Question

EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt $\mathbf{\text{Na}}{}_{\mathbf{\text{2}}}\mathbf{\text{H}}_{\mathbf{\text{2}}}\mathbf{\text{EDTA}}$, are also used to treat heavy metal poisoning. The equilibrium constant for the following reaction is $\mathbf{6}\mathbf{.7}\mathbf{\times }{\mathbf{10}}^{\mathbf{21}}$:

${\mathbf{\text{Pb}}}^{\mathbf{\text{2+}}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}{\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{EDTA}}}^{\mathbf{\text{2-}}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{\rightleftharpoons }{\mathbf{\text{PbEDTA}}}^{\mathbf{\text{2-}}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}{\mathbf{\text{2H}}}^{\mathbf{\text{+}}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}$

Calculate $\mathbf{\left[}{\mathbf{\text{Pb}}}^{\mathbf{\text{2+}}}\mathbf{\right]}$at equilibrium in a solution originally $\mathbf{\text{0.0050M}}$in ${\mathbf{\text{Pb}}}^{\mathbf{\text{2+}}}$,$\mathbf{\text{0.075M}}$ in ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{EDTA}}}^{\mathbf{\text{2-}}}$, and buffered at $\mathbf{\text{pH}}\mathbf{=}\mathbf{\text{7.00}}$.

Short answer

The equilibrium concentration is zero (or close to zero) because the calculation goes beyond the calculator limitation.

Step-by-step solution

Concept Introduction:

When a reversible chemical process is in chemical equilibrium, there is no net change in the amounts of reactants and products. When the products of a chemical reaction interact with the original reactants once they are formed, the reaction is reversible.

Information Provided:

First, write down all information given by exercise as below.

$$\begin{gathered} \text{K}=6.7\times {10}^{21} \\ {c}_1=0.0050\text{M} \\ {c}_2=0.075\text{M} \\ \text{pH}=7 \end{gathered}$$

Equilibrium Equation:

The equilibrium concentration of ${\text{H}}^{\text{+}}$will be determined. After that place an equation for the equilibrium constant. Product equilibrium concentrations will be put in the fraction's numerator and reactant equilibrium concentrations will be put in the denominator.

$\text{pH}=-\text{log}\left(\left[{\text{H}}^{\text{+}}\right]\right)$

$\begin{array}{rcl}\left[{\text{H}}^{\text{+}}\right]& =& {\text{10}}^{\text{-pH}}\\ & =& {\text{10}}^{\text{-7}}\end{array}$

$\begin{array}{rcl}\text{K}& =& \frac{\left[{\text{PbEDTA}}^{\text{2-}}\right]\cdot {\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}{\left[{\text{Pb}}^{\text{2+}}\right]\cdot \left[{\text{H}}_{\text{2}}{\text{EDTA}}^{\text{2-}}\right]}\\ & =& \frac{\text{x}\cdot {\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}{\left({\text{c}}_{\text{1}}\text{-x}\right)\cdot \left({\text{c}}_{\text{2}}\text{-x}\right)}\\ & =& \frac{\text{x}\cdot {\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}{{\text{c}}_{\text{1}}{\text{c}}_{\text{2}}{\text{-c}}_{\text{1}}{\text{x-c}}_{\text{2}}{\text{x+x}}^{\text{2}}}\end{array}$

Here, $x$is equilibrium concentration of ${\text{PbEDTA}}^{\text{2-}}$. Equilibrium concentrations of reactants will be initial concentrations minus equilibrium concentration of product $\left(\text{x}\right)$that was formed from them. Now, continue with solving the previous equation.

$$\begin{gathered} \text{K}=\frac{\text{x}\cdot {\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}}{{\text{c}}_{\text{1}}{\text{c}}_{\text{2}}{\text{-c}}_{\text{1}}{\text{x-c}}_{\text{2}}{\text{x+x}}^{\text{2}}} \\ {\text{Kc}}_{\text{1}}{\text{c}}_{\text{2}}{\text{-Kc}}_{\text{1}}{\text{x-Kc}}_{\text{2}}{\text{x+Kx}}^{\text{2}}\text{-}{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}\text{x}=\text{0} \\ {\text{Kx}}^{\text{2}}\text{-x}\left({\text{Kc}}_{\text{1}}{\text{+Kc}}_{\text{2}}\text{+}{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}\right){\text{+Kc}}_{\text{1}}{\text{c}}_{\text{2}}=\text{0} \end{gathered}$$

Equilibrium value calculation:

Now, solve the quadratic equation as below.

$\begin{array}{rcl}{\text{x}}_{\text{1,2}}& =& \frac{\left({\text{Kc}}_{\text{1}}{\text{+Kc}}_{\text{2}}\text{+}{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}\right)\text{+-}\sqrt{{\left({\text{Kc}}_{\text{1}}{\text{+Kc}}_{\text{2}}\text{+}{\left[{\text{H}}^{\text{+}}\right]}^{\text{2}}\right)}^{\text{2}}{\text{-4KKc}}_{\text{1}}{\text{c}}_{\text{2}}}}{\text{2K}}\\ & =& \text{0.075M,0.005M}\end{array}$

Only take into consideration the second result because the first one is against logic. Therefore, the equilibrium concentration will be,

$\begin{array}{rcl}\left[{\text{Pb}}^{\text{2+}}\right]& =& {\text{c}}_{\text{1}}\text{-x}\\ & =& \text{0.005M-0.005M}\\ & =& \text{0}\end{array}$

Hence, the equilibrium concentration would not be exactly zero because this calculation goes beyond a calculator limitation but, it would be close to zero.