Q19.150 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Ethylenediaminetetraacetic acid (abbreviated ${\mathbf{H}}_{\mathbf{4}}\mathbf{EDTA}$) is a tetraprotic acid. Its salts are used to treat toxic metal poisoning by forming soluble complex ions that are then excreted. Because ${\mathbf{EDTA}}^{\mathbf{4}\mathbf{-}}$also binds essential calcium ions, it is often administered as the calcium disodium salt. For example, when ${\mathbf{Na}}_{\mathbf{2}}\mathbf{Ca}$ (EDTA) is given to a patient ${\mathbf{\left[}\mathbf{Ca}\mathbf{\right(}\mathbf{EDTA}\mathbf{\left)}\mathbf{\right]}}^{\mathbf{2}\mathbf{-}}$, the  ions react with circulating  ions and the metal ions are exchanged:

$\begin{array}{l}{\mathbf{\left[}\mathbf{Ca}\mathbf{\right(}\mathbf{EDTA}\mathbf{\left)}\mathbf{\right]}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{P}{\mathbf{b}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\\ {\mathbf{\left[}\mathbf{Pb}\mathbf{\right(}\mathbf{EDTA}\mathbf{\left)}\mathbf{\right]}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{C}{\mathbf{a}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{ }{\mathbf{K}}_{\mathbf{c}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\end{array}$

A child has a dangerous blood lead level of $\mathbf{120}\mathbf{\mu g}\mathbf{/}\mathbf{100}\mathbf{mL}$. If the child is administered $\mathbf{100}\mathbf{ }\mathbf{mL}$ of $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{ }\mathbf{M}\mathbf{ }{\mathbf{NaNa}}_{\mathbf{2}}$ (EDTA), assuming the exchange reaction and excretion process are  $\mathbf{100}\mathbf{\%}$efficient, what is the final concentration of ${\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}$in  $\mathbf{\mu g}\mathbf{/}\mathbf{100}\mathbf{mL}$ blood? (Total blood volume is $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{L}$.)

Step-by-Step Solution

Answer

The concentration of ${\mathrm{Pb}}^{2+}$ remaining in blood is $3.9126\times {10}^{-9} \mathrm{\mu g}/100\mathrm{mL}$

Step 1: Definition of Concept.

Capacity: The amount of heat energy required to raise the temperature of a body by a certain amount is known as heat capacity. The amount of heat in joules required to raise the temperature 1 Kelvin is known as heat capacity (symbol: C) in SI units.

Step 2: Find the final concentration of Pb 2 + .

Considering the given information:

${\left[\mathrm{Ca}\right(\mathrm{EDTA}\left)\right]}^{2-}\left(\mathrm{aq}\right)+\mathrm{P}{\mathrm{b}}^{2+}\left(\mathrm{aq}\right)\rightleftarrows \left[\mathrm{Pb}{\left({\mathrm{EDTA}}^{2-}\right]}^{2-}\left(\mathrm{aq}\right)+\mathrm{C}{\mathrm{a}}^{2+}\left(\mathrm{aq}\right)\right.$

${\mathrm{K}}_{\mathrm{c}}=2.5\times {10}^7$

Given that sample of blood contains $120\mathrm{\mu g}/100\mathrm{mL}$ of ${\mathrm{Pb}}^{2+}$.

The total amount of blood in the body is $1.5 \mathrm{L}$.

Amount of ${\mathrm{Na}}_2\mathrm{Ca}$(EDTA) added is $100\mathrm{mL} 0.10\mathrm{M}$.

As a result of the addition, the total volume of blood is now (1.5+.1) $\mathrm{L}=1.6 \mathrm{L}$.

Molecular weight of ${\mathrm{Pb}}^{2+}=207.2\mathrm{g}/\mathrm{mol}$.

$1\mathrm{g}={10}^6\mathrm{\mu g}$

The concentration of  ${\mathrm{Pb}}^{2+}$ ,

$\begin{array}{c}\left[{\mathrm{Pb}}^{2+}\right]=\frac{(120 \mathrm{\mu g}\times 1.5 \mathrm{L})}{0.1 \mathrm{L}}\times \frac{{10}^{-6} \mathrm{g}}{1 \mathrm{gg}}\times \frac{1 \mathrm{mol}}{207.2 \mathrm{g}}\times \frac{1}{1.6 \mathrm{L}}\\ (\mathrm{molarity}\left.=\frac{\mathrm{no}. \mathrm{of} \mathrm{moles} \mathrm{taken}}{\mathrm{volume} \mathrm{of} \mathrm{solution} \mathrm{per} \mathrm{litre}}\right)\\ \left[{\mathrm{Pb}}^{2+}\right]=5.43\times {10}^{-6} \mathrm{M}\end{array}$

$\begin{array}{rcl}{\left[\mathrm{Ca}\right(\mathrm{EDTA}\left)\right]}^{2-}& =& \frac{\mathrm{volume} \mathrm{of} \mathrm{added} \mathrm{solution} \mathrm{x} \mathrm{strength} \mathrm{of} \mathrm{added} \mathrm{solution}}{\mathrm{volume} \mathrm{of} \mathrm{blood}}\\ {\left[\mathrm{Ca}\right(\mathrm{EDTA}\left)\right]}^{2-}& =& =\frac{0.1\mathrm{L}\times 0.10\mathrm{M}}{1.6\mathrm{L}}\\ & =& 6.25\times {10}^{-3}\mathrm{M}\end{array}$

According to the equation above,

Initially,

$\begin{array}{c}{\left[\mathrm{Ca}\right(\mathrm{EDTA}\left)\right]}^{2-}=6.25\times {10}^{-3}\mathrm{M}\\ \left[{\mathrm{Pb}}^{2+}\right]=5.43\times {10}^{-6}\mathrm{M}\\ {\left[\mathrm{Pb}\right(\mathrm{EDTA}\left)\right]}^{2-}=0\mathrm{M}\\ \left[{\mathrm{Ca}}^{2+}\right]=0\mathrm{M}\end{array}$

At equilibrium, the amount of  $\left[{\mathrm{Pb}}^{2+}\right]$ left in the reaction is 'x’.

$\begin{array}{c} {\left[\mathrm{Ca}\right(\mathrm{EDTA}\left)\right]}^{2-}=\left(6.25\times {10}^{-3}-\left(5.43\times {10}^{-6}-\mathrm{x}\right)\right)\\ =\left(6.24\times {10}^{-3}+\mathrm{x}\right)\end{array}$

Remaining $\left[{\mathrm{Pb}}^{2+}\right]=\left(5.43\times {10}^{-6}-\mathrm{x}\right)$

$\begin{array}{c}{\left[\mathrm{Pb}\right(\mathrm{EDTA}\left)\right]}^{2-}=\left(5.43\times {10}^{-6}-\mathrm{x}\right)\\ \left[{\mathrm{Ca}}^{2+}\right]=\left(5.43\times {10}^{-6}-\mathrm{x}\right)\end{array}$

By substituting the values in the ${\mathrm{K}}_{\mathrm{c}}$ equation,

$\begin{array}{l}{\mathrm{K}}_{\mathrm{c}}=\frac{{\left[\mathrm{Pb}\right(\mathrm{EDTA}\left)\right]}^2\times \left[{\mathrm{Ca}}^{2+}\right]}{{\left[\mathrm{Ca}\right(\mathrm{EDTA}\left)\right]}^2\times \left[{\mathrm{Pb}}^{2+}\right]}\\ {\mathrm{K}}_{\mathrm{c}}=\frac{\left(5.43\times {10}^{-6}-\mathrm{x}\right)\times \left(5.43\times {10}^{-6}-\mathrm{x}\right)}{\left(6.24\times {10}^{-3}+\mathrm{x}\right)\times \mathrm{x}}\end{array}$

$\{\mathrm{x}<<1$,so for making calculations easy neglect X}

Given,${\mathrm{K}}_{\mathrm{c}}=2.5\times {10}^7$

$\therefore \mathrm{X}=1.89\times {10}^{-16}\mathrm{M}$

As a result, the remaining concentration of ${\mathrm{Pb}}^{2+}$ in blood is,

$\begin{array}{c}{\mathrm{Pb}}^{2+}=\frac{\left(1.89\times {10}^{-16}\mathrm{mol}\times 100\mathrm{ml}\right)}{1000\mathrm{ml}}\times \frac{207.2\mathrm{g}}{1\mathrm{mol}}\times \frac{1\mathrm{\mu g}}{{10}^{-6}\mathrm{g}}\\ =3.9126\times {10}^{-9}\mathrm{\mu g}\end{array}$

Hence, the concentration of ${\mathrm{Pb}}^{2+}$remaining in blood is $3.9126\times {10}^{-9}\mathrm{\mu g}/100\mathrm{mL}$.