solubility product of silver iodate is 1.9×10^-15 at 25°c what is the molar solubility of silver iodide in water
.................AgIO3 ==> Ag^+ + IO3^-
I..............solid..............0...........0
C.............solid..............x............x
E.............solid..............x.............x
Ksp = (Ag^+)(IO3^-)
Substitute the E line into Ksp expression and solve for x = (Ag^+) = (IO3^=) = (AgCl)
Post your work if you get stuck.
To find the molar solubility of silver iodide (AgI) in water, we need to consider the solubility product (Ksp) of silver iodate (AgIO3) because silver iodide is not very soluble in water.
Given that the solubility product (Ksp) of silver iodate (AgIO3) is 1.9×10^-15, we can use the following balanced equation for the dissolution of AgIO3:
AgIO3 ⇌ Ag+ + IO3-
From the balanced equation, we can determine that the concentration of Ag+ in the solution is equal to the molar solubility of AgIO3 (since the stoichiometric coefficient is 1:1). Therefore, the molar solubility of AgI can be calculated using the Ksp of AgIO3 by considering its dissociation in water:
Ksp = [Ag+][IO3-]
Since the stoichiometry of AgI is different from AgIO3, we cannot directly equate the solubility of one to the other. However, we can assume that x mol of AgI will dissolve, resulting in x mol of Ag+ and x mol of IO3-.
Therefore, we can write the Ksp expression for AgI as:
Ksp(AgI) = [Ag+][I-]
Since the solubility product constants for AgIO3 and AgI are equal, we can equate the two expressions:
Ksp(AgIO3) = Ksp(AgI)
[Ag+][IO3-] = [Ag+][I-]
We are given the Ksp value for AgIO3 (1.9×10^-15), so we can substitute it in:
1.9×10^-15 = [Ag+][I-]
Since [Ag+] and [I-] both have concentrations of x, we can write:
1.9×10^-15 = x^2
Taking the square root of both sides, we get:
x = √(1.9×10^-15) = 1.38×10^-8
Therefore, the molar solubility of AgI in water is approximately 1.38×10^-8 mol/L.