Selective Precipitation - Video Tutorials & Practice Problems
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concept
Selective Precipitation
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Now, selected precipitation represents a process of separating ions out of a solution by using reagents that form a precipitate or solid with the ions. Now, a reagent is just another ion that binds to the dissolved ion and precipitates out of a solution. Now, a successful precipitation of a selected ion depends on the solubility of its salt and this is connected to its KSP value. Now, here we're gonna say when Cuba is greater than KSP, a precipitation is successful. So let's say we have here hypothetical equation of an ionic solid, it breaks up into its ions. A positive and B minus. Now, if Q is, let's say this is from negative infinity to positive infinity. If Q is less than KSPQ will shift in the four direction to get to KSP. So our chemical reaction shifts in the four direction we'd be moving away from solid towards ions making more of them. So no precipitate would form. If Q were equal to KSP, then we are at equilibrium and still no precipitate would form. It's not until we get to Q being greater than KSP. That a precipitate can form. Because again, Q will shift to KSP. So it shifts in the reverse direction. So our chemical reaction shifts in the reverse direction towards a solid that shows us that a precipitate form. Now, using this logic, when asked to separate precipitate an ion from a mixture of ions is based on deferring KSP values. So we'll look at the mixture of ions and we'll see what kind of solids can form and we'll see which one forms fastest based on our giving KSP values. So we're gonna put to practice this conceptual idea of selective precipitation.
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example
Selective Precipitation Example
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You were told that a sample of a solution contains 0.405 molar of chromate ion and 0.628 molar of sulfide ions. These two ions can be precipitated with the use of lead two fluoride, which ion will precipitate precipitate out first. And at which concentration. So here, the KSP of lead two chromate is 2.0 times 10 to the negative 16 and the KSP of lead to so 5 to 7.0 times 10 to the negative 29. Well, here we're going to say that this is the smaller KSP, the lower KSP, the less soluble you will be. And therefore the faster you can precipitate, we're gonna say lead two sulfide contains the sulfide ions. So we'd say that the sulfide ion will precipitate out first. Now, if we want to think about this visually, we're gonna say we have floating around in our solution sulfide ions and chromate ions and I come in and I dissolve or pour in lead to SULU. What's going to happen here is that since lead to sulfide has a lower KSP, it's gonna be the one that forms this solid here. First, So we're gonna have lead to sulfide precipitating out first. Now, how do we determine its concentration? Well, since sulfide precipitates out first, we're gonna use its KSP. So we're gonna have PB solid. And with KSP, we talk about how this ionic compound breaks up into its ions to be PB two plus aqueous plus S two minus aqueous. We have initial change equilibrium with the nice chart we look at. Um so we ignore solids and liquids. So the reactant is a solid. So we ignore it. We're going to say here, we don't have any initial amount of lead, but we know that initially, we, this, this is our concentration of sulfide ions. So that's what we're gonna plug in 0.628 molar, they're both products we're making them. So they're both plus X and then bring down everything. Now, remember when dealing with KSP, if you have an actual or real number in front of your X variable, you can ignore the X variable. That's because X is gonna be so small that it's not gonna have a great enough impact on changing this 0.628 value. Now, we're gonna say KSP equals products, so equals lead to ion times sulfide ion. Here, we're going to say that KSP four lead to sulfide is 7.0 times 10 to the negative 29 light two is equal to X. Sulfide is 0.628 divide both sides by 0.628 X year is 1.115 times 10 to the negative 28 more. This number here represents the concentration at which our sulfide ion will begin to precipitate. So this will be our final answer.
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Problem
Problem
Solution contains [Cu2+] = 0.035 M, [Sr2+] = 0.054 M, [Al3+] = 0.23 M. Cu2+ can be separated by selective precipitation using NaOH. What is the minimum concentration of NaOH needed to start precipitation of Cu2+? (Ksp = 2.2 × 10−20 of Cu(OH)2).
A
1.57× 10−19 M
B
7.93 × 10−10 M
C
3.96 × 10−10 M
D
1.19 × 10−9 M
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