Precipitation: Ksp vs Q - Online Tutor, Practice Problems & Exam Prep

Understanding the relationship between the solubility product constant (K_sp) and the reaction quotient (Q) is essential for predicting whether a precipitate will form in a solution. K_sp indicates the maximum solute concentration that can dissolve in a solvent to reach equilibrium. When Q is less than K_sp, the solution is unsaturated, and no precipitate forms as the reaction shifts forward to dissolve more solute. If Q equals K_sp, the solution is saturated, and the system is at equilibrium with no change. However, when Q is greater than K_sp, the solution is supersaturated, and the reaction shifts in reverse to form a precipitate. This understanding is crucial for fields like chemistry and environmental science, where controlling precipitation reactions is often necessary.

Solution Saturation

Now recall that K_sp helps to determine how soluble an ionic solid can be in a solvent and equilibrium. Also recall that Q, which represents our reaction quotient, is a ratio of products to reactants at a particular time. Now here we're going to say comparing K_sp to Q will determine if a solid, otherwise known as a precipitate, is likely to form. Here we use the idea of solution saturation to determine this idea.

Now solution saturation is just the amount of solute that has been dissolved in a solvent, and we're going to see the degree of solution saturation can be determined by the relative value of K_sp to Q. Now we know that K_sp deals with an ionic solid and we're talking about how it breaks up into its ions. So let's say our hypothetical ionic solid is AB solid. It breaks up into A⁺ aqueous plus B^- aqueous. Now Q could be less than K_sp, Q could be greater than K_sp, or Q could be equal to K_sp.

Now if we take a look here at this number line, let's say that this is from negative infinity to positive infinity because we know that K_sp can be a number that's very small. Or here we could have Q be a larger number than K_sp, something greater than one. Now before the equivalence point, Q is less than K_sp. Before the equivalence point, we haven't reached our maximum saturation level or we haven't reached our solution maximum solution saturation. So we are unsaturated. Remember Q will shift to get to K_sp.

Here Q will shift in the forward direction to K_sp. The direction Q shifts is the same direction our chemical reaction will shift, so it shifts in the forward direction making more ions. So here no precipitate is formed. But let's say that Q is greater than K_sp. Well, that would be after equilibrium. But again Q will shift to get to K_sp, get to equilibrium. So here shift in the reverse direction. If it shifts in the reverse direction, my chemical reaction also shifts in the reverse direction and look where we're heading. We're heading towards a solid to precipitate.

That's why a precipitate forms and this would represent a super saturated solution. We've gone beyond our maximum solution saturation level now. If Q is equal to K_sp, then we are at equilibrium and we represent a saturated solution. The maximum amount of solute or ionic solids is dissolved within our solvent and again no precipitate would form. So just remember, in terms of Q versus K_sp, a precipitate can successfully be created when Q is greater than K_sp because that represents a super saturated solution.

Precipitation: Ksp vs Q Example

Here it says barium sulfate precipitate out when 8.2 * 10 to the -7 molar barium carbonate is mixed with 5.7 * 10 to the -6 molar strontium sulfate. Here the K_sp of barium sulfate is given as 1.1 * 10 to the -10. All right, so they give us K_sp of barium sulfate. So that is our point of interest. That's what we're going to focus on.

Breaking up into ions. We have barium salt, barium sulfate solid. We see it breaking up into ions. It breaks up into one barium ion plus one sulfate ion. Now here if we're dealing with K_sp, we're dealing with an ice chart. So initial change equilibrium with a nice chart. We ignore solids and liquids, so this solid will be ignored. Now this barium sulfate is not breaking up in pure water, it's breaking up in a solution composed of these two compounds.

Then mixing together allows the barium ion here and the sulfate ion here to combine together to give us our solid. So we have two common ions involved. We have barium ion, its concentration is this 8.27 * 10 to the -7 molar barium ion and we have this as the concentration of our sulfate ion. This is important because these two ions are common to our ice chart equation. So basically we have double common ion effect.

So this is 8.27 * 10 to the -7 molar initially and this 5.7 * 10 to -6 molar initially. They're both products, we're making them. We bring down everything. Now when you have initial amounts for both of your ions, this helps us to find. Q is just like K_sp equals products here, so it equals barium ion times sulfate ion. Remember, if you have an actual real number, you could ignore the X variables when it comes to K_sp and Q.

So we plug these numbers in for our two ions and then we're going to find out what Q is. So when I do that I find Q is 4.674 * 10 to the -12. I know what K_sp I know Q is. So now I can compare them to each other K_sp is to the -10, Q is to the -12. Q was smaller, so Q is going to shift in the forward direction to get to K_sp. If Q moves in the forward direction, I get to K_sp. Then our chemical reaction also moves in the forward direction.

Where are we moving? We're moving away from our solid towards our ions, our barium sulfate is becoming more soluble. So we're not making any precipitate, we're making more ions. So here you're going to say no, barium sulfate will not precipitate out.

Two mixtures are added into one flask at 25 °C, one mixture contains 0.55 mL of 0.75 M BaF₂ and another 0.25 mL of 1.3 M Mg(OH)₂. K_sp of Magnesium Fluoride, MgF₂, is 7.4 x 10⁻⁹. Identify the correct option.

a) MgF₂ solid will form

b) MgF₂ solid forms, along with Mg⁺² and F⁻ ions

c) solution is unsaturated, precipitate does not form

d) solution is saturated, precipitate forms

• A. MgF₂ solid will form
• B. MgF₂ solid forms, along with Mg⁺² and F⁻ ions
• C. solution is unsaturated, precipitate does not form
• D. solution is saturated, precipitate forms

Here’s what students ask on this topic:

What happens when $q$ is less than $K_{\mathrm{sp}}$?

When $q$, which stands for the reaction quotient, is less than the solubility product constant ($K_{\mathrm{sp}}$), it indicates that the ionic product of the dissolved ions in the solution is less than the equilibrium concentration of these ions in a saturated solution. In simpler terms, the solution is not saturated with the solute, and more solute can dissolve.

This situation can occur when you have a solution where the solute has not fully dissolved, or when you add more solvent to a saturated solution. Since $q$ is less than $K_{\mathrm{sp}}$, the system has not reached equilibrium, and the forward reaction, where the solid solute dissolves to form ions, is favored. This means that if you add more of the solid solute to the solution, it will continue to dissolve until the concentration of the dissolved ions reaches a point where $q$ equals $K_{\mathrm{sp}}$, at which point the solution will be saturated and equilibrium will be established.

Does a precipitate form when $Q$ is greater than $K_{\mathrm{sp}}$?

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Yes, when the reaction quotient (Q) is greater than the solubility product constant (K<sub>sp</sub>), a precipitate will form. This is because the Q value represents the current state of the ionic concentrations in the solution, and the K<sub>sp</sub> value represents the maximum product of the ionic concentrations that can exist in a saturated solution without forming a precipitate.

When Q > K<sub>sp</sub>, it indicates that the concentrations of the ions in the solution exceed the solubility limit, meaning the solution is supersaturated. As a result, the excess ions will start to come together to form a solid, which is the precipitate, in order to reduce the ion concentrations in the solution and return to a state of equilibrium where Q equals K<sub>sp</sub>. This process is known as precipitation, and it continues until the concentrations of the ions in the solution have decreased enough that Q is no longer greater than K<sub>sp</sub>.

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What happens when the concentrations of the ions in solution are high and $Q$ is larger than $K{\mathrm{sp}}_{}$?

When the concentrations of ions in a solution are high enough that the reaction quotient (Q) is larger than the solubility product constant (K_sp), the solution is supersaturated with respect to the particular ionic compound. This means that there are more dissolved ions in the solution than the solubility equilibrium can support.

Under these conditions, the solution is unstable, and a precipitation reaction is likely to occur to restore equilibrium. Excess ions will start to come together to form solid precipitate until the concentration of the dissolved ions decreases enough that Q equals K_sp. At this point, the system will be at equilibrium, and the rate at which the solid dissolves will be equal to the rate at which the solid precipitates out of the solution. This process is governed by Le Chatelier's Principle, which states that a system at equilibrium will adjust concentrations to counteract changes (such as an increase in ion concentration).