Calculate the molar solubility of AgI in pure water.
molar solubility=9.2*10^-9 M
0.14 M NaCN; Kf for Ag(CN)2- is 3.0*10^2
To calculate the molar solubility of AgI in pure water, we need to use the solubility product constant (Ksp) of AgI. The Ksp expression for AgI is as follows:
AgI (s) ⇌ Ag⁺ (aq) + I⁻ (aq)
The Ksp expression can be written as:
Ksp = [Ag⁺] * [I⁻]
Since AgI is a sparingly soluble salt, we can assume that the concentration of Ag⁺ and I⁻ ions formed is equal to the molar solubility of AgI represented by "x". Therefore, the Ksp expression can be simplified as:
Ksp = x * x
Given that the value of Ksp for AgI is not provided, it means we need to follow an alternative approach to calculate the molar solubility.
Let's consider the information about NaCN and the formation of Ag(CN)₂⁻ complex. The complex can be represented as follows:
Ag⁺ (aq) + 2CN⁻ (aq) ⇌ Ag(CN)₂⁻ (aq)
The formation constant for this complex is denoted by Kf and given as 3.0*10².
To determine the concentration of Ag⁺ in the solution, we need to take into account that Ag⁺ reacts with CN⁻ ions.
Since the concentration of CN⁻ is given as 0.14 M, we can assume that the concentration of Ag⁺ formed by the reaction between Ag⁺ and CN⁻ ions is equal to 2 times the concentration of CN⁻, as the stoichiometric coefficient of CN⁻ in the reaction is 2.
Therefore, [Ag⁺] = 2 * [CN⁻]
Substituting the given concentration of CN⁻, we get:
[Ag⁺] = 2 * 0.14 = 0.28 M
Now, we can assume that the concentration of Ag⁺ comes from AgI that dissolved in water. Therefore, we can equate the concentration of Ag⁺ to the molar solubility of AgI (x):
0.28 M = x
Hence, the molar solubility of AgI in pure water is 0.28 M.