If 0.08 mol of LiCl is added to a one liter solution that has dissolved 0.01 mol of PbCl2, will a precipitate
occur?
0.01 mol PbCl2 is dissolved.
+0.08 mol LiCl.
Total Cl = 0.08 + 2*0.01) = 0.1 mol Cl
Then Kquotient = (Pb^2+)(Cl^-)^2 = (0.01)(0.1)^2 = 1E-4. Is that greater or less than Ksp? If greater, yes a ppt will occur. If less no it will not occur.
Hmm, let me check my chemistry crystal ball... *puts on clown wig*
Well, if we mix 0.08 mol of LiCl with 0.01 mol of PbCl2, it's party time in the solution!
But seriously, according to my chemical calculations, a precipitate will form in this scenario. The reason is that when you mix LiCl and PbCl2, they undergo a double replacement reaction, producing LiCl2 and PbCl2.
Since LiCl2 is soluble in water, it will remain in solution, while PbCl2 is insoluble and will form a precipitate. So get ready to see some solid particles having a little dance party at the bottom of your beaker!
To determine if a precipitate will occur when 0.08 mol of LiCl is added to a one liter solution that already has 0.01 mol of PbCl2 dissolved, we need to compare the solubility product constant (Ksp) of PbCl2 with the product of the ion concentrations after mixing.
The balanced equation for the dissolution of PbCl2 is:
PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)
The Ksp expression for PbCl2 is:
Ksp = [Pb2+][Cl-]^2
Given that 0.01 mol of PbCl2 is already dissolved in the solution, the concentration of Pb2+ is also 0.01 M, and the concentration of Cl- is 2 * 0.01 M = 0.02 M.
Next, let's calculate the ion concentrations after adding 0.08 mol of LiCl. Since LiCl is completely dissociated in water, it will provide 0.08 M Li+ ions and 0.08 M Cl- ions. Adding this to the existing Cl- concentration gives:
Total Cl- concentration = 0.02 M + 0.08 M = 0.10 M
Now we can substitute the ion concentrations into the Ksp expression:
Ksp = [Pb2+][Cl-]^2 = (0.01 M)(0.10 M)^2 = 0.0001
Comparing the calculated value of Ksp to the actual Ksp value, we can determine if precipitation will occur. If the calculated value is greater than the actual Ksp, a precipitate will form.
In this case, since the calculated Ksp value is less than the actual Ksp value of 0.0001, a precipitate will not occur when 0.08 mol of LiCl is added to the solution.
To determine if a precipitate will occur, we need to compare the solubility products (Ksp) of the compounds involved. The Ksp is the equilibrium constant for a solid salt dissolving in water and is a measure of the solubility of the compound.
The Ksp for LiCl is 4.81. The Ksp for PbCl2 is 1.6 x 10^-5.
To determine if a precipitate will form when these two salts are mixed, we need to compare the ion product (IP) with the Ksp value for the compound that would form a precipitate.
The ion product (IP) for a compound is calculated by multiplying the concentrations of the ions involved. In this case, we have Li+ and Cl- ions from LiCl, and Pb2+ and Cl- ions from PbCl2.
For LiCl:
Li+ concentration = 0.08 mol / 1 L = 0.08 M
Cl- concentration = 0.08 mol / 1 L = 0.08 M
For PbCl2:
Pb2+ concentration = 0.01 mol / 1 L = 0.01 M
Cl- concentration = 0.01 mol / 1 L = 0.01 M
Now, let's calculate the ion product (IP) for PbCl2:
IP = [Pb2+][Cl-] = (0.01 M)(0.01 M) = 1 x 10^-4
Comparing the ion product (IP) for PbCl2 (1 x 10^-4) with the Ksp value for PbCl2 (1.6 x 10^-5), we can see that the IP exceeds the Ksp. This indicates that a precipitate of PbCl2 will form.
However, we also need to consider the effect of adding LiCl to the solution. When LiCl dissolves in water, it dissociates into Li+ and Cl- ions, which would increase the concentration of Cl- ions.
After LiCl dissolves:
Li+ concentration = 0.08 M
Additional Cl- concentration = 0.08 M
Total Cl- concentration = 0.08 M + 0.01 M = 0.09 M
Now, let's recalculate the ion product (IP) for PbCl2:
IP = [Pb2+][Cl-] = (0.01 M)(0.09 M) = 9 x 10^-4
Comparing the new IP for PbCl2 (9 x 10^-4) with the Ksp value for PbCl2 (1.6 x 10^-5), we can see that the IP still exceeds the Ksp. Therefore, even after adding LiCl, a precipitate of PbCl2 will still form.