Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties. a. \(\mathrm{Ag}_{3} \mathrm{PO}_{4}, K_{\mathrm{sp}}=1.8 \times 10^{-18}\) b. \(\mathrm{CaCO}_{3}, K_{\mathrm{sp}}=8.7 \times 10^{-9}\) c. \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}, K_{\mathrm{sp}}=1.1 \times 10^{-18}\left(\mathrm{Hg}_{2}^{2+}\right.\) is the cation in solution.)

Step 1: Write the balanced dissolution equation and the Ksp expression The balanced equation for the dissolution of Ag3PO4 is: \[ Ag_3PO_4(s) \rightleftharpoons 3Ag^+(aq) + PO_4^{3-}(aq) \] The expression for the solubility product constant (Ksp) is: \[ K_{sp} = [Ag^+]^3[PO_4^{3-}] \] Step 2: Set up the equilibrium table Let s be the solubility of Ag3PO4 in moles per liter. The equilibrium table for the dissolution of Ag3PO4 is: | | Initial | Change | Equilibrium | | Ag3PO4 (s) | - | - | - | | Ag+ (aq) | 0 | +3s | 3s | | PO4^3- (aq) | 0 | +s | s | Step 3: Solve for solubility (s) Substitute the equilibrium concentrations into the Ksp expression: \[1.8 \times 10^{-18} = (3s)^3(s)\] Solve for s: \[s = 1.91 \times 10^{-6}\,M\] The solubility of Ag3PO4 is \(1.91 \times 10^{-6}\,moles/L\).

Step 1: Write the balanced dissolution equation and the Ksp expression The balanced equation for the dissolution of CaCO3 is: \[ CaCO_3(s) \rightleftharpoons Ca^{2+}(aq) + CO_3^{2-}(aq) \] The expression for the solubility product constant (Ksp) is: \[ K_{sp} = [Ca^{2+}][CO_3^{2-}] \] Step 2: Set up the equilibrium table Let s be the solubility of CaCO3 in moles per liter. The equilibrium table for the dissolution of CaCO3 is: | | Initial | Change | Equilibrium | | CaCO3 (s) | - | - | - | | Ca^2+ (aq) | 0 | +s | s | | CO3^2- (aq) | 0 | +s | s | Step 3: Solve for solubility (s) Substitute the equilibrium concentrations into the Ksp expression: \[8.7 \times 10^{-9} = (s)(s)\] Solve for s: \[s = 9.35 \times 10^{-5}\,M\] The solubility of CaCO3 is \(9.35 \times 10^{-5}\,moles/L\).

Step 1: Write the balanced dissolution equation and the Ksp expression The balanced equation for the dissolution of Hg2Cl2 is: \[ Hg_2Cl_2(s) \rightleftharpoons Hg_2^{2+}(aq) + 2Cl^-(aq) \] The expression for the solubility product constant (Ksp) is: \[ K_{sp} = [Hg_2^{2+}][Cl^-]^2 \] Step 2: Set up the equilibrium table Let s be the solubility of Hg2Cl2 in moles per liter. The equilibrium table for the dissolution of Hg2Cl2 is: | | Initial | Change | Equilibrium | | Hg2Cl2 (s) | - | - | - | | Hg2^2+ (aq) | 0 | +s | s | | Cl^- (aq) | 0 | +2s | 2s | Step 3: Solve for solubility (s) Substitute the equilibrium concentrations into the Ksp expression: \[1.1 \times 10^{-18} = (s)(2s)^2\] Solve for s: \[s = 2.96 \times 10^{-6}\,M\] The solubility of Hg2Cl2 is \(2.96 \times 10^{-6}\,moles/L\).