The molar solubility of silver iodide in a 0.186 M potassium iodide solution is __________ M.
.....AgI ==> Ag^+ + I^-
I....solid...0......0
C....solid...x......x
E....solid...x......x
....KI ==> K^+ + I^- (100% ionized)
I..0.186...0.....0
C.-0.186..0.186..0.186
E...0.....0.186..0.186
Ksp = (Ag^+)(I^-)
Ksp = look up
(Ag^+) = x from above
(I^-) = 0.186 + x. 0.186 from the KI and x from the AgI.
Solve for x.
0.186M(KI) => 0.186(K^+) + 0.186(I^-)
------- AgI <=> Ag^+ + I^-
C(eq) ----------x-----0.186M
Ksp = [Ag^+][I^-] = (x)(0.186M)
Ksp(AgI) = 8.3x10^-17 (fm table of Ksp values) = (x)(0.186M)
x = Solubility in presence of 0.186M(KI) = (8.3x10^-17/0.186)M = 4.5x10^-16M
In pure water Solubility = Sqr-Root(Ksp) = 9.1x10^-9M
Here's one for the ole gray matter... What's the solubility of AgI in the presence of 1.0M NH3? (Kf(Ag(NH3)2^+) = 1.7x10^+7, Ksp = 8.3x10^-17). Check out Complex-Ion equilibrium.
To determine the molar solubility of silver iodide (AgI) in a potassium iodide (KI) solution, we can use the concept of the common ion effect. According to the common ion effect, the solubility of a salt decreases when a common ion is present in the solution.
In this case, both AgI and KI contain iodide ions (I⁻). Therefore, the concentration of iodide ions in the solution will affect the solubility of AgI.
Let's assume the molar solubility of AgI in the absence of any other iodide ion is x mol/L.
Now, adding the potassium iodide (KI) to the solution will increase the concentration of iodide ions by 0.186 M.
The solubility equilibrium equation for AgI is written as:
AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)
Using the stoichiometry of the balanced equation, the concentration of iodide ions will be equal to the solubility of AgI.
Therefore, we can write the expression for the solubility of AgI in terms of iodide ion concentration as:
[Ag⁺][I⁻] = x M
Since the concentration of iodide ions in the solution is increased by 0.186 M, we have:
[I⁻] = x + 0.186 M
Substituting this into the solubility expression, we get:
[Ag⁺](x + 0.186) = x M
To simplify, we can ignore the concentration of silver ions (Ag⁺) because it is much lower compared to that of the iodide ions. This is due to the low solubility of AgI.
Therefore, we can approximate the expression as:
0.186x ≈ x
Simplifying this equation, we find:
0.186 ≈ 1
This is not possible since the values are not equal. Therefore, the assumption of ignoring the concentration of silver ions is invalid.
Hence, the molar solubility of silver iodide in a 0.186 M potassium iodide solution is not possible to determine with the provided information.
To determine the molar solubility of silver iodide (AgI) in a solution of potassium iodide (KI), you need to consider the common ion effect.
The common ion effect occurs when a compound with a common ion is added to a solution, which reduces the solubility of a slightly soluble salt. In this case, the common ion is the iodide ion (I-).
To calculate the molar solubility, we need to consider the solubility product constant (Ksp) of silver iodide (AgI). The Ksp expression for AgI is:
Ksp = [Ag+][I-]
Since silver iodide is a sparingly soluble salt, it dissociates according to the equation:
AgI (s) ⇌ Ag+ (aq) + I- (aq)
Let's assume the molar solubility of silver iodide in the 0.186 M KI solution is x. Since KI is a strong electrolyte, it dissociates completely into K+ and I- ions. Therefore, the concentration of the iodide ion (I-) in the solution will be the sum of the initial concentration of iodide ions from KI (0.186 M) and the concentration produced by the dissociation of AgI (x). So, [I-] = 0.186 M + x.
As per the dissociation equation of AgI, the concentration of silver ions ([Ag+]) is also equal to x.
Since the stoichiometry of AgI is 1:1, we can write the expression for the solubility product constant (Ksp):
Ksp = [Ag+][I-] = (x)(0.186 + x)
Now, we can substitute the given value of Ksp for AgI into the equation and solve for x:
Ksp = (x)(0.186 + x)
To find the value of x, you need to know the specific value of the Ksp for AgI. If you have that value, you can solve the above equation using algebraic techniques such as factoring, quadratic equations, or graphical methods.
Unfortunately, without the specific Ksp value for AgI, we cannot calculate the molar solubility.