You have amisture that is 0.100mol/L in each of the following ions: Cl-, Br-, and I-.
(a) Which substance will precipitate first?
AgCl, AgBr, AgI?
(b)what is the concentration of Ag+ required to start the precipitation of the first substance?
ksp (AgCl)= 1.77*10^-10
ksp (AgBr)= 5.35*10^-13
ksp (AgI)= 8.51*10^-17
Could you please help me understand how to do this problem? I am so lost.
Look up Ksp for AgCl, AgBr and AgI.
The salt with the smallest Ksp will ppt first.(this comparison is true only when the salts have the same ratio. In this case all are 1:1).
The following shows you how to calculate Ag^+ at the time EACH ppts. However, AgCl will not ppt first and that's the one you are calculate. AgI ppts first.
......AgCl(s) ==> Ag^+(aq) + Cl^-(aq)
I.......solid......0..........0
C.......solid......x..........x
E.......solid......x..........x
Ksp = whatever = (Ag^+)(Col^-)
Ksp = (x)(x)
Solve for x = (Ag^+).
i got 4.46*10^-4??
No, that isn't right; however, I may have led you down the wrong path. I see at the top that the solution is 0.1M in I^- so
AgI==> Ag^+ + I^-
........x.....x
Ksp = (Ag^+)(I^-)
Ag^+ is x from AgI.
I^= 0.1M from the problem.
8.51E-17 = (0.1)(x)
Solve for x.
Of course! I'd be happy to help you understand how to solve this problem step by step.
(a) To determine which substance will precipitate first, we need to compare the solubility products (Ksp) of each compound with the concentrations of their respective ions in the solution. The compound with a smaller Ksp value compared to the product of its ion concentrations will precipitate first.
The Ksp expression for each compound is as follows:
Ksp (AgCl) = [Ag+][Cl-]
Ksp (AgBr) = [Ag+][Br-]
Ksp (AgI) = [Ag+][I-]
Given that the concentration of each ion in the solution is 0.100 mol/L, we can plug these values into the Ksp expressions to compare the products:
For AgCl: [Ag+][Cl-] = (0.100 mol/L)(0.100 mol/L) = 0.010 mol^2/L^2 = 1.0 × 10^-2 mol^2/L^2
For AgBr: [Ag+][Br-] = (0.100 mol/L)(0.100 mol/L) = 0.010 mol^2/L^2 = 1.0 × 10^-2 mol^2/L^2
For AgI: [Ag+][I-] = (0.100 mol/L)(0.100 mol/L) = 0.010 mol^2/L^2 = 1.0 × 10^-2 mol^2/L^2
Now, let's compare these products with the given Ksp values:
Ksp (AgCl) = 1.77 × 10^-10
Ksp (AgBr) = 5.35 × 10^-13
Ksp (AgI) = 8.51 × 10^-17
Comparing the products to the Ksp values, we can see that the product for AgCl (1.0 × 10^-2 mol^2/L^2) is significantly greater than the Ksp value for AgCl (1.77 × 10^-10). Similarly, the product for AgBr (1.0 × 10^-2 mol^2/L^2) is also greater than the Ksp value for AgBr (5.35 × 10^-13). However, the product for AgI (1.0 × 10^-2 mol^2/L^2) is much smaller than the Ksp value for AgI (8.51 × 10^-17).
Based on these comparisons, we can conclude that AgI will precipitate first since its product is significantly lower than its Ksp value, indicating that the compound is less soluble than AgCl and AgBr.
Therefore, the substance that will precipitate first is AgI.
(b) To determine the concentration of Ag+ required to start the precipitation of the first substance (AgI), we need to calculate the solubility of AgI. The solubility (s) of a compound with a Ksp value can be calculated by taking the square root of the Ksp value:
s = √Ksp (AgI) = √(8.51 × 10^-17) = 9.22 × 10^-9 mol/L
The solubility of AgI is approximately 9.22 × 10^-9 mol/L.
However, since the concentration of Ag+ required to start precipitation is half of the solubility, the concentration of Ag+ required is approximately 4.61 × 10^-9 mol/L.
Therefore, the concentration of Ag+ required to start the precipitation of the first substance is approximately 4.61 × 10^-9 mol/L.