In a solution of .200M NaI, solid AgI was placed in. Calculate the solubility if AgI (Ksp=1.5*10^-16)
Assuming complete dissociation of NaI as it is an ionic compound,
NaI --> Na^+ + I^-
x x (moldm^-3)
and,
AgI <==> Ag^+ + I^-
y y (moldm^-3)
y=solubility of AgI
So the total [I^-]=(x+y) moldm^-3
But, Ksp AgI = [Ag^+][I^-]=y(x+y) mol^2dm^-6 = 1.5*10^-16 mol^2dm^-6
Hope this helps..
(Please note that I haven't mentioned the state of the above compounds)
Thank you so much!
To calculate the solubility of AgI in a solution of NaI, we need to determine the concentration of Ag+ and I- ions in the solution by using the Ksp expression for AgI.
The balanced equation for the dissociation of AgI in water is:
AgI (s) ⇌ Ag+ (aq) + I- (aq)
The Ksp expression for AgI is:
Ksp = [Ag+][I-]
First, we need to find the concentration of Ag+ ions in the solution. Since NaI is a soluble ionic compound, it dissociates completely into its constituent ions when dissolved in water:
NaI (aq) → Na+ (aq) + I- (aq)
Therefore, the concentration of I- ions in the solution is equal to the concentration of NaI, which is 0.200 M.
Now, let's assume that the concentration of AgI that dissolves is "x" M. Consequently, the concentration of Ag+ ions in the solution will also be "x" M.
Using the Ksp expression, we plug in the known values:
1.5 × 10^-16 = (x)(0.200)
Simplifying the equation:
x = (1.5 × 10^-16) / (0.200)
x ≈ 7.5 × 10^-16 M
Therefore, the solubility of AgI in the 0.200 M NaI solution is approximately 7.5 × 10^-16 M.