91Ó°ÊÓ

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.

Considering the given information:

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

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

Given that sample of blood contains $120\mathrm{μ²\mu }/100\mathrm{mL}$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 ${\mathrm{Pb}}^{2+}=207.2\mathrm{g}/\mathrm{mol}$.

$1\mathrm{g}={10}^6\mathrm{μ²\mu }$

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

$\begin{array}{c}\left[{\mathrm{Pb}}^{2+}\right]=\frac{(120\mathrm{μ²\mu }×1.5\mathrm{L})}{0.1\mathrm{L}}×\frac{{10}^{-6}\mathrm{g}}{1\mathrm{gg}}×\frac{1\mathrm{mol}}{207.2\mathrm{g}}×\frac{1}{1.6\mathrm{L}}\\ (\mathrm{molarity}\left.=\frac{\mathrm{no}.\mathrm{of}\mathrm{moles}\mathrm{taken}}{\mathrm{volume}\mathrm{of}\mathrm{solution}\mathrm{per}\mathrm{litre}}\right)\\ \left[{\mathrm{Pb}}^{2+}\right]=5.43×{10}^{-6}\mathrm{M}\end{array}$

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

According to the equation above,

Initially,

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

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

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

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

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

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

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

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

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

$∴\mathrm{X}=1.89×{10}^{-16}\mathrm{M}$

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

$\begin{array}{c}{\mathrm{Pb}}^{2+}=\frac{\left(1.89×{10}^{-16}\mathrm{mol}×100\mathrm{ml}\right)}{1000\mathrm{ml}}×\frac{207.2\mathrm{g}}{1\mathrm{mol}}×\frac{1\mathrm{μ²\mu }}{{10}^{-6}\mathrm{g}}\\ =3.9126×{10}^{-9}\mathrm{μ²\mu }\end{array}$

Hence, the concentration of ${\mathrm{Pb}}^{2+}$remaining in blood is $3.9126×{10}^{-9}\mathrm{μ²\mu }/100\mathrm{mL}$.