Final Answer:
a) The solubility of AgSeCN in solutions with a cation concentration of 0.030 M is approximately 1.33 x 10⁻⁷ M.
b) The solubility of RaSO4 in solutions with a cation concentration of 0.030 M is approximately 6.6 x 10⁻⁶ M.
Step-by-step explanation:
The solubility of salts depends on the equilibrium constant of the dissolution reaction, often represented as Ksp. For both AgSeCN and RaSO4, the dissolution reactions are as follows:
a) AgSeCN ⇄ Ag^+ + SeCN^-
b) RaSO_4 ⇆ s Ra^2+ + SO_4^2-
The equilibrium expression for these reactions is given by the product of the ion concentrations, which can be expressed as follows:
a) K_sp = [Ag^+][SeCN^-]
b) K_sp = [Ra^2+][SO_4^2-]
Given that we know the concentration of one of the ions (Ag^+ or Ra^2+) and want to find the solubility of the salt, we can set up an expression in terms of the unknown ion concentration. Solving for this concentration gives us the solubility.
For AgSeCN, if we let x be the concentration of Ag^+ and SeCN^-, the equilibrium expression becomes K_sp = x * x, and solving for x yields the solubility.
For RaSO4, the expression becomes K_sp = (0.030 + x) * x), where 0.030 M is the initial concentration of Ra^{2+}.
After solving these expressions, we find that the solubility of AgSeCN is approximately 1.33 x 10⁻⁷ M, and the solubility of RaSO4 is approximately 6.6 x 10⁻⁶ M in solutions with a cation concentration of 0.030 M.