}\\ \text{Initial(M):}\text{1}{\text{.0×10}}^{\text{-3}}\text{2}{\text{.0×10}}^{\text{-3}}\text{0}\\ \text{Change(M):}\text{-x}\text{-x}\text{+x}\end{array}}\\ \text{Equilibrium(M): 1}{\text{.0×10}}^{\text{-3}}\text{-x}\text{2}{\text{.0×10}}^{\text{-3}}\text{-x}\text{x}\end{array}}\\ {\text{K}}_{\text{f}}\text{=}\frac{{{\text{[Pb(EDTA)}}^{\text{2-}}\text{]}}_{\text{eq}}}{{\text{[Pb]}}^{\text{2+}}{}_{\text{eq}}{\text{ [EDTA}}^{\text{4-}}{\text{]}}_{\text{eq}}}\text{=1}{\text{.0×10}}^{\text{18}}\\ \text{1}{\text{.0×10}}^{\text{18}}\text{=}\frac{\text{x}}{\text{(1}{\text{.0×10}}^{\text{-3}}\text{-x}\text{)(2}{\text{.0×10}}^{\text{-3}}\text{-x)}}\\ \text{1}{\text{.0×10}}^{\text{18}}\text{=}\frac{\text{x}}{\text{(2}{\text{.0×10}}^{\text{-6}}\text{)-(3}{\text{.0×10}}^{\text{-3}}{\text{)x+x}}^{\text{2}}}\\ \text{0}\text{=}\text{(1}{\text{.0×10}}^{\text{18}}{\text{)x}}^{\text{2}}\text{-(3}{\text{.0×10}}^{\text{15}}\text{)x+(2}{\text{.0×10}}^{\text{12}}\text{)}\\ \text{use the quadratic equation to solve x:}\\ \text{x}\text{=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}\\ \text{=}\frac{\text{(3}{\text{.0×10}}^{\text{15}}\text{)±(1}{\text{.0×10}}^{\text{15}}\text{)}}{\text{(2}{\text{.0×10}}^{\text{18}}\text{)}}\\ \text{=}\text{2}{\text{.0×10}}^{\text{-3}}\text{or}\text{1}{\text{.0×10}}^{\text{-3}}\end{array}$

The equilibrium concentration of ${\text{[Pb]}}^{\text{2+}}$ is $\text{1}{\text{.0×10}}^{\text{-3}}\text{-x}$, the first answer is not possible. Thus, the calculated equilibrium concentration of ${\text{[Pb(EDTA)}}^{\text{2-}}\text{]}$ is $\text{0}\text{.0 M}$

Conclusion

The concentration of lead ion ${\text{[Pb]}}^{\text{2+}}$ in the given equilibrium is calculated.