A solution contains 0.0390 M Pb2 (aq) and 0.0390 M Sr2 (aq). If we add SO42–(aq), what will be the concentration of Pb2 (aq) when SrSO4(s) begins to precipitate?
I am just completely lost on this one.
Hey Dr. Bob. I'm not sure where you got the 1.6E-8 from?
Also I found the Ksp values from my text. It says Pb = 2.53E–8 and Sr = 3.44E–7. Could you re-write with those values so that I can understand it better please? I really appreciate your help.
I'm stuck.
To determine the concentration of Pb2+ when SrSO4(s) begins to precipitate, we need to consider the solubility product constant (Ksp) of SrSO4.
The balanced chemical equation for the precipitation reaction is:
Sr2+ (aq) + SO42- (aq) → SrSO4(s)
The Ksp expression for SrSO4 can be written as:
Ksp = [Sr2+][SO42-]
Since we are given the initial concentrations of both Pb2+ and Sr2+ as 0.0390 M, we can assume that the initial concentration of SO42- is 0 M.
At equilibrium, the concentration of Sr2+ will decrease by x (the molar solubility of SrSO4) and the concentration of SO42- will increase by x. Thus, the equilibrium expression becomes:
Ksp = (0.0390 - x)(x)
Given that Ksp for SrSO4 is 3.8 x 10^-7, we can solve for x:
3.8 x 10^-7 = (0.0390 - x)(x)
Now, we solve this quadratic equation using the quadratic formula or the method of factoring. For simplicity, let's use the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / 2a
Substituting a = 1, b = -0.039, and c = 3.8 x 10^-7 into the formula, we get:
x = [-(-0.039) ± √((-0.039)^2 - 4(1)(3.8 x 10^-7))] / (2)(1)
Simplifying the equation:
x = (0.039 ± √(0.001521 - 1.52 x 10^-6)) / 2
Now, we evaluate the two possible values of x:
x1 = (0.039 + √(0.001521 - 1.52 x 10^-6)) / 2
x2 = (0.039 - √(0.001521 - 1.52 x 10^-6)) / 2
Since we are interested in the concentration of Pb2+ when SrSO4(s) begins to precipitate, we consider the value of x at which the concentration of Sr2+ reaches zero. Therefore, the relevant value is x2.
Substituting the values into the equation:
x2 = (0.039 - √(0.001521 - 1.52 x 10^-6)) / 2
Evaluating this expression will provide the concentration of Pb2+ at the point of precipitation.
........PbSO4 ==> Pb^2+ + SO4^2-
........SrSO4 ==> Sr^2+ + SO4^2-
You know Ksp for each and you know metal is 0.039. I like to solve for OH^- needed to ppt each. Using my values for Ksp (which won't be the same as yours probably since texts differ),
SO4^2- for Pb = 4.1E-7 M.
SO4^2- for Sr = 8.2E-6 M.
This shows that if you 0.039 M solution of the two metals then add SO4^2- drop wise, the Pb ppts first. It will continue to ppt until the sulfate concn is what? When SO4^2- exceeds 8.2E-6, then Ksp for SrSO4 is exceeded and SrSO4 will start; i.e., (0.039)(8.2E-6) =or> 3.2E-7.
So what will that do to PbSO4.
Ksp PbSO4 = (Pb^2+)(SO4^2-)
1.6E-8 = (Pb^2+)(8.2E-6)
Solve for (Pb^2+). I get something around 0.002 M for Pb^2+. By the way this shows you can't separate (by pptn anyway) two metals with Ksp values so close together, especially if you want to do a CLEAN separation.