An aqueous solution that 2.10 M in AgNO3 is slowly added from a buret to an aqueous solution that is 1.20×10^−2 M in Cl- and 0.260 M in I- .
When the second ion begins to precipitate, what is the remaining concentration of the first ion?
You should have typed in the Ksp for AgCl and AgI. The Ksp I look up probably aren't the same as in your text. I will use approximations.
Ksp for AgI = about 10^-17.
Ksp for AgCl = about 10^-10
Therefore, AgI will ppt first. AgCl will ppt second.
What will be the (Ag^+) when AgCl just starts to ppt.
That will be (Ag^+) = Ksp/(Cl^-) = 10^-10/0.012 = ?
Now what will be the (I^-) at this point? That will be (I^-) = Ksp/(Ag^+) just calculated.
Post your work if you still need help.
I got it thanks
But how did you get the (Cl^-)?
To determine the remaining concentration of the first ion when the second ion begins to precipitate, we need to use the concept of chemical reactions and the solubility product constant (Ksp).
First, let's write the balanced chemical equation for the precipitation reaction that occurs between the Ag+ ion and the Cl- ion:
Ag+(aq) + Cl-(aq) → AgCl(s)
According to the equation, 1 mole of Ag+ will react with 1 mole of Cl- to produce 1 mole of solid AgCl.
Now, let's find the concentration of Ag+ ions when AgCl starts to precipitate. The solubility product constant (Ksp) expression for AgCl is given by:
Ksp = [Ag+][Cl-]
Since we want to find the concentration of Ag+ when Cl- begins to precipitate, we can assume that the concentration of Cl- is equal to its solubility concentration in AgCl, which is 1.20×10^−2 M.
Now, we'll plug in the known values into the Ksp expression and solve for [Ag+]:
Ksp = [Ag+][Cl-]
Ksp = [Ag+](1.20×10^−2)
Ag+ = Ksp / (1.20×10^−2)
To find the Ksp value for AgCl, we need to consult a solubility table or use the Ksp value provided in the question. Assuming the Ksp value for AgCl is 1.82×10^−10 (as an example):
Ag+ = (1.82×10^−10) / (1.20×10^−2)
Ag+ ≈ 1.52×10^−8 M
Therefore, the remaining concentration of Ag+ ions when the second ion begins to precipitate is approximately 1.52×10^−8 M.