- #1

TheSodesa

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## Homework Statement

**1. The question**

EthyleneDiamineTetra-acetate(##EDTA^{4-}##) is used in chemical analysis as a complex-former. ##EDTA^{4-}## is also used to treat heavy metal poisoning, as it forms water soluble complexes with them which can then be easily removed by the body. The reaction of ##EDTA^{4-}## with lead is:

$$Pb^{2+}(aq) + EDTA^{4-}(aq) \rightleftharpoons PbEDTA^{2-}(aq), K = 1.1 \times 10^{18}$$

0.010 mol of ##Pb(NO_3)_2##-solution is added to 1 liter ##(dm^3)## of buffered water solution (pH = 13), whose ##Na_4 EDTA## concentration is 0.050M. Does the added lead precipitate in the form of ##Pb(OH)_2##-salt?

\begin{array}{|l|c|}

Some \ values\\ \hline

K_s(Pb(OH)_2) & 1.2 \times 10^{-15}\\

K & 1.1\times 10^{18}\\

n_0 (Pb(NO_3)_2) & 0.010 mol\\

[Na_4 (EDTA)]_0 & 0.050M\\

pH & 13

\end{array}

## Homework Equations

If our chemical reaction is

$$aA + bB \rightleftharpoons cC + dD $$

then our equilibrium constant

$$K = \frac{[A]^a [ B ]^b}{[C]^c[D]^d} $$

The solubility constant would be

$$K_s = [ C ]^c [D]^d$$

since the activity ##a_s## of the pure solid on the left side of the reaction equation is 1.

## The Attempt at a Solution

I managed to solve this, I just need a second opinion on whether my solution is correct.

The lead is going to precipitate if the reaction quotient

$$Q > K_s$$

To see if this holds, I need to find out the concentration of lead in the solution. I'm going to assume that the added ##Pb(NO_3)_2## dissolves completely and then reacts with the ##EDTA^{4-}##, before the remaining lead from

*this*reaction then reacts with the ##OH^-##-ions. Below is a table of the first reaction.

\begin{array}{|l|c|}

concentration & Pb^{2+} & EDTA^{4-} & PbEDTA^{2-}\\ \hline

beginning & 0.01 & 0.05 & 0\\

end & 0.01-x & 0.05-x & x

\end{array}

We can then solve for ##x## using ##K##:

\begin{align*}

K

&= \frac{x}{(0.01-x)(0.05-x)}\\

\iff\\

&(0.01-x)(0.05-x)K = x\\

\iff\\

&\frac{1}{2000}K-0.06Kx + Kx^2 = x\\

\iff\\

& Kx^2 -(0.06K+1)x + \frac{1}{2000}K = 0

\end{align*}

Solving for ##x##:

\begin{align*}

x

&= \frac{(0.06K+1) \pm \sqrt{(0.06K+1)^2 - 4K\times \frac{K}{2000}})}{2K}\\

&=

\begin{cases}

1/20\\

1/100

\end{cases}

\end{align*}

Now ##1/20 = 0.05## won't do, since ##(0.01-x)## would be negative. Plugging in ##x = 0.01## gives us the concentration of lead after the reaction which is ##[Pb]_1 = (0.01 - 0.01) = 0##. Therefore at these concentrations there is no lead left over to react with the ##OH^-##-ions, and there can be no precipitate.

This is a desirable result, considering the use of EDTA in medicine, but this seems too easy. I must be missing something or making wrong assumptions.

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