A solution contains 0.0200 M Pb2 (aq) and 0.0200 M Sr2 (aq). If we add SO42–(aq), what will be the concentration of Pb2 (aq) when SrSO4(s) begins to precipitate?
Ksp = (Pb^2+)(SO4^2-) = ?
Ksp = (Sr^2+)(SO4^2-) = ?
Substitute 0.02M for Pb^2+ and solve for SO4^2-. That will be the sulfate concn when PbSO4 starts to ppt since the Ksp for PbSO4 is smaller than that of SrSO4. The problem doesn't ask for that but it's nice to know.
Substitute 0.02M for Sr^2+ into that Ksp and solve for SO4^2-. That will be the sulfate concn when SrSO4 just begins to ppt. Take that sulfate concn, plug it into Ksp for PbSO4 and solve for Pb. That is the Pb^2+ concn when SrSO4 begins to ppt.
Thank you so much!
Well, I'd say it's time for a chemistry joke! Why did the scarecrow win an award? Because he was outstanding in his field!
Now, let's get back to your question. When SrSO4(s) begins to precipitate, it means that all the available Sr2+ ions have reacted with SO42- ions to form a solid compound. In other words, the concentration of Sr2+ ions in solution will decrease to zero.
Since Pb2+ ions are not involved in the precipitation reaction, their concentration will remain unchanged. Therefore, the concentration of Pb2+ ions will still be 0.0200 M even when SrSO4(s) begins to precipitate.
I hope this answers your question and brings a smile to your face!
To find the concentration of Pb2+ (aq) when SrSO4(s) begins to precipitate, we need to determine the solubility product constant (Ksp) of SrSO4 and compare it with the concentrations of Pb2+ and Sr2+ in the solution.
The balanced equation for the formation of SrSO4(s) is:
Sr2+ (aq) + SO42- (aq) → SrSO4(s)
The Ksp expression for SrSO4 is given by:
Ksp = [Sr2+][SO42-]
Given:
[PB2+] = 0.0200 M
[Sr2+] = 0.0200 M (initially)
However, we need to consider that when SrSO4(s) precipitates, it will reduce the concentration of Sr2+ in the solution and cause the concentration of Pb2+ to increase due to a common ion effect.
Let's assume x M is the concentration of Pb2+ (aq) at equilibrium. At equilibrium, the concentration of Sr2+ will decrease by x M.
The Ksp expression at equilibrium can be written as:
Ksp = (0.0200 - x)(x)
Since the concentration of Pb2+ increases by x M, the expression becomes:
Ksp = (0.0200 + x)(x)
Given that the Ksp of SrSO4 is 3.54 × 10^-7 (from reference sources), we can solve for x:
3.54 × 10^-7 = (0.0200 + x)(x)
Expanding and rearranging the equation:
x^2 + 0.0200x - 3.54 × 10^-7 = 0
Solving this quadratic equation will give us the value of x, which represents the concentration of Pb2+ when SrSO4(s) begins to precipitate.
To find the concentration of Pb2+(aq) when SrSO4(s) begins to precipitate, we need to determine the solubility product constant (Ksp) of SrSO4(s) and the common ion effect.
Step 1: Write the balanced equation for the precipitation reaction:
SrSO4(s) ⇌ Sr2+(aq) + SO42-(aq)
The Ksp expression for the reaction is:
Ksp = [Sr2+][SO42-]
Step 2: Calculate the concentration of the common ion, SO42-(aq), when precipitation begins.
Since the solution already contains 0.0200 M of Pb2+(aq) and 0.0200 M of Sr2+(aq), the initial concentration of SO42-(aq) is zero.
Step 3: Use the Ksp expression to find the concentration of Pb2+(aq) when precipitation begins.
At the point of precipitation, we assume that the concentration of Sr2+(aq) decreases by x, and the concentration of SO42-(aq) also increases by x. Since the concentration of SO42-(aq) was initially zero, it becomes x when precipitation begins. Therefore, the concentration of SO42-(aq) is also x.
Using the Ksp expression for SrSO4, we have:
Ksp = [Sr2+][SO42-]
Ksp = (0.0200 - x)(x)
Step 4: Solve for x using the given Ksp value.
The Ksp value for SrSO4 is typically given in tables or can be found in the textbook or online sources. Let's assume the Ksp value is 3.2 × 10^(-7) (this is a hypothetical value).
3.2×10^(-7) = (0.0200 - x)(x)
Solving this quadratic equation will give us the value of x, which represents the increase in the concentration of SO42-(aq) when precipitation begins.
Step 5: Calculate the concentration of Pb2+(aq) when precipitation begins.
The concentration of Pb2+(aq) when SrSO4(s) begins to precipitate will be 0.0200 - x.
By solving the equation in step 4, you will find the value of x, and then you can subtract that value from 0.0200 M to obtain the concentration of Pb2+(aq) when precipitation begins.