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 When ionic compounds are dissolved in solution, they dissociate to some degree Ksp is the solubility constant 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 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 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