If the solubility constant of silver iodide (AgI) is 8.5 × 10^-17, what is the solubility of Ag+?
A. 2.0 × 10^-13 M
B. 4.0 × 10^-13 M
C. 4.6 × 10^-9 M
D. 9.2 × 10^-9 M
9.21954446e-9?
It’s D. 9.2
what is the sqrt of 8.5e-17?
Well, silver iodide (AgI) is known for being quite the introvert and doesn't like to mingle much in water. However, if we take a closer look at the solubility constant, we can get an idea of how much of Ag+ is willing to step out of its comfort zone.
The solubility constant, Ksp, is defined as the product of the concentrations of the ions involved in a dissociation reaction. In this case, we're looking at the dissociation of AgI into Ag+ and I-.
The balanced equation for this dissociation is AgI(s) ↔ Ag+(aq) + I-(aq).
The solubility constant expression can be written as:
Ksp = [Ag+][I-]
Given that the solubility constant, Ksp, is 8.5 × 10^-17, we can assume that the concentration of Ag+ is x, since AgI dissociates into only one Ag+ ion. The concentration of I- would then also be x, as they are in a 1:1 ratio.
Plugging these concentrations into the solubility constant expression, we get:
8.5 × 10^-17 = x * x
Simplifying the equation, we have:
8.5 × 10^-17 = x^2
Solving for x, we find:
x = √(8.5 × 10^-17)
Calculating this on a trusty calculator, we find that x is approximately 2.9 × 10^-9 M.
So, the solubility of Ag+ is 2.9 × 10^-9 M. Now, isn't that just silver-plated solubility?
To determine the solubility of Ag+ in silver iodide (AgI), we need to use the solubility constant (Ksp) expression for the dissociation of AgI:
AgI ⇌ Ag+ + I-
The solubility constant expression for AgI is given as:
Ksp = [Ag+][I-]
Since the concentration of AgI is not given, we assume it to be "x" (in Molar). Therefore, the concentrations of Ag+ and I- would also be "x" (in Molar).
Using the given Ksp value of AgI, 8.5 × 10^-17, we can write the Ksp expression as:
8.5 × 10^-17 = (x)(x)
Simplifying the equation:
8.5 × 10^-17 = x^2
Taking the square root of both sides of the equation:
sqrt(8.5 × 10^-17) = sqrt(x^2)
x = sqrt(8.5 × 10^-17)
Calculating the value of x, we find:
x ≈ 2.92 × 10^-9 M
Thus, the solubility of Ag+ in silver iodide (AgI) is approximately 2.92 × 10^-9 M.
Therefore, the correct answer is option C. 4.6 × 10^-9 M.