Chapter 23, Problem 23.90QA

Interpretation Introduction

To find:

The concentration ratio of $\raisebox{1ex}{$ \left[P{b}^{+2}\right] \left(aq\right)$}\!\left/ \!\raisebox{-1ex}{$[P{b}^{+2}-EDTA]$}\right.$

Answer to Problem 23.90QA

Solution:

The concentration ratio of free $P{b}^{2+}\left(aq\right)$ in the blood to the much toxic $Pb{\left(EDTA\right)}^{-2}$ complex is $2.0 \times {10}^{-11}$.

Explanation of Solution

1) Concept:

The complexation reaction of free lead in the blood with the EDTA solution to form $Pb{\left(EDTA\right)}^{-2}$ complex is

$P{b}^{+2} \left(aq\right)+EDT{A}^{-4} \left(aq\right) \to Pb{\left(EDTA\right)}^{-2} \left(aq\right)$

The formation constant of the $Pb{\left(EDTA\right)}^{-2}$ complex is written as

$K_f= \frac{\left[Pb{\left(EDTA\right)}^{-2}\right]}{\left[P{b}^{+2}\right]\left[EDT{A}^{-4}\right]}$

Since we need to find the ratio $\frac{\left[P{b}^{+2}\right]}{\left[Pb{\left(EDTA\right)}^{-2}\right]}$, we can calculate it from the above relation as we have been provided with the value of formation constant of $Pb{\left(EDTA\right)}^{-2}$ and concentration of $EDTA, i.e., \left[EDT{A}^{-4}\right]$.

2) Formula:

$K_f= \frac{\left[Pb{\left(EDTA\right)}^{-2}\right]}{\left[P{b}^{+2}\right]\left[EDT{A}^{-4}\right]}$

3) Given:

i) Formation constant $K_f=2.0 \times {10}^{18} M^{-1}$

ii) Concentration of EDTA solution $\left[EDT{A}^{-4}\right]=2.5 \times {10}^{-8} M$

4) Calculations:

We know that formation constant of $Pb{\left(EDTA\right)}^{-2}$ complex is

$K_f= \frac{\left[Pb{\left(EDTA\right)}^{-2}\right]}{\left[P{b}^{+2}\right]\left[EDT{A}^{-4}\right]}$

$\therefore \frac{\left[Pb{\left(EDTA\right)}^{-2}\right]}{\left[P{b}^{+2}\right]}= K_f \times \left[EDT{A}^{-4}\right]$

Thus, the concentration ratio of free lead in the blood to the much less toxic $Pb{\left(EDTA\right)}^{-2}$ complex is

$\frac{\left[P{b}^{+2}\right]}{\left[Pb{\left(EDTA\right)}^{-2}\right]}= \frac{1}{K_f \times \left[EDT{A}^{-4}\right]}=\frac{1}{2.0 \times {10}^{18} M^{-1} \times 2.5 \times {10}^{-8} M}$

$\therefore \frac{\left[P{b}^{+2}\right]}{\left[Pb{\left(EDTA\right)}^{-2}\right]}=\frac{1}{5.0 \times {10}^{10} }=2\times {10}^{-11}$

Therefore, the concentration ratio is $2\times {10}^{-11}$.

Conclusion:

The formation constant for $Pb{\left(EDTA\right)}^{-2}$ complex ion is quite large, so that a small concentration of EDTA is enough to form the complex, which lowers the toxicity due to free lead ions in the body.