A solution contains 1.4 * 10^−5 mol of KBr and 0.14 mol of KCl per liter. AgNO3 is gradually added to this solution. Which forms first, solid AgBr or solid AgCl? Or do they co-precipitate?
I used Ksp for AgCl = 1.77E-10 and for AgBr = 5.35E-13. Your text may have different numbers.
Ksp AgCl = (Ag^+)(Cl^-). You know Cl^ from the problem. Calculate (Ag^+). I get approx 1E-9 M but you need a better answer than that.
Do the same for Ag^+ from AgBr. I get approx 4E-8 M. Again you need a better answer.
Now visualize your solution and you add the AgNO3 drop wise and with stirring. Which Ag^+ is reached first? Obriously the AgCl (the smaller number) will be reached first so AgCl will ppt first.
This is not part of the problem but some questions continue with, "What is the(Cl^-) when the first ppt of AgBr appears?" That is AgBr begins to ppt when (Ag^+) = approx 4E-8 so plug that number back in for Ag^+ to Ksp for AgCl and solve for Cl^-. I get about 5E-3. When enough AgNO3 has been added so that the Cl^- has been reduced from 0.14 to about 0.005 M, then both AgCl and AgBr will ppt together.
To determine whether solid AgBr or solid AgCl will form first, or if they will co-precipitate, we need to compare the solubility products (Ksp) of AgBr and AgCl.
The solubility product (Ksp) is the equilibrium constant for a solid dissolving in a solution. It can be calculated using the following formula:
Ksp = [Ag+][Br-] for AgBr
Ksp = [Ag+][Cl-] for AgCl
Given that the solution contains 1.4 * 10^−5 mol of KBr and 0.14 mol of KCl per liter, we can determine the concentration of Ag+ and Br- (from KBr) and Ag+ and Cl- (from KCl).
Since KBr dissociates into one K+ ion and one Br- ion, the concentration of Ag+ ion will be equal to the concentration of Br- ion (since they will react to form AgBr).
Similarly, since KCl dissociates into one K+ ion and one Cl- ion, the concentration of Ag+ ion will be equal to the concentration of Cl- ion (since they will react to form AgCl).
Therefore, the concentrations of Ag+ and Br- are:
[Ag+] = [Br-] = 1.4 * 10^−5 M
And the concentrations of Ag+ and Cl- are:
[Ag+] = [Cl-] = 0.14 M
To determine which solid will precipitate first or if they will co-precipitate, we need to compare the solubility products (Ksp) of AgBr and AgCl.
The solubility product (Ksp) of AgBr is:
Ksp(AgBr) = [Ag+][Br-] = (1.4 * 10^−5)(1.4 * 10^−5)
The solubility product (Ksp) of AgCl is:
Ksp(AgCl) = [Ag+][Cl-] = (0.14)(0.14)
Comparing the two Ksp values, we can observe that Ksp(AgBr) is considerably smaller than Ksp(AgCl) since Ksp(AgCl) is based on the higher concentration of Ag+. Therefore, AgCl has a higher solubility and is less likely to precipitate compared to AgBr.
Hence, solid AgBr will form first while solid AgCl will most likely remain in solution.
To determine whether solid AgBr or solid AgCl forms first, or if they co-precipitate, we need to compare the solubility products (Ksp) of AgBr and AgCl.
1. Calculate the concentration of Ag+ ions in the solution:
- Ag+ ions are formed when AgNO3 dissociates in water.
- The number of moles of AgNO3 added to the solution is given by the equation: AgNO3(mol) = 0.14 mol KCl
- Since KCl dissociates completely, the concentration of Ag+ ions contributed by KCl is 0.14 M.
2. Compare the solubility products (Ksp) of AgBr and AgCl:
- The Ksp for AgBr is 5.0 × 10^−13.
- The Ksp for AgCl is 1.8 × 10^−10.
3. Determine which product will precipitate first:
- Compare the product of the concentration of Ag+ ions in the solution with the Ksp values:
- For AgBr: (0.14 M)^1 x (1.4 × 10^−5 M)^1 = 1.96 × 10^−6
- For AgCl: (0.14 M)^1 x (0.14 M)^1 ≈ 0.02
- As we can see, the product of the concentrations for AgBr is significantly lower than AgCl.
- Therefore, solid AgCl will precipitate first as it exceeds its solubility product (Ksp) value.
- Solid AgBr will not precipitate because its product of concentrations is much lower than its Ksp value.
In conclusion, solid AgCl will precipitate first, while AgBr will remain in solution as ions. They do not co-precipitate together.