The equilibrium constant for the following reaction is \(1.0 \times 10^{23}:\) $$\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q) \rightleftharpoons \mathrm{CrEDTA}^{-}(a q)+2 \mathrm{H}^{+}(a q)$$ 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 used to treat heavy metal poisoning. Calculate \(\left[\mathrm{Cr}^{3+}\right]\) at equilibrium in a solution originally \(0.0010 \mathrm{M}\) in \(\mathrm{Cr}^{3+}\) and \(0.050 M\) in \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) and buffered at \(\mathrm{pH}=6.00\).

First, we need to write down the balanced chemical equation given in the exercise: \[\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{EDTA}^{2-}(a q) \rightleftharpoons \mathrm{CrEDTA}^{-}(a q)+2 \mathrm{H}^{+}(a q)\] Then, we write the expression for the equilibrium constant, \(K_{eq}\): \[K_{eq} = \frac{[\mathrm{CrEDTA}^{-}][\mathrm{H^{+}}]^{2}}{[\mathrm{Cr}^{3+}][\mathrm{H}_{2}\mathrm{EDTA}^{2-}]}\] The equilibrium constant value is given as \(1.0 \times 10^{23}\).

In order to keep track of the concentrations of each species during the reaction, we can set up a RICE table: | | Cr³⁺ | H₂EDTA²⁻ | CrEDTA⁻ | H⁺ | |------|------|--------|--------|----| | R | | | | | | I | 0.0010 | 0.050 | 0 | \(10^{-6}\) | | C | -x | -x | +x | +2x | | E |0.0010 - x | 0.050 - x | x | \(10^{-6}\) + 2x | Here, RICE stands for Reaction, Initial, Change, and Equilibrium. The initial concentrations of Cr³⁺ and H₂EDTA²⁻ are given as 0.0010 M and 0.050 M, respectively. The initial concentration of CrEDTA⁻ is 0 M, as it has not formed yet. Since the pH is given as 6.00, we know that the initial concentration of H⁺ = \(10^{-6}\) M. During the reaction, the concentrations of the reactants Cr³⁺ and H₂EDTA²⁻ will decrease by an amount x and the concentrations of the products CrEDTA⁻ and H⁺ will increase by x and 2x, respectively.

Now we can substitute the equilibrium concentrations given in the RICE table into the expression for the equilibrium constant: \begin{align*} 1.0 \times 10^{23} &= \frac{(x)[10^{-6} + 2x]^{2}}{(0.0010 - x)(0.050 - x)} \end{align*} Since \(K_{eq}\) is very large, we can assume that the reaction will go almost to completion, so x is close to the initial concentration of Cr³⁺ (0.0010 M). Therefore, we can simplify the expression by assuming that \(0.0010 - x \approx 0.0010\) and \(0.050 - x \approx 0.050\): \begin{align*} 1.0 \times 10^{23} &\approx \frac{(x)[10^{-6} + 2x]^{2}}{(0.0010)(0.050)} \end{align*}

Now we need to solve the simplified equation for x: \begin{align*} x[10^{-6} + 2x]^{2} &= 1.0 \times 10^{23} \times 0.0010 \times 0.050 \\ x(10^{-6} + 2x)^{2}& \approx 5.0 \times 10^{20} \end{align*} To simplify the equation, we can also assume that \(2x \ll 10^{-6}\) since the value of x will be very small compared to \(10^{-6}\). This results in: \begin{align*} x(10^{-6})^2 &\approx 5.0 \times 10^{20} \\ x &\approx \frac{5.0 \times 10^{20}}{(10^{-6})^{2}} \\ x &\approx 5.0 \times 10^{-8} \end{align*}

We know that the equilibrium concentration of Cr³⁺ is given by \(0.0010 - x\), so: \[[\mathrm{Cr}^{3+}]_{eq} = 0.0010 - 5.0 \times 10^{-8} \approx 0.0010\ \mathrm{M}\] Thus, the equilibrium concentration of Cr³⁺ is approximately 0.0010 M.