Student Doctor Network Forums > Pre-Medical Forums > MCAT Discussions > MCAT Study Question Q&A > What is considered a Pure liquid when doing solubility formulas? PDA View Full Version : What is considered a Pure liquid when doing solubility formulas? yaba04-07-2008, 09:36 AMWhen I get a formula like AgX + NH3 --> AgNH3 How do I know if any of the reactants are pure liquids or not? This is in regards to writting the mass action formula to this problem. Thanks guys/girls any help would be appreciated. Kaustikos04-07-2008, 09:41 AMWhen I get a formula like AgX + NH3 --> AgNH3 How do I know if any of the reactants are pure liquids or not? This is in regards to writting the mass action formula to this problem. Thanks guys/girls any help would be appreciated. Do you mean which are solvents and which are solutes? Or if they are in liquid form or gas? Because they generally tell you what they are (subscript) in the formula. edit - nvm. The above answer suffices yaba04-07-2008, 10:13 AMDo you mean which are solvents and which are solutes? Or if they are in liquid form or gas? Because they generally tell you what they are (subscript) in the formula. edit - nvm. The above answer suffices Well Im talking in regards to finding for example the solubility product constant, If they give me the concentrations for the reactant and the products I can technically figure out the Ksp because its [products]/[reactants], but you omit any pure solids or liquids from the equation, right? So if thats the case, then if there is any solids or liquids in the formula at all, can I omit them? Or are only certain solids and liquids considered pure? Im sorry if I sound confusing its only because im confused my self, thanks QuaerensIntelle04-07-2008, 11:33 AMIf it has an L, it means it's a pure liquid. (aq) means that it's a solute dissolved in a solvent. If you see anything with an (s) or an (l), you omit it. yaba04-07-2008, 12:44 PMIf it has an L, it means it's a pure liquid. (aq) means that it's a solute dissolved in a solvent. If you see anything with an (s) or an (l), you omit it. Awesome thanks DrMattOglesby04-10-2008, 06:14 AMWell Im talking in regards to finding for example the solubility product constant, If they give me the concentrations for the reactant and the products I can technically figure out the Ksp because its [products]/[reactants], but you omit any pure solids or liquids from the equation, right? So if thats the case, then if there is any solids or liquids in the formula at all, can I omit them? Or are only certain solids and liquids considered pure? Im sorry if I sound confusing its only because im confused my self, thanks are you sure the Ksp = [products]/[reactants] ? i thought that was Keq. BloodySurgeon04-10-2008, 06:45 AMare you sure the Ksp = [products]/[reactants] ? i thought that was Keq. They are all pretty much the same thing... Kp, Ksp, and Keq... just what they represent is different. For the Ksp it uses molar solubilities and the reactant is not apart of the equation because it is the solid therefore omitted. No matter what way you want to think of it, it all comes out the same. But the most general formula for K itself is (products)/(reactants) because it IS the equilibrium constant and has both the product and reactant. Agent4704-10-2008, 07:30 AMThey are all pretty much the same thing... Kp, Ksp, and Keq... just what they represent is different. For the Ksp it uses molar solubilities and the reactant is not apart of the equation because it is the solid therefore omitted. No matter what way you want to think of it, it all comes out the same. But the most general formula for K itself is (products)/(reactants) because it IS the equilibrium constant and has both the product and reactant. Ksp for Solubility is different!!! Kep=Ksp..Ksp will never have an denominator. Example: AaBb(s)=aA + bB Ksp=[A]^a[B]^b assuming that A and B are both ions! Just watch out on that one! BloodySurgeon04-10-2008, 08:03 AMI know that there is no denominator. That is why I said it will not show up because the reactant is a solid and solids/liquids do not show up in the Keq. I just want to get the point out that they all are "K's" in a sense. If you want ignore all i said or take it... it makes no difference... you get the same answer anyways. BloodySurgeon04-10-2008, 08:05 AMIonic compounds normally dissociate into their constituent ions when they dissolve in water. For example, for calcium sulfate: CaSO4(s) ----> Ca2+(aq) + SO4^2- As for the previous example, the equilibrium expression is: K = [Ca2+][SO4 2-] / [CaSO4] where K is called the equilibrium (or solubility) constant and curly brackets indicate activity. The activity of a pure solid is, by definition, equal to one. When the solubility of the salt is very low the activity coefficients of the ions in solution will also be equal to one and this expression reduces to the solubility product expression: K = [Ca2+][SO4 2-] Agent4704-10-2008, 10:09 AMIonic compounds normally dissociate into their constituent ions when they dissolve in water. For example, for calcium sulfate: CaSO4(s) ----> Ca2+(aq) + SO4^2- As for the previous example, the equilibrium expression is: K = [Ca2+][SO4 2-] / [CaSO4] where K is called the equilibrium (or solubility) constant and curly brackets indicate activity. The activity of a pure solid is, by definition, equal to one. When the solubility of the salt is very low the activity coefficients of the ions in solution will also be equal to one and this expression reduces to the solubility product expression: K = [Ca2+][SO4 2-] Exactly what I wanted to Say!:D