Aqueous Equilibria Chapter 15 DE Chemistry Dr. Walker

Common Ion Effect When multiple compounds with a common ion are in solution, equilibrium is shifted and the pH is affected. HF (aq) and NaF (aq) HF (aq) H+ (aq) + F-(aq) NaF (aq) Na+ (aq) + F-(aq) Presence of flouride ion shifts equilibrium to the left, which lowers [H+], raising the pH

Common Ion Effect When multiple compounds with a common ion are in solution, equilibrium is shifted and the pH is affected. NH4OH and NH4Cl (aq) NH3 (aq) + H2O NH4+ + OH- (aq) NH4Cl(aq) NH4+ (aq) + Cl- (aq) Presence of ammonium ion shifts equilibrium to the left, which lowers [OH-], lowering the pH

Example Calculate the pH of a solution containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF.

Example Calculate the pH of a solution containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF. Notice that we have a source of F- from two different places, the dissociation of NaF and the dissociation of HF. The presence of F- from NaF will keep some HF from dissociating. NaF F- + Na+ HF F- + H+ Add F-, shift equilibrium left, lower [H+]

Example Calculate the pH of a solution containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF. HF F- + H+ From 1.0 M NaF! HF F- H+ I 1.0 C -x +x E 1.0 - x 1.0 + x + x

Example Calculate the pH of a solution containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF. pH = -log (7.2 x 10-4) = 3.14 Compare to a 1.0 M solution of only HF, the pH = 1.57 Pretty big difference!!

Buffered Solutions A solution that resists a change in pH when either hydroxide ions or protons are added. Buffered solutions contain either: A weak acid and its salt A weak base and its salt

Buffered Solutions HC2H3O2 C2H3O2- + H+

Buffered Solutions HC2H3O2 C2H3O2- + H+ HC2H3O2 C2H3O2- H+ I 0.5 C -x C -x +x E 0.5 - x 0.5 + x + x

Buffered Solutions HC2H3O2 C2H3O2- + H+

A Different Scenario The previous example used a buffer system where the concentrations were equal. This doesn’t always happen The Henderson-Hasselbalch equation helps us solve for this scenario as long as our given concentrations are significantly larger than the pKa. pKa = - log Ka

Unequal Buffering Example

Unequal Buffering Example We can still use the standard method (ICE) to determine our pH, but Henderson-Hasselbalch is actually less work here. Lactic acid is our “acid” and sodium lactate is our “base”

Solubility Equilibria Many ionic compounds dissociate completely in aqueous solutions Nitrates, Group 1 cations, most sulfates, halides Others are technically “insoluble” Some of the material still dissolves, just a small fraction of it Similar to the dissociation of strong vs. weak acids

Solubility Equilibria “Soluble” compounds Similar to strong acids Dissociate completely Large K value, not really an equilibrium process “Insoluble” compounds Similar to weak acids Dissociate partially Mostly reactant at equilibrium, small K value Reactant NOT in equilibrium expression, since it’s a solid

Solubility Equilibria When ionic compounds are dissolved in solution, they dissociate to some degree Ksp is the solubility constant Higher Ksp = more soluble Remember, pure liquids and solids are NOT included in the law of mass action These problems work the same as problems for equilibrium, weak acids/bases, and buffers

Solving Solubility Problems What is the solubility for the salt AgI at 25C, Ksp = 1.5 x 10-16 AgI(s)  Ag+(aq) + I-(aq) I C E O O +x +x x x

Solving Solubility Problems For the salt AgI at 25C, Ksp = 1.5 x 10-16 AgI(s)  Ag+(aq) + I-(aq) I C E O O +x +x x x 1.5 x 10-16 = x2 x = solubility of AgI in mol/L = 1.2 x 10-8 M

Solving Solubility Problems For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5 PbCl2(s)  Pb2+(aq) + 2Cl-(aq) I C E O O +x +2x x 2x 1.6 x 10-5 = (x)(2x)2 = 4x3 x = solubility of PbCl2 in mol/L = 1.6 x 10-2 M

Solving Solubility Problems What is the solubility for the salt For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5 PbCl2(s)  Pb2+(aq) + 2Cl-(aq) I C E O O +x +2x x 2x

Solving Solubility with a Common Ion For the salt AgI at 25C, Ksp = 1.5 x 10-16 What is its solubility in 0.05 M NaI? AgI(s)  Ag+(aq) + I-(aq) I C E O 0.05 +x 0.05+x x 0.05+x 1.5 x 10-16 = (x)(0.05+x)  (x)(0.05) x = solubility of AgI in mol/L = 3.0 x 10-15 M