Skip to main content
Pearson+ LogoPearson+ Logo
Start typing, then use the up and down arrows to select an option from the list.

Selective Precipitation

Jules Bruno
Was this helpful?
Hey, guys. So here we're looking at this question that split into two parts. So here they tell us first, the concentration of Claude and iodine ions present in the solution Our 20.73 Moeller and 0.60 Mueller feature the ions respectively. The use of silver can be used to precipitate these two ions. Here they give us the K SP for silver chloride as 1.77 times 10 to the negative 10 and silver iodide is 8.51 times 10 to the negative 17 First question just asks us which ion will precipitate out first All you have to say here is the higher K sp then the more soluble the compound And remember, the more soluble you are Thus less likely you are to form a precipitate or solid. So we could say the K SP of Silver client is ah larger value into the negative 10 Whereas silver iodide is smaller because it took the negative because silver iodide has a smaller K SP. That would mean that the iodide ion would precipitate out first. So remember the higher your Caspi, the more soluble, which means less likely to former precipitate first. Now that we know that we can answer part B And what concentration will this ion from part begin to precipitate out? So we know we're dealing with iodine, so I died, actually, so that means that we're dealing with silver iodide and because it has a K S P value attached to it, we're gonna talk about how it breaks up into its ions. Now, how do we determine what the concentration will be once it starts to precipitate out? We determined that by the fact that they tell us the concentration of iodine in the very beginning, So we're dealing with a nice chart here, and that concentration would represent the initial concentration here. So that means we're dealing with the common iron effect here. We don't have a common ion for silver. So in zero the world products or making them so this is plus X plus X plus X and then 0.60 plus X. Now we're gonna say here K SP just equals products. Okay. S P equals silver times. I died. So K s P for silver iodide is 8. times 10 to the negative. 17. So already equilibrium is X. I died. Remember, When we have a real number, we could ignore the X variable. So this is 0.60 Then all we gotta do your solve for X. So divide both sides by 0. So X here equals 1.42 times 10 to the negative. 15 Mueller. This represents the Saudi ability off our entire ionic compound. And in fact, it also answers the question. This will be the concentration in which the ion begins to precipitate out. So this would be our final answer.