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### Course: MCAT>Unit 10

Lesson 3: Solubility equilibria

# The common-ion effect

The solubility of an ionic compound is decreased by the presence of a common ion (an ion that is also present in the compound); this is known as the common-ion effect. Created by Jay.

## Want to join the conversation?

• how do you know if it is exceptable to treat x as zero?
• in this example, .1 is so big compared to the ksp that x will not make a big difference to .1 and .1-2x will pretty much just be .1 so now you dont have to solve with weird quadratic stuff and it's just way simpler and you have less of a chance of making a mistake this way
• What will change the Ksp value, if common ions do not?
• The Ksp is temperature dependent, so heating or cooling the solution can affect the Ksp. For some substances, Ksp is higher at higher temperatures, for other substances Ksp is higher at lower temperatures, for some substances there isn't a great deal of temperature dependence for the Ksp, for some substances there is a very complicated relationship between temperature and Ksp.
• Oh! So is this how those Grow A Crystal kits work? They put a powdered form of the ionic compound of the crystal into water, and then add a much more soluble salt with a common ion, which will force the crystal to grow. Is that correct? It's really cool!

Can you grow any ionic network crystal this way? It would be really great to grow a ruby or an emerald by mixing my own solutions.
• It's possible to create gemstones using the common-ion effect I would think. My biggest concern with this method is that, would it create large enough crystals and how many defects would it have? Because with gemstone creation it's not enough just to know their chemical composition, a lot of work goes into how they are made since that effects the quality of the crystals produced.

Hope that helps.
• He says that due to common ion effect the reaction shifts in the backward direction to form more PbCl2.But why cant it form more KCl?
• This is an example of Le Châtelier's Principle.
If you add more Cl⁻, the position of equilibrium shifts in a direction that gets rid of the excess Cl⁻.
That is, it forms more PbCl₂.
• How does Qsp relate to Kip?
• I'm confused about how 0.016 M was found at . Please correct me if I am wrong but, if there were no common ions in the solution, then the Ksp would equal 1.7*10^-5= [x]*[2x]^2=x*4x^2=4x^3. I ended up with x=1.41666*10^-6, which should be equal to the molar solubility of the salt.
• So you're correct that if there were no chloride common ions in solution and we were solving for the molar solubility the equlibium equation would be:
1.7x10^(-5) = 4x^(3). Solving for x, we'd need to divide both sides by 4 first.
4.25x10^(-6) = x^(3) Then we'd need to take a cube root of both sides.
1.6198x10^(-2) = x, which agrees with Jay's answer when rounded for sig figs. So if I had to guess, I'd say you didn't do the cube root correctly.

Hope that helps.
• Wouldn't the existing concentration of Cl- be 0.200M in the ICE table? I have a very similar example in my equilibrium package in which the solvent, CaCl2, is 0.150M. in the ICE table when PbCl2 is dissociated into its ions (PbCl2 (s) <---------> Pb (aq) + 2Cl (aq)), the 2Cl initial concentration is 0.300M. It was doubled from the common ion's 0.150M to 0.300M due to the coefficient of 2. In this example that is not so. I'm confused.
• In the example in this video, the initial Cl- concentration is only 0.100 M, which comes entirely from the KCl. Initially, you assume no dissociation of the PbCl2 has occurred.

If you were using CaCl2, rather than KCl, then you have double the number of chlorines so then the initial molarity of Cl- would be doubled.
• How is the common ion effect different from simply increasing the concentration of a reactant by manually adding it?
• I mean, that’s still the common ion effect. The common ion effect says the solubility of a salt is decreased by the presence of constituent ions in the solution.

If the constituent ions are already present in the solution, then the salt will dissolve less so than if it were in pure water. If the salt has already dissolved as much as its equilibrium will allow, and then we add some additional constituent ions, this is essentially a stress on equilibrium and it will shift the reaction to the left thereby decreasing the solubility of the salt. What would happen in that case is that a precipitate would begin to form (or increase the size of an already existing solid).

In both cases we reduce the solubility of the salt with the presence of constituent ions. The only difference is what order we add them to the solution.

Hope that helps.
(1 vote)
• Could you just find the solubility for PbCl2 as if there is no KCl in the solution and then multiply by the molarity of the common ion (in this case Cl-) to get the new solubility?
(1 vote)
• No, because all the concentrations change when you add a common ion.