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.
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- how do you know if it is exceptable to treat x as zero?(14 votes)
- 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(18 votes)
- What will change the Ksp value, if common ions do not?(5 votes)
- 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.(24 votes)
- 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?(6 votes)
- 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.(3 votes)
- How does Qsp relate to Kip?(4 votes)
- By analogy to Qeq, Qsp will tell you how 'close' you are to saturating a solution.
see this section: https://www.khanacademy.org/science/chemistry/chemical-equilibrium/factors-that-affect-chemical-equilibrium/(2 votes)
- 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.(3 votes)
- 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.(3 votes)
- 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.(0 votes)
- 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.(8 votes)
- 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.(3 votes)
- Was it just a coincidence that the molarity of the new solvent was also .1, the same factor that the molar solubility differed from the solubility in water? Or is there some sort of connection to the solubility in water and the molarity in another solution like that?(2 votes)
- If KCl is already in the solution, why aren't we including the concentration of K+ in the equation for Ksp?(1 vote)
- I suspect because K+ is a spectator ion and does not participate in the overall reaction, which is why its concentration is omitted.(1 vote)
- i am looking for a video that explains the factors afecting solubility of compounds,atoms,ions
and also a video on factors affecting reactivity of metals
plz give me links. i am unable to find on K.A.(1 vote)
- [Instructor] The presence of a common ion can affect a solubility equilibrium. For example, let's say we have a saturated solution of lead II chloride. Lead II chloride is a white solid, so here's the white solid on the bottom of the beaker. And the solid's at equilibrium with the ions in solution. So that would be Pb2+ and Cl-. Notice how the mole ratio is one-to-two of Pb2+ to 2Cl-. So if we have two Pb2+ ions in our diagram, there should be twice as many chloride anions. At equilibrium, the rate of dissolution is equal to the rate of precipitation. Therefore, the concentration of ions in solution remains constant. So our system is at equilibrium. And let's add some solid potassium chloride. Potassium chloride is a soluble salt. So it will dissociate and turn into K+ and Cl- in solution. Adding a source of chloride anion means the system is no longer at equilibrium. So let me write in here, not at equilibrium at that moment in time. So the system was at equilibrium and a stress was added to the system. In this case, the stress was increased chloride anion. So there's an increase in the concentration of Cl-. According to Le Chatelier's principle, the system will move in the direction that decreases the stress. So if the stress is increased concentration of chloride anion, the system will move to the left to get rid of some of that extra chloride anion. When the system moves to the left, Pb2+ ions will combine with chloride anions to form PbCl2. And we can see that down here in the diagram. So imagine, say this Pb2+ ion combined with these two chloride anions to form some more of the white solid. Looking at the third diagram, the amount of white solid has increased from the second diagram, and we've lost this Pb2+ ion and these two chloride anions. And the amount of our precipitate PbCl2 will keep forming until equilibrium is reached. Let's just say this third diagram, it does represent the system at equilibrium. So I'll write on here, at equilibrium. This is an example of the common ion effect. For this problem, the common ion is the chloride anion, because there were two sources of it. One was from the disillusion of PbCl2. If we had dissolved some solid to make a saturated solution, the source of these chloride anions would be from PbCl2. And the second source is from the added KCl, which of course dissolved to form chloride anion. So the chloride anion is the common ion. And we use Le Chatelier's principle to predict the system will move to the left to get rid of the extra chloride anion. When the system moved to the left, we formed more of the solid PbCl2. And that's why this amount got bigger over here. So if we compare the first diagram with the third diagram, the first diagram has more of the lead II chloride in solution. And the third diagram has less of it. Therefore, the addition of the common ion of chloride anion, that decreased the solubility of lead II chloride. So the common ion effect says that the solubility of a slightly soluble salt, like lead II chloride, is decreased by the presence of a common ion. Another way to think about this is using the reaction quotient, Q. For the diagram on the left, we're at equilibrium. Therefore the reaction quotient Qsp is equal to the Ksp value for lead II chloride, which means the system is at equilibrium. Adding chloride anion increases the value for Qsp. So now Qsp is greater than Ksp and the system is not at equilibrium. In order to decrease the value for Q, the system needs to move to the left. And the system will continue to move to the left until Qsp is equal to Ksp again and the system is at equilibrium. A shift to the left means an increase in the amount of PbCl2 which therefore decreases the solubility of PbCl2. But it doesn't change the value for Ksp. Ksp for PbCl2 stays the same at the same temperature. Next, let's see how the presence of a common ion affects the molar solubility of lead II chloride. And to do that, let's calculate the molar solubility of lead II chloride at 25 degrees Celsius in a solution that is 0.10 molar in potassium chloride. The Ksp value for lead II chloride at 25 degrees Celsius is 1.7 times 10 to the negative fifth. To help us calculate the molar solubility, we're going to use an ICE table, where I stands for the initial concentration, C is the change in concentration, and E is the equilibrium concentration. First, let's say that none of the lead II chloride has dissolved yet. And if that's true, the concentration of lead II plus ions would be zero, and the concentration of chloride anions from the lead II chloride would also be zero. However, there's another source of chloride anions, and that's because our solution is 0.10 molar in KCl. KCl is a soluble salt. So KCl associates completely to turn to K+ and Cl-. Therefore, if the concentration of KCl is 0.10 molar, that's also the concentration of Cl- from the KCl. So we can add here plus 0.10 molar. And think about that as being from our KCl. So there are two sources. There's going to be two sources of chloride anions here. And so the Cl anion and is our common ion. The other source of chloride anion is PbCl2 when it dissolves. So some of the PbCl2 will dissolve, we don't know how much, so I like to write -X in here. And if some of that dissolves, the mole ratio of PbCl2 to Pb2+ is a one-to-one mole ratio. So if we're losing X for PbCl2, we're gaining X for Pb2+. And looking at our mole ratios, if we're gaining X for Pb2+ and it's a one-to-two mole ratio, we would write in here +2X for the chloride anion. So for the equilibrium concentration of Pb2+, it would be zero plus X, or just X. And for the equilibrium concentration of the chloride anion, it would be 0.10 plus 2X. So the 0.10 came from the potassium chloride, and the 2X came from the dissolution of lead II chloride. Next, we need to write a Ksp expression, which we can get from the disillusion equation. So, Ksp is equal to the concentration of lead II plus ions raised to the first power times the concentration of chloride anions. And since there's a two as a coefficient in the balanced equation, we need to raise that concentration to the second power. Pure solids are left out of equilibrium constant expressions. So we don't write anything for PbCl2. Next, we plug in our equilibrium concentrations. So for lead II plus, the equilibrium concentration is X. And for the chloride anion, the equilibrium concentration is 0.10 plus 2X. We also need to plug in the Ksp value for lead II chloride. Here we have the Ksp value plugged in. X and 0.10 plus 2X. And let's think about 0.10 plus 2X for a second here. With a very low value for Ksp, 1.7 times 10 to the negative fifth, that means that not very much of the PbCl2 will dissolve. And if that's true, X is a pretty small number. And if X is a small number, 2X is also pretty small. So we're going to make an approximation and say that 0.10 plus a pretty small number is approximately equal to just 0.10. And that's going to make the math easier on us. So instead of writing 0.10 plus 2X squared, we just have 0.10 squared. Solving for X, we find that X is equal to 0.0017, which we could just write as 1.7 times 10 to the negative third molar. It's okay to write molar here because this X value represents the equilibrium concentration of Pb2+. And if that's the equilibrium concentration of Pb2+, that's also the concentration of lead to chloride that dissolved. So this number, this concentration, is the molar solubility of lead II chloride in a solution at 25 degrees where the solution is 0.10 molar in KCl. Most textbooks leave this -X out of their ICE tables because the concentration of a solid doesn't change. I like to just leave it in here though to remind me that X represents the molar solubility of the slightly soluble salt. Finally, if we'd calculate the molar solubility of lead II chloride without the presence of a common ion, this 0.10 would have been gone from everything. And doing the math that way, we would have found that the molar solubility at 25 degrees Celsius and using this value for the Ksp, the molar solubility comes out to 0.016 molar. So comparing these two smaller solubilities, 0.016 molar versus 0.0017, that's approximately a factor of 10. Therefore the addition of a common ion decreased the solubility by approximately a factor of 10. So doing the common ion effect in a quantitative way also shows a decrease in the solubility of a slightly soluble salt because of the presence of a common ion.