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Chapter 20: Problem 84

EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt $\mathrm{Na}_{2} \mathrm{H}_{2} \mathrm{EDTA}$ , are also used to treat heavy metal poisoning. The equilibrium constant for the following reaction is \(6.7 \times 10^{21} :\) Calculate \(\left[\mathrm{Pb}^{2+}\right]\) at equilibrium in a solution originally 0.0050 \(\mathrm{M}\) in \(\mathrm{Pb}^{2+}, 0.075 M\) in \(\mathrm{H}_{2} \mathrm{EDTA}^{2-},\) and buffered at \(\mathrm{pH}=7.00 .\)

Short Answer

Expert verified
The equilibrium concentration of \(\mathrm{Pb}^{2+}\) in the given solution is \(4.95 \times 10^{-3} \text{ M}\).

Step by step solution

01

Write down the initial concentrations

The initial concentrations are given as follows: \(\left[\mathrm{Pb}^{2+}\right]_i = 0.0050 \text{ M}\) \(\left[\mathrm{H}_{2} \mathrm{EDTA}^{2-}\right]_i = 0.075 \text{ M}\)
02

Write the ICE table

Let \(x\) be the change in molar concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) at equilibrium. Then the ICE table looks like: $\qquad\qquad\begin{array}{|c|c|c|c|c|} \hline \textbf{ } &\textbf{Pb}^{2+} & \textbf{H}_{2} \mathrm{EDTA}^{2-} & \textbf{PbEDTA}^{4-} & \textbf{H}^{+} \\ \hline \textbf{Initial}& 0.0050 & 0.075 & 0 & \\ \hline \textbf{Change}& -x & -x & x & 2x \\ \hline \textbf{Equilibrium}& 0.0050-x & 0.075-x & x & 2x \\ \hline \end{array}$
03

Write the expression for the equilibrium constant

From the given data, we have the equilibrium constant, \(K = 6.7 \times 10^{21}\). The expression for the equilibrium constant is: \[K = \frac{[\mathrm{PbEDTA}^{4-}][\mathrm{H}^+]^2}{[\mathrm{Pb}^{2+}][\mathrm{H}_{2} \mathrm{EDTA}^{2-}]} \]
04

Substitute the equilibrium concentrations in the constant expression

Substitute the equilibrium concentrations from the ICE table into the expression for the equilibrium constant: \[6.7 \times 10^{21} = \frac{x(2x)^2}{(0.0050-x)(0.075-x)}\]
05

Solve for x

Solve the equation to find the equilibrium concentration of \(\mathrm{Pb}^{2+}\): Since the equilibrium constant is very large value, we can make the following assumption: The change in concentrations of the reactants, i.e. \(x\), is very small compared to the initial concentrations, so we can approximate that \(0.0050-x \approx 0.0050\) and \(0.075-x \approx 0.075\). Thus, the equation simplifies to: \[6.7 \times 10^{21} = \frac{x(2x)^2}{(0.0050)(0.075)}\] Solve for \(x\): \(x = 5.27 \times 10^{-6}\) Since \(x\) represents the equilibrium concentration of \(\mathrm{PbEDTA}^{4-}\) and the decrease in the concentration of \(\mathrm{Pb}^{2+}\): \(\left[\mathrm{Pb}^{2+}\right]_{eq} = 0.0050-x\)
06

Calculate the equilibrium concentration of \(\mathrm{Pb}^{2+}\)

Substitute the value of \(x\) in the expression for equilibrium concentration of \(\mathrm{Pb}^{2+}\): \(\left[\mathrm{Pb}^{2+}\right]_{eq} = 0.0050 - 5.27 \times 10^{-6} = 4.95 \times 10^{-3} \text{ M}\) So, at equilibrium, the concentration of \(\mathrm{Pb}^{2+}\) is \(4.95 \times 10^{-3} \text{ M}\).

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