Calculate the concentration (in M) of Tl^+ when TlI just begins to precipitate from a solution that is 0.0315 M in I^−. (Ksp = 5.54 ✕ 10^−8)
Do I just set up an ICE table for this? So it would be 5.54X10^-8 = (.0315+x)x?
That is almost completely right and I'm being picky. You will get the correct answer your way BUT the problem tells you that Tl^+ is x. The problem tells you that the concn of I^- is 0.0315 M; i.e., THAT IS THE CONCN OF I^- so you plug in 0.0315 for I^- and not 0.0315 + x.
Perfect! Thank you for your help! :)
Yes, setting up an ICE (initial, change, equilibrium) table is a good approach to solve this problem. However, your setup for the ICE table is not correct.
In this problem, TlI (thallium iodide) is the compound that is forming a precipitate. The balanced equation for the dissociation of TlI in water is:
TlI (s) ⇌ Tl^+ (aq) + I^- (aq)
Given that the solution is already 0.0315 M in I^-, we can assume that the concentration of I^- remains constant throughout the reaction. Let's denote the concentration of Tl^+ as x.
The solubility product constant (Ksp) expression for this reaction is:
Ksp = [Tl^+][I^-]
Substituting the given values, we have:
5.54 × 10^-8 = x × 0.0315
Rearranging the equation to solve for x, we get:
x = (5.54 × 10^-8) / 0.0315
Calculating this expression, we find:
x ≈ 1.75 × 10^-6
Therefore, the concentration of Tl^+ when TlI just begins to precipitate is approximately 1.75 × 10^-6 M.