Selective Precipitation - Video Tutorials & Practice Problems
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1
concept
Selective Precipitation
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Now selective 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 q is greater than Ksp, a precipitation is successful. So let's say we have here a 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 k s p, q will shift in the forward direction to get to k s p, so our chemical reaction shifts in the forward 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 that a precipitable 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 given Ksp values. So we're going to put to practice this conceptual idea of selective precipitation.
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example
Selective Precipitation Example
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You are told that a sample of a solution contains 0.405 molar of chromate ion and 0.628 molar of sulfide ions. These 2 ions can be precipitated with the use of lead 2 fluoride. Which ion will precipitate out first and at which concentration? So here, the Ksp of lead 2 chromate is 2.0 times 10 to the negative 16, and the Ksp of lead 2, so on 5 is 7.0 times 10 to the negative 29. Well, here we're going to say that this is the smaller Ksp. The lower your KSP, the less soluble you will be, and therefore the faster you can precipitate. We're gonna say lead 2 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 saphloride. What's going to happen here is that since lead to sulfide has a lower Ksp, it's going to 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 going to 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 p b 2+aqueouspluss2minusaqueous. We have initial change equilibrium. With a nice chart, we look at, 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 2 ion times sulfide ion. Here we're going to say that k s p for lead 2 sulfide is 7.0 times 10 to the negative 29. Lead 2 is equal to x, sulfide is 0.628. Divide both sides by 0.628. X here is 1.1 15 times 10 to the negative 28 molar. 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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