Question

Write the balanced chemical equation describing the dissolving of each of the following sparingly soluble salts in water. Write the expression for \(K_{\mathrm{sp}}\) for each process. a. \(\operatorname{AgIO}_{3}(s)\) b. \(\operatorname{Sn}(\mathrm{OH})_{2}(s)\) c. \(\mathrm{Zn}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) d. \(\mathrm{BaF}_{2}(s)\)

Short answer

The balanced chemical equations and \(K_\text{sp}\) expressions for each sparingly soluble salt are: a. AgIO\(_3\): \(AgIO_{3}(s) \rightleftharpoons Ag^{+}(aq) + IO_{3}^{-}(aq)\), \(K_\text{sp} = [Ag^{+}][IO_{3}^{-}]\) b. Sn(OH)\(_2\): \(Sn(OH)_{2}(s) \rightleftharpoons Sn^{2+}(aq) + 2OH^{-}(aq)\), \(K_\text{sp} = [Sn^{2+}][OH^{-}]^2\) c. Zn\(_3\)(PO\(_4\))\(_2\): \(Zn_{3}(PO_{4})_{2}(s) \rightleftharpoons 3Zn^{2+}(aq) + 2PO_{4}^{3-}(aq)\), \(K_\text{sp} = [Zn^{2+}]^3[PO_{4}^{3-}]^2\) d. BaF\(_2\): \(BaF_{2}(s) \rightleftharpoons Ba^{2+}(aq) + 2F^{-}(aq)\), \(K_\text{sp} = [Ba^{2+}][F^{-}]^2\)

Step-by-step solution

Write the balanced chemical equations for each salt dissolving in water

For each salt, break down the compound into its respective ions and assign their coefficients. a. AgIO\(_3\): \( AgIO_{3}(s) \rightleftharpoons Ag^{+}(aq) + IO_{3}^{-}(aq) \) b. Sn(OH)\(_2\): \( Sn(OH)_{2}(s) \rightleftharpoons Sn^{2+}(aq) + 2OH^{-}(aq) \) c. Zn\(_3\)(PO\(_4\))\(_2\): \( Zn_{3}(PO_{4})_{2}(s) \rightleftharpoons 3Zn^{2+}(aq) + 2PO_{4}^{3-}(aq) \) d. BaF\(_2\): \( BaF_{2}(s) \rightleftharpoons Ba^{2+}(aq) + 2F^{-}(aq) \)

Write the expression for \(K_\text{sp}\) for each process

To write the \(K_\text{sp}\) expressions, the solubility product constant is equal to the product of the concentrations of the ions in the solution raised to the power of their coefficients. Do this for each salt. a. AgIO\(_3\): \[ K_\text{sp} = [Ag^{+}][IO_{3}^{-}] \] b. Sn(OH)\(_2\): \[ K_\text{sp} = [Sn^{2+}][OH^{-}]^2 \] c. Zn\(_3\)(PO\(_4\))\(_2\): \[ K_\text{sp} = [Zn^{2+}]^3[PO_{4}^{3-}]^2 \] d. BaF\(_2\): \[ K_\text{sp} = [Ba^{2+}][F^{-}]^2 \]

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is essential when exploring the behavior of various substances in chemical reactions. It's a state in which the rate of the forward reaction equals that of the reverse reaction, meaning that the concentrations of reactants and products remain constant over time, but not necessarily equal.

Importantly, this concept applies to both soluble and insoluble compounds in water. In the case of insoluble or sparingly soluble substances, such as the ones provided in the exercise, chemical equilibrium is crucial in understanding their behavior. The equations we've written represent the equilibrium state where the solid compound is in balance with its dissolved ions.

In the tutorial provided, the balanced chemical equations show how sparingly soluble salts reach equilibrium when dissolving in water. This information provides a foundation for calculating the solubility product constant, a numerical expression of the solubility equilibrium for sparingly soluble compounds.

Solubility Equilibrium

The concept of solubility equilibrium is specifically tied to the dissolution process of sparingly soluble salts, like in our textbook examples. It signifies a dynamic equilibrium between the solid phase (undissolved salt) and the aqueous phase (dissolved ions) when the salt is mixed with water.

At this equilibrium point, the rate at which the salt dissolves is equal to the rate at which the dissolved ions come together to form the solid salt, resulting in no net change in concentration of either form. The solubility product constant, K_sp, is a key player here. It quantifies this equilibrium state and predicts the extent of solubility for a compound.

To provide more clarity for those who may still require additional understanding, K_sp values are determined empirically and are specific to each compound at a given temperature. These constants help us to understand and predict the solubility of compounds in solution, guiding chemists in various applications, such as predicting the formation of precipitates.

Ionic Compounds in Water

Ionic compounds in water can generally be classified as soluble or insoluble, although some fall in between as sparingly soluble. When an ionic compound dissolves in water, it dissociates into its constituent ions, a process that’s essential for a wide range of chemical reactions.

Our textbook exercises featured several sparingly soluble ionic salts that, when in water, may partially dissolve to produce a solution containing dissociated ions. The degree of this dissolving action is captured by the solubility product constant. Solubility is influenced by factors such as temperature, the presence of other ions in solution, and the nature of the solvent.

It’s also worth noting that the interplay between solubility and the solubility product constant is a great application of Le Châtelier's Principle. When the equilibrium of a sparingly soluble salt is disturbed, such as by adding more of one of the ions already present in solution, the system will adjust to counteract the change. This could result in a precipitate forming if the product of the ion concentrations exceeds the K_sp, demonstrating the salt's limited solubility.