Suppose a solution contains 0.21 M Pb2 and 0.41 M Al3 . Calculate the pH range that would allow Al(OH)3 to precipitate but not Pb(OH)2.
Why do you never post a step by step solution so others can follow. instead its just brief explanation
Oops, sorry. The Ksp for Al(OH)3 is given as 4.6×10^–33 and the Ksp for Pb(OH)2 is given as 1.43×10^–20.
What's the best way to calculate the answer with those values? I don't understand.
Ah, a question about precipitation and pH. Well, buckle up because I'm about to break it down with a touch of humor!
When it comes to precipitation, it's always good to keep an eye on those tricky hydroxide ions (OH-). They can cause quite a commotion in the solution! In this case, we want Al(OH)3 to form and kick out Pb(OH)2. So, let's find that sweet pH range!
To figure out when the precipitation party starts, we need to determine the solubility product constants (Ksp) for both compounds. The Ksp for Pb(OH)2 is about 1.6 x 10^-20, while for Al(OH)3, it's around 3 x 10^-35.
Now, let's focus on the lead (Pb2) first. To keep it from precipitating into Pb(OH)2, we want the concentration of hydroxide ions (OH-) to stay below the critical value. The formula for the solubility product is Ksp = [Pb2+][OH-]^2. By plugging in our Ksp value and rearranging the equation a bit, we find [OH-] < (Ksp/[Pb2+])^0.5.
For aluminum (Al3), things get a bit trickier. Its solubility product equation is Ksp = [Al3+][OH-]^3. Since we don't want Al(OH)3 to precipitate, we want the concentration of hydroxide ions to be higher than (Ksp/[Al3+])^(1/3). If we simplify it, [OH-] > (Ksp/[Al3+])^(1/3).
Now, let's merge these two inequalities and solve for pH. We want the pH to be within the range that satisfies both conditions: [OH-] < (Ksp/[Pb2+])^0.5 and [OH-] > (Ksp/[Al3+])^(1/3).
So, the pH range that would allow Al(OH)3 to precipitate but not Pb(OH)2 is like finding the perfect balance between two friends. Imagine a range that starts with everyone whispering "hi" at pH = -log[OH-] < (Ksp/[Pb2+])^0.5 and ends with "hellooo" at pH = -log[OH-] > (Ksp/[Al3+])^(1/3). Find that sweet spot, and voila! You've got your answer.
But hey, remember to always double-check your calculations, and don't forget to invite me to your next chemistry-themed party!
To determine the pH range that allows Al(OH)3 to precipitate but not Pb(OH)2, we need to consider the solubility product constants (Ksp) for both compounds.
Step 1: Write the balanced chemical equations for the precipitation of Al(OH)3 and Pb(OH)2.
Al(OH)3(s) ⇌ Al3+(aq) + 3OH^-(aq)
Pb(OH)2(s) ⇌ Pb2+(aq) + 2OH^-(aq)
Step 2: Write the solubility product expressions for both compounds.
For Al(OH)3: Ksp = [Al3+][OH^-]^3
For Pb(OH)2: Ksp = [Pb2+][OH^-]^2
Step 3: Find the concentration of OH^- ions.
In order to precipitate Al(OH)3 but not Pb(OH)2, the concentration of OH^- ions should be large enough to exceed the solubility product of Al(OH)3, but not the solubility product of Pb(OH)2.
Since we have a solution containing 0.21 M Pb2+ and 0.41 M Al3+, we can assume that these ions will dissociate completely, as they are strong electrolytes. Therefore, the concentration of Pb2+ and Al3+ is already known.
Step 4: Substitute the known concentrations into the solubility product expressions and solve for [OH^-].
For Al(OH)3: Ksp = [Al3+][OH^-]^3
0.41M x [OH^-]^3 = Ksp for Al(OH)3
For Pb(OH)2: Ksp = [Pb2+][OH^-]^2
0.21M x [OH^-]^2 = Ksp for Pb(OH)2
Step 5: Find the value of [OH^-] for both compounds.
Using the known Ksp values for each compound, and the concentrations of the respective ions, you can solve the equations to find the value of [OH^-].
For Al(OH)3: Substitute the Ksp value of Al(OH)3 and the concentration of Al3+:
0.41M x [OH^-]^3 = Ksp of Al(OH)3
Solve for [OH^-].
For Pb(OH)2: Substitute the Ksp value of Pb(OH)2 and the concentration of Pb2+:
0.21M x [OH^-]^2 = Ksp of Pb(OH)2
Solve for [OH^-].
Step 6: Calculate the pH range.
Once you have the value of [OH^-] for both compounds, you can calculate the corresponding pH values using the equation:
pOH = -log[OH^-]
pH = 14 - pOH
The pH range that allows Al(OH)3 to precipitate but not Pb(OH)2 is the pH range where the pOH value for Al(OH)3 is greater than the pOH value for Pb(OH)2. Convert the pOH values to pH values to obtain the pH range.
Pb(OH)2 ==> Pb^2+ + 2OH^-
Al(OH)3 ==> Al^3+ + 3OH^-
Calculate OH^- needed for a 0.21M Pb^2+ solution to ppt Pb(OH)2,
Calculate OH^- needed for a 0.41M Al^3+ solution to ppt Al(OH)3.
You will want the (OH^-) to be less than you calculate to prevent pptn of Pb(OH)2. Convert that OH to pH so you will want the pH to be greater than that. When you don't show the Ksp values you are using we can't calculate anything and hope to get the right answer since you are probably not using the same values for Ksp that we are. Textbooks differ in the tables they use.