calculate the concentration of a silver ion when the solubility produc constant of AgI is 10 to the power of negative 16
To calculate the concentration of a silver ion (Ag+) when the solubility product constant of AgI is 10^-16, we need to use the concept of solubility product and the stoichiometry of the equilibrium reaction. The solubility product expression for AgI is:
Ksp = [Ag+][I-]
Here, [Ag+] represents the concentration of silver ion and [I-] represents the concentration of iodide ion. Since AgI is a very soluble compound, we can assume that the concentration of iodide ion is negligible compared to [Ag+] (because AgI mostly dissociates into Ag+ and I- ions). Thus, we can represent the expression as:
Ksp = [Ag+]^2
Now, substitute the given value of the solubility product constant (Ksp = 10^-16) into the equation:
10^-16 = [Ag+]^2
To solve for [Ag+], take the square root of both sides of the equation:
√(10^-16) = √[Ag+]^2
10^-8 = [Ag+]
Therefore, the concentration of silver ion ([Ag+]) is 10^-8.
To calculate the concentration of a silver ion (Ag+) when the solubility product constant (Ksp) of silver iodide (AgI) is given, you need to know the molar solubility of AgI. The molar solubility is defined as the number of moles of a solute that dissolves in one liter of the solution at equilibrium.
Let's denote the molar solubility of AgI as "x". In the equilibrium equation for the dissolution of AgI, it can be represented as follows:
AgI ↔ Ag+ + I-
From the equation, we can see that for every AgI molecule that dissolves, one silver ion (Ag+) is produced. Therefore, the concentration of Ag+ is also equal to the molar solubility "x" of AgI at equilibrium.
By using the solubility product constant equation for AgI:
Ksp = [Ag+][I-]
We can substitute the values into the equation:
10^(-16) = x * x
Solving the equation, we find that the molar solubility "x" is equal to 10^(-8) mol/L.
Therefore, the concentration of silver ion (Ag+) is 10^(-8) mol/L.
AgI ==> Ag^+ + I^-
Ksp = (Ag^+)(I^-)
Ag^+ = x
I^- = x
solve for x.