Solubility Equilibria Will it all dissolve and if

Solubility Equilibria Will it all dissolve, and if not, how much?

n n n All dissolving is an equilibrium. If there is not much solid it will all dissolve. As more solid is added the solution will become saturated. Solid dissolved The solid will precipitate as fast as it dissolves. Equilibrium

General equation n n n M+ stands for the cation (usually metal). Nm- stands for the anion (a nonmetal). Ma. Nmb(s) a. M+(aq) + b. Nm- (aq) K = [M+]a[Nm-]b/[Ma. Nmb] But the concentration of a solid doesn’t change. Ksp = [M+]a[Nm-]b Called the solubility product for each compound.

Common misconceptions n n n Solubility is not the same as solubility product. Solubility product is an equilibrium constant. it doesn’t change except with temperature. Solubility is an equilibrium position for how much can dissolve. A common ion can change this.

Calculating Ksp 1. The solubility of iron(II) oxalate Fe. C 2 O 4 is 65. 9 mg/L. Determine Ksp (mm of Fe. C 2 O 4 is 143. 91 g/mol) 2. The solubility of Li 2 CO 3 is 5. 48 g/L Determine Ksp (mm of Li 2 CO 3 is 73. 89 g/mol)

Fe. C 2 O 4(s) Fe 2+(aq) + C 2 O 4 Ksp = [Fe 2+] [C 2 O 4 2 -] 65. 9 mg/L = 7. 49 x 10 -4 mol/L At equilibrium: [Fe 2+] = 7. 49 x 10 -4 mol/L [C 2 O 4 2 -]= 7. 49 x 10 -4 mol/L Ksp = [7. 49 x 10 -4 mol/L] = 5. 6 x 10 -7 M 2 2 2 - (aq)

Li 2 CO 3(s) 2 Li+(aq) + CO 3 Ksp = [Li+]2 [CO 3 2 -] 5. 48 g/L= 7. 28 x 10 -2 mol/L At equilibrium: [Li+] = 1. 46 x 10 -1 mol/L [CO 3 2 -]= 7. 28 x 10 -2 mol/L Ksp = [1. 46 x 10 -1 mol/L ] 2[7. 28 x 10 -2 mol/L ] = 1. 54 x 10 -3 M 3 2 - (aq)

Calculating Solubility n n n The solubility is determined by equilibrium. Its an equilibrium problem. Calculate the solubility of Sr. SO 4, with a Ksp of 3. 2 x 10 -7 in M and g/L. n (mm of Sr. SO 4 is 183. 68 g/mol) n n Calculate the solubility of Ag 2 Cr. O 4, with a Ksp of 9. 0 x 10 -12 in M and g/L. (mm of Ag 2 Cr. O 4 is 331. 73 g/mol)

Sr. SO 4(s) Sr 2+(aq) + SO 4 Ksp = [Sr 2+] [SO 4 2 -] 2 - (aq) = 3. 2 x 10 -7 M 2 Let x(M) rep. the solubility or the amount of Sr 2+ present at equilibrium At equilibrium: [Sr 2+] = x [SO 4 2 -]= x Ksp = [x] x 2 = 3. 2 x 10 -7 M 2 = 5. 7 x 10 -4 mol/L = 1. 05 x 10 -1 g/L

Ag 2 Cr. O 4(s) 2 Ag+(aq) + Cr. O 4 Ksp = [Ag+]2 [Cr. O 4 2 -] 2 - (aq) = 9. 0 x 10 -12 M 3 Let x(M) rep. the solubility or the amount of Cr. O 42 - present at equilibrium At equilibrium: [Cr. O 42 -] = x [Ag+]= 2 x Ksp = [2 x] 2 x 4 x 3 x = 9. 0 x 10 -12 = 3 5. 7 x 10 -4 mol/L = 1. 43 x 10 -4 mol/L = 4. 74 x 10 -2 g/L

n Homework: p 486 #1 -4 p. 493 # 4 -9

Common Ion Effect n n If we try to dissolve a solid in a solution with either the cation or anion already present, less will dissolve. Calculate the solubility of Sr. SO 4, with a Ksp of 3. 2 x 10 -7 in M and g/L in a) water and b) solution of 0. 010 M Na 2 SO 4. n n (mm is 183. 68 g/mol) Calculate the solubility of Sr. SO 4, with a Ksp of 3. 2 x 10 -7 in M and g/L in a) water and b) solution of 0. 010 M Sr(NO 3. )2 n (mm is 183. 68 g/mol)

Common ion causes a decrease in solubility as per Le Chatelier’s principle! Sulphate is the common ion n a) solved before…. answer was = 5. 7 x 10 -4 mol/L = 1. 05 x 10 -1 g/L b) Ksp = [Sr+2][SO 4 -2] n n Ksp = [x][0. 01 +x] 3. 2 x 10 -7 = [x][0. 01 +x] x= 3. 2 x 10 -5 M (using quadratic or simplifying for x is small when added to 0. 01) Solubility is now 3. 2 x 10 -5 M or 5. 88 x 10 -3 g/L which is much less than in a)

Common ion causes a decrease in solubility as per Le Chatelier’s principle! Strontium is the common ion n a) solved before…. answer was = 5. 7 x 10 -4 mol/L = 1. 05 x 10 -1 g/L b) Ksp = [Sr+2][SO 4 -2] n n Ksp = [0. 01 + x][x] 3. 2 x 10 -7 = [0. 01 + x][x] x= 3. 2 x 10 -5 M (using quadratic or simplifying for x is small when added to 0. 01) Solubility is now 3. 2 x 10 -5 M or 5. 88 x 10 -3 g/L which is much less than in a)

p. H and solubility n n n Why are hydroxides more soluble in acid? For other anions, if they come from a weak acid (CO 32 -, PO 43 -, CH 3 COO, etc) they are more soluble in a acidic solution than in water. Ca. C 2 O 4 Ca+2 + C 2 O 4 -2 H+ + C 2 O 4 -2 HC 2 O 4 Reduces C 2 O 4 -2 in acidic solution.

Precipitation n n Ion Product, Q =[M+]a[Nm-]b If Q>Ksp a precipitate forms. If Q<Ksp No precipitate. If Q = Ksp equilibrium. A solution of 750. 0 m. L of 4. 00 x 10 -3 M Ce(NO 3)3 is added to 300. 0 m. L of 2. 00 x 10 -2 M KIO 3. Will Ce(IO 3)3(S) (Ksp= 1. 9 x 10 -10 M)precipitate and if so, what is the concentration of the ions?

A precipitate ONLY forms if the salt formed by double displacement is slightly soluble and the concentration of its ions reaches saturation point n In the solubility charts, the salt would be identified as insoluble or a precipitate n The precipitate forms if the concentration of its ions reaches the minimum Ksp value n Calculate the concentration of the ions of the possible precipitate after the two aqueous reactant solutions are mixed (and thus are diluted each of them in the process of mixing)

Calculate the trial ion product – the Q and compare to the Ksp n 750. 0 m. L of 4. 00 x 10 -3 M Ce(NO 3)3 plus 300. 0 m. L of 2. 00 x 10 -2 M KIO 3 n Total is 1050. 0 m. L or 1. 050 L n c 1 V 1=c 2 v 2 for Ce+3 n (0. 004 M)(750)= c 2 (1050) n c 2= 0. 00286 M n c 1 V 1=c 2 v 2 for IO 3 -1 n (0. 02 M)(300)=c 2(1050) n c 2=0. 00571 M n Q= [Ce+3][IO 3 -1]3=(0. 00286)(0. 00571)3 -10

Ce(IO 3)3(s) = Ce+3 + 3 IO 3 -1 (other salt is potassium nitrate which is soluble no exception…it would appear as aqueous (aq) in bce) n Q= [Ce+3][IO 3 -1]3 n Q=(0. 00286 M)(0. 00571 M)3 n Q= 5. 3 x 10 -10 M 4 n …Q is greater than Ksp (1. 9 x 10 -10) n …a precipitate forms

Relative solubilities n n Visual Ksp comparison will only allow us to compare the solubility of solids that fall apart into the same number of ions. The bigger the Ksp of those the more soluble. If they fall apart into different number of pieces you have to do the math and solve for solubility to compare Example: two binary salts have a Ksp of 4. 1 x 10 -5 and 7. 6 x 10 -8. Which is less soluble and will precipitate first?

Selective Precipitations n n Used to separate mixtures of metal ions in solutions. Add anions that will only precipitate certain metals at a time. Used to purify mixtures. Often use H 2 S because in acidic solutions Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate.

Selective precipitation AP n n n Follow the steps first with insoluble chlorides (Ag, Pb, Ba) Then sulfides in Acid. Then sulfides in base. Then insoluble carbonate (Ca, Ba, Mg) Alkali metals and NH 4+ remain in solution.

Complex ion Equilibria AP n n n A charged ion surrounded by ligands. Ligands are Lewis bases using their lone pair to stabilize the charged metal ions. Common ligands are NH 3, H 2 O, Cl-, CNCoordination number is the number of attached ligands. Cu(NH 3)4+2 has a coordination # of 4

AP The addition of each ligand has its own equilibrium n n n Usually the ligand is in large excess. And the individual K’s will be large so we can treat them as if they go to equilibrium. The complex ion will be the biggest ion in solution.

AP n n n Calculate the concentrations of Ag+, Ag(S 2 O 3)-, and Ag(S 2 O 3)2 -3 in a solution made by mixing 150. 0 m. L of Ag. NO 3 with 200. 0 m. L of 5. 00 M Na 2 S 2 O 3 Ag+ + S 2 O 3 -2 Ag(S 2 O 3)K 1=7. 4 x 108 Ag(S 2 O 3)- + S 2 O 3 -2 Ag(S 2 O 3)2 -3 K 2=3. 9 x 104

AP 500 m. L of a 0. 20 M solution of Ag. NO 3 is mixed with 200 m. L of a 0. 10 M Sr(NO 3)2. Al 2(CO 3)3 solid is added slowly. Which will precipitate first? How many milligrams of solid will just begin precipitation? Ksp Ag 2 CO 3 = 8. 46 x 10 -12 Ksp Sr. CO 3 = 5. 6 x 10 -10