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Chapter 16: Problem 92

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 \(0.0010 M\) in \(\mathrm{Cr}^{3+}\) and \(0.050 M\) in $\mathrm{H}_{2} \mathrm{EDTA}^{2-}\( and buffered at \)\mathrm{pH}=6.00.$

Short Answer

Expert verified
The equilibrium concentration of Cr³⁺ in the given reaction is approximately 0.0010 M.

Step by step solution

01

Write down the equilibrium constant expression

For the given reaction: K = \(\frac{[CrEDTA^-][H^+]^2}{[Cr^{3+}][H_2EDTA^{2-}]}\)
02

Set up an ICE table to determine concentrations

We can set up an ICE table as follows: Cr^3+ + H₂EDTA²⁻ ↔ CrEDTA⁻ + 2H⁺ Initial: 0.0010 M 0.050 M 0 M Given Change: -x -x +x +2x Equilibrium: 0.0010-x 0.050-x x 2x
03

Calculate H⁺ concentration from given pH

Since pH = 6.00: -log[H^⁺] = 6.00 We can solve for the H⁺ concentration: H⁺ = \(10^{-6}\) M As it is a buffered solution, [H⁺] doesn't change much during the reaction. Thus, [H⁺] ≈ 2x ≈ \(10^{-6}\) M.
04

Substitute concentrations into the equilibrium expression

Substitute these concentrations into the equilibrium expression: \(1.0 \times 10^{23}\) = \(\frac{x(10^{-6})^2}{(0.0010-x)(0.050-x)}\)
05

Solve for x (concentration of CrEDTA⁻ and Cr³⁺ at equilibrium)

Assuming x is very small compared to 0.0010 and 0.050, we can drop x from the denominator: \(1.0 \times 10^{23}\) = \(\frac{x(10^{-6})^2}{(0.0010)(0.050)}\) Solve for x: x ≈ \(5.0 \times 10^{-20}\) M
06

Calculate equilibrium concentration of Cr³⁺

At equilibrium, [Cr³⁺] = 0.0010 - x: [Cr³⁺] ≈ 0.0010 M - \(5.0 \times 10^{-20}\) M ≈ 0.0010 M

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