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Chapter 22: Problem 24

The trichloride ion, \(\mathrm{Cl}_{3}^{-1}\), is not very stable in aqueous solution. The equilibrium constant for the following dissociation reaction is 5.5 at \(25^{\circ} \mathrm{C}\) : $$\mathrm{Cl}_{3}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Cl}_{2}(\mathrm{aq})$$ (a) Draw a Lewis structure for the \(\mathrm{Cl}_{3}^{-}\) ion and predict the geometry. (b) Calculate the equilibrium concentration of \(\mathrm{Cl}_{3}^{-}\) if 0.0010 moles each of \(\mathrm{KCl}\) and \(\mathrm{Cl}_{2}\) are dissolved in water at \(25^{\circ} \mathrm{C}\) to make \(1.0 \mathrm{L}\) of solution.

Short Answer

Expert verified
The predicted geometry of \(\mathrm{Cl}_{3}^{-}\) is linear. The equilibrium concentration of \(\mathrm{Cl}_{3}^{-}\) can be calculated by solving the given equilibrium expression.

Step by step solution

01

Draw the Lewis structure of \(\mathrm{Cl}_{3}^{-}\)

Start by counting the total number of valence electrons. Chlorine has 7 valence electrons, but since there are three, we have 21. The extra electron given by the ion charge leads to 22 electrons. As per the octet rule, each Cl atom needs 8 electrons. Hence, each of the two chlorine atoms forms a single bond with the central chlorine atom to complete its octet, taking up 4 electrons. The remaining 18 electrons are distributed as lone pairs. 12 go to the two bonding chlorine atoms (6 each), and the remaining 6 go to the central atom. Thus, the central chlorine atom has a formal negative charge, in accordance with the -1 charge of the overall ion.
02

Predict the geometry of \(\mathrm{Cl}_{3}^{-}\)

The geometry of a molecule/ion can be predicted using the VSEPR (Valence Shell Electron Pair Repulsion) theory that essentially posits that electron groups (including lone pair electrons and bonding electrons) tend to repel each other and will thus orient themselves as far apart as possible. In Cl3-, the central Cl has 5 electron groups (3 lone pairs and 2 atoms), implying a 'Trigonal Bipyramidal' shape for the 5 electron groups. However, when considering atoms only, they occupy 'Linear' positions.
03

Calculate initial concentrations

Cl3- has zero initial concentration as it has yet to form. For Cl- and Cl2, after KCl and Cl2 dissolve, their concentrations will be the amount of moles in 1 L, hence their initial concentrations are both 0.0010 M.
04

Write the expression for the equilibrium concentrations

Consider the dissociation reaction at equilibrium. Cl3- will form changing its initial concentration by +x. As for Cl- and Cl2, they will decrease by x each from the initial concentration. Hence, the equilibrium concentrations are x M for Cl3-, and (0.0010-x) M for both Cl- and Cl2. x is the variable we need to solve for.
05

Use the given equilibrium constant to solve for x

Given the equilibrium constant (K) as 5.5, we plug the equilibrium concentrations into the expression for K, which will be \([Cl^-][Cl_2]/[Cl_3^-] = K\). Substituting the concentrations gives \((0.0010-x)*(0.0010-x)/x = 5.5\). Solving this equation for x leads to an equilibrium concentration of Cl3-

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