Suppose I take a 1.0 L saturated, room temp solution of lead(2)bromide and dissolve 1.3g of potassium bromide in the solution. Will lead(2) bromide precipitate out? If yes how much( in grams)? If no, why not?
I found the ksp for PbBr2 at room temp, and used it to find the Molarity of PbBr2 saturated, .0116M. I know I have to use the ICE box method but I'm just confused as to what the reactants/products are and how to set it up. Thanks'!
OK. We'll take your number of 0.116M for the solubility of KBr in a saturated solution. I don't know what value you used for Ksp but I used one from the web and ended up with 0.0118M so the answers are pretty close. You probably used a value for Ksp close to what I used of 6.6E-6.
(KBr) = 1.3/119 = approx 0.0109M and it ionizes 100% so Br^- will be 0.0109.
I tried the easy way on this as
Ksp = (Pb^2+)(Br^-)^2
Ksp = (x)(0.0109)^2
and x = approx 0.055 which obviously is not right because the solubility can't be greater when adding a common ion. Therefore, I think you have to go the long route which is as follows:
............PbBr2 ==> Pb^2+ + 2Br^-
I...........solid..0.0116...2*0.0116
add..........................0.0109
C...........solid......-x....-2x
E...........solid..0.0116-x..0.0341-2x
Substitute the E line into Ksp expression and solve for x = molarity PbBr2 that ppts.
You get a cubic equation which I solved with the help of an on-line calculator. Here is one you can use.
http://www.1728.org/cubic.htm
Post your work if you get stuck.
To determine whether lead(2) bromide (PbBr2) will precipitate out or not, we need to compare the ion product (Q) with the solubility product (Ksp) of PbBr2.
First, let's write the balanced chemical equation for the dissolution of PbBr2 and the dissociation of KBr:
PbBr2(s) ⇌ Pb2+(aq) + 2Br-(aq)
KBr(s) ⇌ K+(aq) + Br-(aq)
The concentration of Pb2+ ions in solution is determined by the solubility of PbBr2. At room temperature, we assume the solution is saturated, so the concentration of Pb2+ can be calculated from the given Ksp value.
Ksp of PbBr2 at room temperature is 6.60 x 10^-6.
Next, let's determine the initial concentrations of Pb2+(aq) and Br-(aq) ions.
Given:
Volume of solution = 1.0 L
Mass of KBr added = 1.3 g
Molar mass of KBr = 119 g/mol
Molar mass of PbBr2 = 367 g/mol
Now, we can calculate the molarity (M) of Br- ions added to the solution:
Molarity (M) = moles of solute / volume of solution in L
Molarity of KBr = 1.3 g / (119 g/mol) / 1.0 L
We also know that two Br- ions are formed from each formula unit of KBr dissolved, so we multiply the molarity of KBr by 2 to get the initial concentration of Br- ions.
Now, let's set up the ICE (Initial, Change, Equilibrium) box for the dissociation of PbBr2:
PbBr2(s) ⇌ Pb2+(aq) + 2Br-(aq)
| Initial | Change | Equilibrium
Pb2+| 0 | +x | x
Br- | 2y | +2x | 2y + 2x
Since we're not given the initial concentrations of Pb2+ and Br- ions directly, we represent them as 0 and 2y, respectively. The change (x) represents the increase in concentration of Pb2+ and Br- ions, and the equilibrium concentration is x for Pb2+ and 2y + 2x for Br-.
Using the solubility product expression, we have:
Q = [Pb2+][Br-]^2 = x(2y + 2x)^2
Comparing Q and Ksp, we can determine if PbBr2 will precipitate or remain in solution.
If Q is greater than Ksp, PbBr2 will precipitate. If Q is less than or equal to Ksp, PbBr2 will remain dissolved.
Now, you can substitute the values into the equation and solve for x:
x(2y + 2x)^2 = Ksp
You can then use the value of x to find the mass of PbBr2 precipitated. The mass of PbBr2 precipitated is equal to the molar mass of PbBr2 multiplied by the concentration of Pb2+ ions precipitated from the solution.
I hope this explanation helps you set up the problem and solve it. Let me know if you need any further assistance!