[Cd(NH_(3))_(4)]Cl_(2);K_(2)[Cd(CN)_(4)] 2. Which of the above mentioned complexes is more stable? 3. Calculate the concentration of the Ag^+ ions in a 0 ,I M solution containing an excess of 1mol/l of NH_(3)

1. The stability of the complexes $[Cd(NH_{3})_{4}]Cl_{2}$ and $K_{2}[Cd(CN)_{4}]$ depends on the ligand field stabilization energy and the overall stability of the coordination compound. In general, the stability of a complex is influenced by factors such as the nature of the metal ion, the type of ligand, the coordination number, and the overall charge of the complex.<br /><br />2. To determine which of the two complexes is more stable, we need to consider the ligand field stabilization energy and the overall stability of the coordination compound. In general, the stability of a complex is influenced by factors such as the nature of the metal ion, the type of ligand, the coordination number, and the overall charge of the complex.<br /><br />3. To calculate the concentration of $Ag^{+}$ ions in a 0.1 M solution containing an excess of 1 mol/L of $NH_{3}$, we can use the solubility product constant (Ksp) for the precipitation of $Ag(NH_{3})_{2}^{+}$.<br /><br />The solubility product constant (Ksp) for the precipitation of $Ag(NH_{3})_{2}^{+}$ is given by:<br /><br />Ksp = [Ag(NH3)2+] [NH3]^2<br /><br />Given that the concentration of $NH_{3}$ is in excess, we can assume that the concentration of $NH_{3}$ remains constant at 1 mol/L. Therefore, the concentration of $Ag^{+}$ ions can be calculated as:<br /><br />[Ag+] = Ksp / [NH3]^2<br /><br />Substituting the given values, we have:<br /><br />[Ag+] = 1.8 x 10^-8 / (1)^2 = 1.8 x 10^-8 M<br /><br />Therefore, the concentration of $Ag^{+}$ ions in the 0.1 M solution containing an excess of 1 mol/L of $NH_{3}$ is 1.8 x 10^-8 M.