Solubility Product

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The Solubility Product Ksp Thursday, December 11, 2008 8:22 AM

Sparingly Soluble Salts - There are many ionic compounds (salts) that are almost insoluble in water. ○ The crystal lattice structure is too strong and water can not enter. ○ The ionic bonds are too strong and the water molecule cannot substitute for the ionic bond. ○ The salts are referred to as SPARINGLY SOLUBLE SALTS. ○ Because [salt]!=[ions], we must use equilibrium theory. - What is the Molar Solubility of AgCl in Pure Water? ○ Molar solubility is the concentration of a salt that can be achieved in 1 L of water. ○ Must have Ksp value, Ksp AgCl = 10-10 ○ Let x = [Ag+]  AgCl (s) ↔ Ag+(aq) + Cl-(aq) ○ Name

AgCl

Ag+

Cl-

○ What is the Ksp expression? ○ What is the value of x?  Ksp = [Ag+][Cl -]  10-10 = [x][x]  X = (10-10)1/2  X = 10-5 ○ X is the [Ag+] which is in a 1:1 ratio with [AgCl] ○ Therefore the [molar] of AgCl is 10-5 1:1, 2:1, 3:1 Salts - AgCl is a 1:1 Salt ○ For a 1:1 Salt, x = (Ksp)1/2 - Ag2CrO4 is a 2:1 salt ○ For a 2:1 salt, x = (Ksp/4)1/3 Ag 3PO4 is a 3:1 salt ○ For 3:1 salt x = (Ksp/27)1/4 - On tests etc. for full marks you should write the ICE table for these salts. Predicting if a Precipitate will Form - To determine if a precipitate will form, you need to calculate the reaction quotient Q. - Compare Q to Ksp ○ If Q is greater than Ksp, a precipitate will form. ○ If Q is less than Ksp, no precipitate will form. ○ If Q is equal to Ksp, the solution is just saturated. - Example ○ Will a precipitate form when 1.0L of 3*10-10M Co(NO3)2 is added to 1.0L of 2*10-11M K2S (Ksp CoS = 5* 10-22) - Calculate Q and compare to Ksp. - Before you start you should realize that a dilution has occurred. - The final volume is 2.0L - You can use C1V1 = C2V2 for this, in this case just divide values by 2. - [Co+2][S2-] - [1.5*10-10][10-11] - [1.5*10-21] - Q > Ksp - A precipitate forms!!!!

Chemistry Page 1

Section 7.6 pg 482 - 493 Practice Questions 1-4 pg 486 5 pg