Ksp: Common Ion Effect - Online Tutor, Practice Problems & Exam Prep

The common ion effect reduces the solubility of an ionic solid in a solution that already contains an ion or ions similar to those in the solid. This is due to Le Chatelier's principle, which states that a system in equilibrium will adjust to minimize the impact of any changes. When an ionic solid is introduced into pure water, it dissolves up to its solubility limit as determined by its solubility product constant (K_SP). However, if the solution already contains common ions, the solid's solubility decreases because the equilibrium shifts to favor the reverse reaction, thus less of the solid dissolves. This effect also applies to acids and bases, where the presence of hydroxide ions or hydrogen ions (H⁺ or H₃O⁺) in a solution decreases the solubility of bases or acids, respectively. Understanding the common ion effect is essential for predicting solubility changes in various chemical contexts.

Common Ion Effect on Solubility

With KSP, we now have to talk about the common ion effect. Now, the common ion effect decreases the solubility of a solid in a solution. Here we're going to say occurs when an ionic solid dissolves in a solution containing ion or ions common to it. Now the decrease insolubility is due to Le Chatelier's principle, which remember states that the chemical reaction will shift either in the reverse direction or forward direction to decrease the disturbance to its equilibrium.

If we take a look here in the first beaker, we have no common onion, meaning that I have taken this barium sulfite and placed it within pure water. It naturally will break up into its ions, so to break up into a barium ion plus a sulfite ion. Now here we're going to say for the less soluble one, what's the difference? Well, I'm taking this barium sulfate sulfite and I'm placing it into that solution, but already dissolve within the solution is some barium and some sulfite. So we're going to write this over here. We have barium solid being thrown into a solution that already possesses Barry myon and sulfide ion up here was being thrown into pure water.

Alright. The way you need to think about this is there is basically a limit on how much of this ionic solid you can get to dissolve. In the pure water. There is none of those these ions present initially, so it's free to dissolve up to its limit based on its KSP value. In the second beaker, though some of these are already floating around, which means that this ionic solid can't basically dissolve to its maximum that it wants based on its KSP. It is limited by the fact there's already some barium and sulfite in the solution, so that means less of these would be created.

So here there are common ions, and because there's already some of these in the solution, the reaction will favor the reverse direction to maintain its equilibrium. And this is a call back to the Schottelers principle. All right, So just keep that in mind. When ionic solid breaks up in pure water, there is no common ion effect involved. The ionic solid is free to dissolve as much as it can based on its KSP value. Within a solution that already has some of its ions present. It's not as free to dissolve as much as it can because it's going to reach its limit sooner. All right, So keep that in mind when talking about the common eye, in effect, and its relation to KSP.

Ksp: Common Ion Effect Example

Here it says determine molar solubility of copper 2 carbonate in .15 molar magnesium carbonate solution. All right, so here they're giving us the K_SP of copper 2 carbonate. So that's what we're going to break up into ions Here, Copper to carbonate breaks up into copper 2 ion plus the carbonate ion.

Step one, we set up the ice chart with solid as the only reactant and we cross out the reactant side because remember in an ice chart we ignore solids and liquids. With a nice chart, we have Initial change Equilibrium Step 2 using initial row place the amount given for the common ion. Alright, so let's go back to this question. It says that this is not breaking up in pure water, it's breaking up in a solution that's .15 molar magnesium carbonate.

So magnesium carbonate is made-up of magnesium ion plus the carbonate ion. Each one has a concentration of 0.15 molar. Now where is the common ion? Well, in our ice chart we're dealing with carbonate. Our solution has carbonate. They have the same ion. So that means initially we're starting out with 0.15 molar. We don't have a common ion for copper 2, so it's 0 initially.

Now remember we lose reactants to make product, so we're going to say using the change row, place a + X for the products. So this is plus X + X. Now using the equilibrium row, set up the equilibrium constant expression with K_SP and solve for X. The variable X and its number can be ignored if it follows a real number. So let's fill out the rest of this ice chart. Bring down everything. This is plus X. This is .15 + X.

Now it's this portion that we're talking about, because if we look, we have .15 + X. The variable X in its number can be ignored if it follows a real number. This X here follows a real number .15. That means we can ignore this X. And that's because a comparison of point 15X will be so small that it's not going to have an impactful change on the .15 number.

Now here we're going to say see K_SP equals products overreactants. Your reactant gets cancelled out because of the solid, so we have copper 2 plus times carbonate here. K_SP is 2.4×10-10 = X times 0.15. Again, we're ignoring the plus X. Divide both sides by .15 so we already have our x, x equals 1.6×10-9 molar. This will be our answer because remember when it comes to the solubility or molar solubility of your ionic compound as a whole, that's just X.

So that's what we just found. And step five, we don't need to do it because it says convert found value of X into appropriate units if necessary. Here they want us to find molar solubility which is typically in units of molarity. So we're already done. So this would be our final answer for the molar solubility of our ionic solid.

Common Ion Effect: Acids & Bases

Now a common ion effect can also occur with acids and bases. Here we can say that the solubility of a base decreases if the solution contains hydroxide ions already.

And we can say that the solubility of an acid also decreases if the solution contains hydronium ion in the form of H+. But also remember we can see hydronium ion in the form of H3O+.

So just remember, common ion effect doesn't only affect ionic solids, it can affect acids and bases as well.

Ksp: Common Ion Effect Example

Here it says find the solubility in grams per milliliter of chromium hydroxide. You were told the K_sp is 6.7 * 10 to the -31 if the solution is buffered at a pH of 8.4 at 25°C. All right, so here we have chromium 3 hydroxide solid breaking up into its ions, which are one chromium 3 ion plus three hydroxyl ions. Because we're talking about K_sp, we set up an ice chart which stands for initial change equilibrium.

Now in step one it says set up an ice chart. What the solid is only reacted cross out the reactant side because in a nice Charlie ignore solids and liquids. The reactant is a solid so it's ignored. Using the initial roll, place the amount for the common ions. Calculate OH^- or H⁺ from the given pH or pOH. This is a base. Base is used pOH initially, so we would change the pH given to US into pOH. pOH equals 14 minus pH, so 14 - 8.4 which is 5.6.

Now if we know the pOH, then we know how much OH^- concentration we have, because OH^- OH^- equals 10 to the negative pOH. So we're going to say OH^- equals 10 to the -5.6, which comes out to be 2.511886 times 10 to the -6. And we're going to say that that amount of OH^- represents the initial amount of our OH^-. Within this question, we're being placed in a solution with that pH, which means it already has that much OH^- present. We don't have any common unanswered chromium 3, so this is 0 again. This is 2.511886 * 10 to the -6.

Now we lose reactants to make products using the change role. Place a + X for the products because we're making them so plus X. There's a three here so this is plus 3X. Bring down everything plus X and then 2.511886 * 10 to the -6 + 3 X. So Step 4. Using the equilibrium row, set up the equilibrium constant expression with K_sp and solve for X. The variable X and its number can be ignored if it follows a real number. If we look here, this is a real number and it's followed by three plus by three X. Because of that, we can ignore this portion.

Now here we have to convert the found value of X into appropriate units if necessary. So let's just solve for X initially. So K_sp equals products, so it's chromium 3 ion times hydroxide ion cubed. Because of the coefficient of three K_sp is 6.7 * 10 to the minus 31, and that equals chromium 3, which is X at equilibrium times 2.511886 * 10 to the -6 ^3. We're going to say here this is 6.7 * 10 to the -31 = X times. So I'm just going to take the cube of this value when I do that. That gives me 1.58 times 10 to the -17. Divide that out from both sides. Now when I do that, that'll give me x, x here will equal 4.227 * 10 to the -14 molar. This will be the molar solubility of my entire ionic compound. So this is for Chromium 3 hydroxide.

Now, they don't want it in molarity, which is moles per liter. They want the solubility in grams per milliliter. So we're going to do some converting here. Now remember, molarity itself is moles over liters, So this number is equivalent to 4.227 * 10 to the -14 moles of Chromium 3 hydroxide per one liter. So here we're going to say that one mole of chromium 3 hydroxide. If you look at its mass, it comes out to 103.02g, and that's the mass of the chromium and the three hydroxides, all their masses together. And then finally, we need to get rid of liters. So we put one liter up here, and that's equal to 1000 milliliters.

Here liters cancel out and what we have at the end will be grams per milliliter. So when I work that out, I get 4.35 * 10 to the -15g per milliliter of Chromium 3 hydroxide. So this would be our final answer for this given question.

Which of the following compounds will become less soluble in a chloride containing solution?

a) Na₂S

b) AgCl

c) BaI₂

d) CaF₂

• A. Na₂S
• B. AgCl
• C. BaI₂
• D. CaF₂

A solution of Ba(OH)₂ has a K_sp of 5.0 x 10⁻³.

i) Determine the pH of this solution.

ii) Determine the pH if Ba(OH)₂ was added to a solution containing 3.2 M of BaF₂ and 0.94 M of Al(OH)₃.

• A. i. 2.30
  ii. 14.03
• B. i. 0.67
  ii. 13.55
• C. i. 13.33
  ii. 14.45
• D. i. 2.30
  ii. 13.65

Here’s what students ask on this topic:

What is the common ion effect?

The common ion effect refers to the phenomenon where the solubility of a salt is reduced when another compound that shares a common ion is added to the same solution. This effect is a consequence of Le Chatelier's principle, which states that a system at equilibrium will react to counteract any changes imposed upon it.

For example, consider a solution of silver chloride (AgCl), which is only slightly soluble in water. If you add sodium chloride (NaCl) to the solution, the concentration of chloride ions (Cl^-) will increase because NaCl is highly soluble and dissociates completely. Since AgCl and NaCl share the common ion Cl^-, the increased concentration of Cl^- from the NaCl will shift the equilibrium of the slightly soluble AgCl to the left, according to the equilibrium equation:

AgCl(s) → Ag⁺(aq) + Cl^-(aq)

As a result, more AgCl will precipitate out of the solution to re-establish equilibrium by reducing the concentration of Cl^- ions. This is the common ion effect in action, where the addition of a common ion decreases the solubility of an ionic compound in a solution.

The addition of which compound will cause a shift in equilibrium because of the common ion effect?

The addition of a compound that contains a common ion will cause a shift in equilibrium due to the common ion effect. This effect is based on Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

For example, consider the equilibrium established in a saturated solution of calcium sulfate (CaSO₄):

CaSO₄ (s) → Ca²⁺ (aq) + SO₄^2- (aq)

If you add a compound like sodium sulfate (Na₂SO₄), which also contains the sulfate ion (SO₄^2-), you introduce more sulfate ions into the solution. According to the common ion effect, the equilibrium will shift to the left to reduce the concentration of the common sulfate ions by precipitating more calcium sulfate. This results in a decrease in the solubility of calcium sulfate in the solution. The common ion effect can be applied to any equilibrium involving ions in solution, where the addition of a common ion will always shift the equilibrium away from the side with the common ion.

Why does the common ion effect decrease solubility?

The common ion effect refers to the decrease in solubility of an ionic compound when a solution already contains one of the ions present in the compound. This occurs due to Le Chatelier's principle, which states that if a dynamic equilibrium is disturbed, the system will adjust to minimize that disturbance.

When you introduce a common ion into the solution, you are effectively increasing the concentration of that ion. For a saturated solution in equilibrium, this addition shifts the equilibrium to favor the formation of the solid precipitate in order to reduce the concentration of the common ion in the solution, thus decreasing the solubility of the compound.

This is because the solubility product (K_sp) of the ionic compound—a constant value at a given temperature—must be maintained. The presence of the additional common ion means that less of the ionic compound can dissolve to reach the required product of ion concentrations dictated by the K_sp. Consequently, the solubility of the compound decreases in the presence of a common ion.