Problem 94 - Chemistry The Central Science

Question

Silver chloride, \(\mathrm{AgCl}(s)\), is an "insoluble" strong electrolyte. (a) Write the equation for the dissolution of \(\mathrm{AgCl}(s)\) in \(\mathrm{H}_{2} \mathrm{O}(l)\) (b) Write the expression for \(K_{c}\) for the reaction in part (a). (c) Based on the thermochemical data in Appendix \(\mathrm{C}\) and Le Châtelier's principle, predict whether the solubility of \(\mathrm{AgCl}\) in \(\mathrm{H}_{2} \mathrm{O}\) increases or decreases with increasing temperature. (d) The equilibrium constant for the dissolution of \(\mathrm{AgCl}\) in water is \(1.6 \times 10^{-10}\) at \(25^{\circ} \mathrm{C}\). In addition, \(\mathrm{Ag}^{+}(a q)\) can react with \(\mathrm{Cl}^{-}(a q)\) according to the reaction $$\mathrm{Ag}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}_{2}^{-}(a q)$$ where \(K_{c}=1.8 \times 10^{5}\) at \(25^{\circ} \mathrm{C}\). Although \(\mathrm{AgCl}\) is "not soluble" in water, the complex \(\mathrm{AgCl}_{2}^{-}\) is soluble. At \(25^{\circ} \mathrm{C},\) is the solubility of AgCl in a \(0.100 M\) NaCl solution greater than the solubility of AgCl in pure water, due to the formation of soluble \(\mathrm{AgCl}_{2}^{-}\) ions? Or is the \(\mathrm{AgCl}\) solubility in \(0.100 \mathrm{M} \mathrm{NaCl}\) less than in pure water because of a Le Châtelier-type argument? Justify your answer with calculations.

Step-by-Step Solution

Answer

The solubility of silver chloride (AgCl) in water increases with increasing temperature due to its endothermic dissolution process. The formation of a soluble complex, AgCl2-, also increases the solubility of AgCl in a 0.100 M NaCl solution compared to pure water. This is because the concentration of the complex AgCl2- (\(2.88\, x\, 10^{-7}\) M) is much greater than the solubility of AgCl in water (\(1.6\, x\, 10^{-10}\) M).

Step 1: Equation for the dissolution of AgCl in water

The dissolution of silver chloride (AgCl) in water can be represented by the following chemical equation: $$\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q) + \mathrm{Cl}^{-}(a q)$$

Step 2: Expression for \(K_{c}\) for the dissolution of AgCl in water

The chemical equation in step 1 represents an equilibrium reaction. We can now write the expression for the equilibrium constant Kc as the product of the concentrations of the products divided by the reactants' concentrations. Since AgCl is a solid, it doesn't appear in the expression as its concentration doesn't vary during the reaction. $$K_{c} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]$$

Step 3: Predict the solubility of AgCl with increasing temperature using Le Châtelier's principle

To predict the solubility's dependence on temperature, we need to evaluate the reaction's enthalpy change (ΔH). In Appendix C, we find that the enthalpy of formation (ΔHf°) of AgCl(s) is -127.1 kJ/mol. Before we can use Le Châtelier's principle, we need to consider the enthalpy of dissolution. We know that the dissolution of AgCl is endothermic, so it absorbs heat in the process (positive ΔHdissolution). Le Châtelier's principle states that a system in equilibrium will counteract any changes in temperature, pressure, or concentration imposed on it. Since the dissolution of AgCl is endothermic, increasing the temperature will favor the dissolution process, and therefore the solubility of AgCl in water will increase.

Step 4: Analyze the effect of a soluble complex in the solubility of AgCl

In the presence of chloride ions in the solution, Ag+ ions may form a soluble complex with Cl- ions as follows: $$\mathrm{Ag}^{+}(a q)+2\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}_{2}^{-}(a q)$$ The equilibrium constant for this reaction is given as \(K_{c}' = 1.8\,x\, 10^{5}\) at 25°C. In a 0.100 M NaCl solution, the concentration of Cl- ions is 0.100 M. Now, let's write the expression for the equilibrium constant of the given reaction: $$K_{c}' = \frac{[\mathrm{AgCl}_{2}^{-}]}{[\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]^{2}}$$ We know the solubility of AgCl in water is \(1.6\, x\, 10^{-10}\) at 25°C. Let's call the solubility 'x': $$K_{c} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] = x(0.100 + x)$$ As x is small compared to 0.100, we can ignore \(x^{2}\), so we have x = \(1.6\, x\, 10^{-10}\). Now, let's find the concentration of AgCl2-: $$K_{c}' = \frac{[\mathrm{AgCl}_{2}^{-}]}{x(0.100)^{2}}$$ $$[\mathrm{AgCl}_{2}^{-}] = K_{c}' * x * (0.100)^{2} = 1.8\, x\, 10^{5} * 1.6\, x\, 10^{-10} * 0.01 = 2.88\, x\, 10^{-7}\, M$$ Since the concentration of the complex AgCl2- (\(2.88\, x\, 10^{-7}\) M) is much greater than the solubility of AgCl in water (\(1.6\, x\, 10^{-10}\) M), it shows that the formation of the soluble complex does, in fact, increase the solubility of AgCl in a 0.100 M NaCl solution compared to pure water.

Key Concepts

Le Châtelier's Principle

Le Châtelier's Principle is a useful guide in predicting how a reaction at equilibrium will respond to changes. Essentially, it states that a system at equilibrium will adjust to counteract any imposed changes in temperature, pressure, or concentration. This principle helps us understand the behavior of substances when conditions are altered, such as changing the temperature or mixing with other substances.

For the dissolution of AgCl, this process is endothermic, meaning it absorbs heat. According to Le Châtelier's Principle, if we increase the temperature, the system will shift to favor the absorption of heat, which in this case, is the dissolution of AgCl. Consequently, the solubility of AgCl in water increases with a rise in temperature.

Equilibrium Expression

An equilibrium expression describes the balance between reactants and products in a reversible reaction. In the example of AgCl dissolving in water, this can be shown as o the equation: \[\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(aq) + \mathrm{Cl}^{-}(aq)\]The equilibrium expression, or constant \(K_c\), is formulated from the concentrations of the dissolved ions because the solid \(\mathrm{AgCl}\) does not appear in the expression as it does not change its concentration. Thus, the expression is:\[K_c = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}]\]

This equilibrium constant helps us understand the extent to which the reaction will proceed under certain conditions, showing the relationship between concentrations at equilibrium.

Enthalpy Change

Enthalpy change, represented by \(\Delta H\), is the heat change that occurs during a chemical reaction at constant pressure. For the dissolution of AgCl, the process is endothermic, indicated by a positive \(\Delta H_{dissolution}\). This signifies that heat is absorbed as AgCl dissolves, and thus, increasing temperature also increases solubility.

The endothermic nature of the reaction means that as heat is added, the system compensates by dissolving more AgCl, thereby increasing its solubility. Such molecular behavior reflects the delicate balance of energy in chemical processes where enthalpy plays a critical role.

Complex Ion Formation

Complex ion formation greatly influences the solubility of certain substances. In the case of AgCl, adding chloride ions from a solution like NaCl can lead to the formation of a complex ion, \(\mathrm{AgCl}_2^{-}\), through the reaction:\[\mathrm{Ag}^{+} (aq) + 2\mathrm{Cl}^{-} (aq) \rightleftharpoons \mathrm{AgCl}_{2}^{-} (aq)\]

The presence of this soluble complex increases the apparent solubility of AgCl. The equilibrium constant \(K'_c\) for this formation is exceptionally high, indicating that the reaction heavily favors the products, and the complex ion forms readily. In a solution with additional chloride ions, like 0.100 M NaCl, more \(\mathrm{Ag}^{+}\) ions link to form \(\mathrm{AgCl}_2^{-}\), effectively increasing the solubility compared to pure water.

Complex ions like \(\mathrm{AgCl}_2^{-}\) demonstrate a chemical strategy for achieving greater solubility by adding certain ions, hence altering the environmental conditions.