Which of the following statements are true? (a) For an insoluble metallic salt, \(K_{\text {sp }}\) is always less than 1 . (b) More \(\mathrm{PbCl}_{2}\) can be dissolved at \(100^{\circ} \mathrm{C}\) than at \(25^{\circ} \mathrm{C}\). One can conclude that dissolving \(\mathrm{PbCl}_{2}\) is an exothermic process. (c) When strips of copper metal are added to a saturated solution of \(\mathrm{Cu}(\mathrm{OH})_{2}\), a precipitate of \(\mathrm{Cu}(\mathrm{OH})_{2}\) can be expected to form because of the common ion effect.

For an insoluble metallic salt, Ksp is always less than 1. Let's break down the concept of Ksp, or solubility product constant. Ksp is an equilibrium constant for the dissolution process of slightly soluble salts in water. The smaller the Ksp, the less soluble the salt is in water. With this information, we can conclude that the statement is true. For an insoluble metallic salt, Ksp is always less than 1 because the salt is less soluble in water, and its solubility equilibria would lie predominantly to the left, which leads to a smaller value of Ksp.

More \(PbCl_2\) can be dissolved at \(100^{\circ}C\) than at \(25^{\circ}C\). One can conclude that dissolving \(PbCl_2\) is an exothermic process. To analyze this statement, we need to consider how solubility is affected by temperature. Generally, for most salts, solubility increases with an increase in temperature in the case of endothermic processes. However, the solubility decreases with temperature increase for exothermic processes. Since the dissolution of \(PbCl_2\) is more at higher temperature (\(100^{\circ}C\)) than at lower temperature (\(25^{\circ}C\)), this indicates that its dissolution process is endothermic, not exothermic. Thus, the statement is false.

When strips of copper metal are added to a saturated solution of \(Cu(OH)_2\), a precipitate of \(Cu(OH)_2\) can be expected to form because of the common ion effect. The common ion effect states that adding an ion that is common to the ions in an existing equilibrium will cause a shift in the equilibrium position to accommodate the additional ion and minimize its effect. In this case, the copper strips will not affect the equilibrium, as there is no common ion between the copper metal and the ions in the solution of \(Cu(OH)_2\). Therefore, the statement is false because the copper strips will not cause any shift in the equilibrium, and no additional precipitate will form due to the common ion effect. In summary: - Statement (a) is true. - Statement (b) is false. - Statement (c) is false.