Selective Precipitation - Online Tutor, Practice Problems & Exam Prep

Selective precipitation is a method used to separate ions from a solution by forming a precipitate with a reagent that binds to the desired ions. The success of this process hinges on the solubility product constant ($K$_sp) of the salt formed. When the reaction quotient ($Q$) exceeds the $K$_sp value, precipitation occurs as the system shifts to restore equilibrium, leading to the formation of a solid. This principle allows for the separation of specific ions from a mixture by strategically choosing reagents that will precipitate the target ion first, based on differing $K$_sp values. Understanding this concept is crucial for executing selective precipitation effectively in separating ionic compounds.

Selective Precipitation

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 K_sp value.

Now here we're going to say when Q is greater than K_sp, a precipitation is successful. So let's say we have here hypothetical equation of an ionic solid. It breaks up into its ions A⁺ and B^-. Now if Q is let's say this is from negative Infinity to positive Infinity. If Q is less than K_sp, Q will shift in the forward direction to get to K_sp. 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 is equal to K_sp, then we are at equilibrium and still no precipitate would form. It's not until we get to Q being greater than K_sp that a precipitate can form, because again Q will shift to K_sp, 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 will form.

Now using this logic, when asked to separate precipitate an ion from a mixture of ions is based on differing K_sp 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 K_sp values. So we're going to put to practice this conceptual idea of selective precipitation.

Selective Precipitation Example

Here we're told that a sample of a solution contains 0.405 molar of chromate ion and .628 molar of sulfide ions. These two ions can be precipitated with the use of lead to fluoride. Which ion will precipitate? Precipitate out first and at which concentration? So here the K_sp of lead to chromate is 2.0×10-16, and the K_sp of lead to sulfide is 7.0×10-29.

Well, here we're going to say that this is the smaller K_sp, the lawyer K_sp, the less soluble you will be and therefore the faster you can precipitate. We're going to say lead to 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 going to 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 fluoride.

What's going to happen here is that since led to sulfite has a lower cast P, it's going to be the one that forms this solid here first. So we're going to 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 K_sp. So we're going to have PB solid. And with K_sp we talk about how this ionic compound breaks up into its ions to be PB²⁺ aqueous plus S^2- aqueous.

We have initial change equilibrium with a nice chart we look at. 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 this is our concentration of sulfide ions. So that's what we're going to 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 K_sp, 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 going to be so small that it's not going to have a great enough impact on changing this .628 value. Now we're going to say K_sp equals products, so it equals lead to ion times sulfide ion. Here we're going to say that K_sp 4 led to sulfide in 7.0×10-29 lead. Two is equal to X Sulfide is 0.628.

Divide both sides by 0.628. X here is 1.115×10-28 molar. This number here represents the concentration at which our sulfide ion will begin to precipitate. So this will be our final answer.

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⁺? (K_sp = 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

Here’s what students ask on this topic:

Which of the statements correctly describe selective precipitation? Can you select all that apply?

Selective precipitation is a process used in chemistry to separate ions from a solution through the addition of a precipitating agent, which forms an insoluble compound with one of the ions. Here are the statements that correctly describe selective precipitation:

1. It involves the addition of a reagent that reacts with a specific ion to form an insoluble precipitate.
2. The precipitate that forms can be removed by filtration, leaving other ions in solution.
3. The choice of precipitating agent is crucial, as it must selectively react with one ion while not affecting others.
4. The process is governed by the solubility product constant (K_sp) of the precipitate; a lower K_sp indicates a less soluble compound, which is more likely to precipitate.
5. Selective precipitation is often used in qualitative analysis to identify the presence of specific ions in a mixture.
6. It can also be used in quantitative analysis by measuring the mass of the precipitate to determine the concentration of the ion in the original solution.

These statements encapsulate the concept and application of selective precipitation in chemical analysis and separation techniques.

What is selective precipitation?

Selective precipitation is a chemical process used to separate different ions from a solution through the addition of a reagent that precipitates one ion while leaving the others in solution. This is possible because different ions have different solubilities in a given solvent. By carefully choosing a reagent and controlling the conditions (such as concentration, pH, and temperature), you can selectively precipitate out a specific ion.

For example, if you have a solution containing both silver ions (Ag⁺) and sodium ions (Na⁺), and you add a reagent like hydrochloric acid (HCl), the silver ions will react with the chloride ions (Cl^-) from the HCl to form silver chloride (AgCl), which is insoluble in water and will precipitate out. The sodium ions, however, will remain in solution because sodium chloride (NaCl) is soluble in water. This technique is widely used in qualitative analysis and in purifying compounds in chemistry.

What is the name of the principle of selective precipitation used to identify the types of ions present in a solution?

The principle you're referring to is called qualitative analysis by selective precipitation. This method is used in chemistry to identify the types of ions present in a solution. It involves adding specific reagents that will react with certain ions to form a precipitate, which is a solid that separates from the solution. Each precipitate is characteristic of particular ions, and by observing the formation of these solids, you can deduce which ions are present. The process typically follows a series of steps, starting with the most insoluble compounds. By carefully choosing the sequence of reagents and conditions, chemists can methodically separate and identify the ions in a mixture. This technique is particularly useful in inorganic chemistry for analyzing the composition of minerals and other chemical samples.

When does selective precipitation occur?

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Selective precipitation occurs in the context of chemical solutions when a particular substance is caused to form a solid precipitate while leaving other substances in solution. This typically happens when the solubility limits of different ions in a solution are exploited. By adjusting the concentration of certain ions or changing the pH of the solution, you can reach a point where one ion exceeds its solubility product and precipitates out of the solution without affecting the others.

For example, if you have a solution containing both silver ions (<sup>Ag+</sup>) and lead ions (<sup>Pb2+</sup>), you can add chloride ions (<sup>Cl-</sup>) to the solution. Silver chloride (<sub>AgCl</sub>) has a much lower solubility than lead chloride (<sub>PbCl2</sub>), so the silver ions will precipitate as <sub>AgCl</sub> when a sufficient amount of chloride is added, while the lead ions remain in solution. This is a common technique in analytical chemistry for separating and identifying ions in a mixture.

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