solubilityproduct

Solubility Product Constant and Equilibrium [|Ksp picture] Solubility Rules Lots of times, when we say a substance is insoluble, we mean it's only very very slighty soluble. NaCl splits up completely into its ions ([|demonstration]), but AgCl only dissolves a little bit ([|another demonstration]). In these cases, the solid will dissolve a little but as ions.
 * Physical Description**

a. To begin, write out the equation (AgCl(s) <>Ag+(aq) + Cl-(aq), PbBr2 <> Pb+(aq) + 2Br-(aq)). Then create an __equilibrium expression__, Ksp for the reaction. Note that only the products will be included since the sole reactant is a solid. (Ksp= [Ag+][Cl-], [Pb²+][Br-]²). Equilibrium expression [|questions] and answers:
 * Introduction to Ksp**

 Thanks to Chemistry 40S for the chart!

b. To find solubility (x), you make what we like to call an ICE ICE BABY chart. ICE stands for initial, change, and equilibrium. To begin, you use your balanced equation and your Ksp value given to solve for your solubiltiy (x). c.To find Ksp from solubility, you create another ICE chart. Once you determine the coefficients of x and the equilibrium expression, you plug in the given solubility value to determine Ksp.

d. When Qk, the solution is above saturation point and a precipitate forms.

The Common Ion Effect:  Decrease in solubility of an ionic [|salt], i.e., one that dissociates in solution into its [|ions] , caused by the presence in [|solution] of another solute that contains one of the same ions as the salt. The common-ion effect is an example of [|chemical equilibrium]. For example, silver chloride, AgCl, is a slightly soluble salt that in solution dissociates into the ions Ag+ and Cl−, the equilibrium state being represented by the equation AgClsolid ↔Ag++Cl−. According to [|Le Châtelier's principle], when a stress is placed on a system in equilibrium, the system responds by tending to reduce that stress. In the system taken as an example, if another solute containing one of those ions is added, e.g., sodium chloride, NaCl, which supplies Cl− ions, the solubility equilibrium of the solution will be shifted to remove more Cl− from the solution, so that at the new equilibrium point there will be fewer Ag+ ions in solution and more AgCl precipitated out as a solid. (submitted by John Bowman) [|scan0001.jpg](Submitted by John Bowman)

a. Most solids are more soluble in hot water, so make sure you check what temperature the Ksp you're using is for. b. If a solid is being dissolved in a solution with an already-present concentration of one of the solid's ions, it will be less soluble. c. Adding a strong acid makes solid more soluble. The H+ from the strong acid will react with the anions of the solid. Pulling those ions out of the system will increase the solubility because the reaction will shift to the right by LeChâtelier's Principle. d. If you want a solution with only Na+ ions in one that is dissolved NaCl, you could add a solution of Ag+ ions that would cause the Cl- ions to precipitate out.
 * Special Conditions**

[|Kf Chart] a. Kf is the formation of products which is the inverse of the Ksp. b. Consider the system AlCl3(s) <---> Al3+ + 3Cl- Ksp= Al3+ + 4OH- <> Al(OH)4- Kf=_ AlCl3(s) + 4OH- <---> Al(OH)4- + 3Cl- K= Kf • Ksp
 * Kf and Complex Ions**
 * Note:** The most common complex ions involve OH- and NH3.

ALWAYS REMEMBER THAT ADDING MORE SOLD WILL NOT AFFECT THE EQUILIBRIUM STATE SINCE SOLIDS AREN'T INCLUDED IN THE K EXPRESSION! 2 NH3(g)
 * (Travis DePriest) Example:** A mixture consisting initially of 3.00 moles NH3, 2.00 moles of N2, and 5.00 moles of H2, in a 5.00 L container was heated to 900 K, and allowed to reach equilibrium. Determine the equilibrium concentration for each species present in the equilibrium mixture.

N2(g) + 3 H2(g) Kc = 0.0076 @ 900 K > || Initial Concentration (M) || 0.600 || 0.400 || 1.00 || > || Change in Conc. (M) || || || || > || Equilibrium Conc. (M) || || || || > || Initial Concentration (M) || 0.600 || 0.400 || 1.00 || > || Change in Conc. (M) || + 2 x || - x || - 3 x || > || Equilibrium Conc. (M) || || || || > || Initial Concentration (M) || 0.600 || 0.400 || 1.00 || > || Change in Conc. (M) || + 2 x || - x || - 3 x || > || Equilibrium Conc. (M) || 0.600 + 2 x || 0.400 - x || 1.00 - 3 x || TRAVIS DEPRIEST []
 * Convert the initial quantities to molarities as shown for NH3. [[image:http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/EquilibriumArt/ICE1.gif width="236" height="43" align="center" caption="Calculating Molarities"]]
 * Create a chart as illustrated below and enter in the known quantities.
 * || NH3 || N2 || H2 ||
 * [|Calculate Qc] and compare to Kc to determine the direction the reaction will proceed. [[image:http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/EquilibriumArt/ICE2.gif width="414" height="80" align="center" caption="Calculation of Q"]]
 * Assign a variable "x" that represents the change in the amount of one of the species. The species with the lowest coefficient in the balanced equation usually is the easiest to handle when it comes to doing the math. Here let "x" = change in the amount of N2.
 * Determine the change in all the other species in terms of "x." Remember the change must be in agreement with the stoichiometry of the balanced equation, in this case 2:1:3. Since the reaction goes in the reverse direction the concentrations of N2 and H2 gases will decrease (note the negative sign) and that of NH3 will increase. Put these quantities into the chart (shown in red).
 * || NH3 || N2 || H2 ||
 * Express the equilibrium concentrations in terms of "x" and the initial amounts (shown in green).
 * || NH3 || N2 || H2 ||
 * Substitute the expressions for the equilibrium concentration into the expression for the equilibrium constant and solve for "x." Once x is known, the equilibrium concentration for each species can be calculated.