the presence of a common ion can affect a solubility equilibrium for example let's say we have a saturated solution of lead to chloride lead to lead to chloride is a white solid so here's the white solid on the bottom of the beaker and the solids at equilibrium with the ions in solutions that would be pb2 plus and cl minus notice how the mole ratio is one to two of pb2 plus to two cl minus so if we have two pb two plus ions in our 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 plus and cl minus 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 minus 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 plus ions will combine with chloride anions to form pbcl2 and we can see that down here in the diagram so imagine say this pb2 plus 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 plus 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 does represent the system at equilibrium so i'll write in 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 was from the dissolution 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 dissolves 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 to 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 to 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 pb cl2 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 two 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 dissociates completely to turn to k plus and cl minus therefore if the concentration of kcl is 0.10 molar that's also the concentration of cl minus 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 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 minus x in here and if some of that dissolves the mole ratio of pbcl2 to pb2 plus is a one to one mole ratio so for losing x for pbcl2 we're gaining x for pb2 plus and looking at our mole ratios if we're gaining x for pv2 plus and it's a 1 to 2 mole ratio we would write in here plus 2x for the chloride anion so for the equilibrium concentration of pb2 plus it would be 0 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 dissolution equation so ksp is equal to the concentration of lead two 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 two plus the equilibrium concentration is x and for the chloride anion the equilibrium concentration is 0.10 plus 2x we're also we also need to plug in the ksp value for lead 2 chloride here we have the ksp value plugged in x and 0.10 plus 2x now 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 .10 plus a pretty small number is approximately equal to just .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 0.0 which we could just write as 1.7 times 10 to the negative third molar it's okay to write molar here because this this x value represents the equilibrium concentration of pb2 plus and if that's the equilibrium concentration of pb2 plus that's also the concentration of lead ii chloride that dissolved so this number this concentration is the molar solubility of lead to chloride in a solution at 25 degrees where the solution is 0.10 molar in kcl most textbooks leave this minus 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 calculate the molar solubility of lead ii chloride without the presence of a common ion this .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 these using this value for the ksp the molar solubility comes out to 0.01 six molar so comparing these two molar solubilities point zero one six molar versus point zero zero one seven there's that's approximately a factor of ten 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 by the presence because of the presence of a common ion