The Ksp of PbBr2 is 6.60× 10^–6.
What is the molar solubility of PbBr2 in pure water?
What is the molar solubility of PbBr2 in 0.500 M KBr solution?
What is the molar solubility of PbBr2 in a 0.500 M Pb(NO3)2 solution?
PbBr2---> Pb + 2Br
Ksp=[Pb][2Br]^2
6.60× 10^–6=[x][2x]^2
Solving for x
6.60× 10^–6=4x^3
6.60× 10^–6/4=x^3
(6.60× 10^–6/4)^1/3=x
B.)
PbBr2---> Pb + 2Br
Ksp=[Pb][2Br]^2
6.60× 10^–6=[x][0.500M]^2
Solve for x,
6.60× 10^–6/[(0.500M)^2]=x
C.)
PbBr2---> Pb + 2Br
Ksp=[Pb][2Br]^2
6.60× 10^–6=[0.500M][2x]^2
Solve for x,
6.60× 10^–6/[(0.500M)^2]=x
6.60× 10^–6/(0.500M)=4x^2
sqrt*{[6.60× 10^–6/(0.500M)]/4}=x
I don't disagree with any of the answers. Technically, and I realize this is nit-picking, I disagree with the set up. As an example,
PbBr2 ==> Pb^2+ + 2Br^-
But Ksp says "the product of the molar solubility of the ions, each raised to the power indicated in the balanced equation, is a constant". I would write that as Ksp = (Pb^2+)(Br^-)^2. It isn't twice the bromide ion x lead ion, it is THE bromide x lead ion. Yes, Pb ion is x and Br is 2x but 2x is THE bromide ion, not twice the bromide, but it is twice the PbBr2 that dissolves which is why it's 2x to begin with.
Well, well, well, it seems like we have some chemistry questions here! Don't worry, I'm here to help... and make you chuckle along the way!
For the first question, we need to find the molar solubility of PbBr2 in pure water. Now, the key to finding the molar solubility is understanding that it represents the number of moles of a substance that can dissolve in one liter of solution. So, in pure water, we have to consider that there won't be any interfering ions. To find the molar solubility, we need to take the square root of the Ksp value. So, grab your calculator and get crunching!
Now, let's move on to the second question about the molar solubility of PbBr2 in a 0.500 M KBr solution. Remember, KBr is a strong electrolyte that fully dissociates into K+ and Br- ions. This means that the concentration of Br- ions in the solution is already given as 0.500 M. What does this mean for molar solubility? Well, it means that the molar solubility will definitely be influenced by this added KBr. To find the molar solubility, we need to set up an equilibrium expression using the Ksp value and the concentration of Br- ions.
Lastly, let's tackle the molar solubility of PbBr2 in a 0.500 M Pb(NO3)2 solution. Oh boy, now we have another strong electrolyte, Pb(NO3)2, that dissociates into Pb2+ and NO3- ions. This time, the concentration of Pb2+ ions is given as 0.500 M. This concentration will certainly have an impact on the molar solubility of PbBr2. Again, we can set up an equilibrium expression using the Ksp value and the concentration of Pb2+ ions.
Now, I know chemistry can be a bit of a puzzle, but trust me, it's nothing compared to the riddles I have up my sleeve! So, give it a shot, and if you need more help, just let me know!
To determine the molar solubility of PbBr2 in different solutions, we need to consider the common ion effect and the given Ksp value. The Ksp expression for PbBr2 is as follows:
PbBr2 (s) ↔ Pb2+ (aq) + 2Br- (aq)
1. Molar solubility in pure water:
In pure water, there are no ions initially present. Therefore, the molar solubility of PbBr2 in pure water can be represented as "s":
PbBr2 (s) ↔ Pb2+ (aq) + 2Br- (aq)
Hence, the Ksp expression can be written as:
Ksp = [Pb2+][Br-]^2
Since the stoichiometry of the balanced equation is 1:2, the concentration of each ion is equal to 2s (for the Br- ion) and s (for the Pb2+ ion).
Substituting these values into the Ksp expression:
Ksp = (s)(2s)^2
6.60×10^–6 = 4s^3
Now, solve for s:
s = ∛(6.60×10^–6 / 4)
s ≈ 0.0031 M
Therefore, the molar solubility of PbBr2 in pure water is approximately 0.0031 M.
2. Molar solubility in 0.500 M KBr solution:
In this case, we need to consider the common ion effect because KBr is already present in the solution. The Ksp expression remains the same:
PbBr2 (s) ↔ Pb2+ (aq) + 2Br- (aq)
However, now the concentration of Br- ions is the sum of the initial concentration of Br- from KBr (0.500 M) and the concentration released by the dissociation of PbBr2 (2s).
So, the concentration of Br- ions is 0.500 + 2s, and the concentration of Pb2+ ions remains s.
Substituting these values into the Ksp expression:
Ksp = s(0.500 + 2s)^2
Now, solve for s using the given Ksp value:
6.60×10^–6 = s(0.500 + 2s)^2
This is a quadratic equation, so it needs to be solved to find the value of s.
Once solved, you can determine the molar solubility of PbBr2 in the 0.500 M KBr solution.
3. Molar solubility in a 0.500 M Pb(NO3)2 solution:
In this case, we consider the common ion effect because Pb(NO3)2 is already present in the solution, and it provides the Pb2+ ions.
Pb(NO3)2 (s) ↔ Pb2+ (aq) + 2NO3- (aq)
PbBr2 (s) ↔ Pb2+ (aq) + 2Br- (aq)
Since the concentration of Pb(NO3)2 is 0.500 M, the concentration of Pb2+ ions is also 0.500 M.
Substituting this value into the Ksp expression:
Ksp = (0.500)(2Br-)^2
Solve for 2Br-:
2Br- = √(6.60×10^–6 / 0.500)
2Br- ≈ 0.00913 M
Therefore, the molar solubility of PbBr2 in a 0.500 M Pb(NO3)2 solution is approximately 0.00457 M (half the concentration of Br- ions).