CHEM 163 Chapter 19 Spring

Buffers Solution that resists pH changes –Ex. Blood (pH ~ 7.4) Acid must neutralize small amounts of base Base must neutralize small amounts of acid Acid and base must not neutralize each other 2 Use conjugate acid-base pairs! CH 3 COOH (aq) + H 2 O (l)CH 3 COO - (aq) + H 3 O + (aq) Added in as salt (NaCH 3 COO) Common-ion effect Ex: acetate

High concentrations of weak acid/conjugate base Add H 3 O + or OH - –Added amounts are relatively small –Cause only small shifts –React with weak acid or conjugate base 3 HA (aq) + H 2 O (l)A - (aq) + H 3 O + (aq) pH depends on [HA]/[A-] ratio HA (aq) + OH- (aq)A - (aq) + H 2 O (l)

Making a buffer 1.Choose the conjugate acid-base pair (pK a ≈ pH) 2.Calculate the ratio of acid-base concentrations 3.Determine the buffer concentration 4.Mix solution; adjust pH 4 Henderson-Hasselbalch equation:

Buffer Properties Buffer Capacity: –Ability to resist pH change –Unrelated to pH of buffer –Dependent on concentration of weak acid/conj base –Highest when [A - ] = [HA] Buffer Range: –pH range over which buffer is effective –Usually within ±1 pH unit of the p Ka of weak acid 5

Sample Problem Make 200. mL of a pH 3.5 citric acid/sodium citrate buffer with an acid concentration of 0.50 M. We are given solid sodium citrate (294 g/mol) and 5.0 M citric acid. The pKa of citric acid is

Measuring pH Acid-Base Titration Curves: pH v. volume titrant Measuring pH: 1.pH meter 2.Acid-base indicators Indicator: Weak organic acid HIn different color than In - Intensely colored (small amount needed) Changes color over ~ 2 pH units 7

Titration Curves: Strong acid – Strong base Low pH (strong acid) Sudden pH rise (6-8 units) Slow pH increase 8  [OH - ] added ≈ [H 3 O + ] init Equivalence point: [OH - ] added = [H 3 O + ] init pH = 7 End point: when indicator changes color

Calculating pH during titration Original solution of strong HA Before the equivalence point –Moles of acid remaining? –Calculate [H 3 O + ] At the equivalence point: pH = 7 After the equivalence point –Excess moles of OH - added? –Calculate [OH - ] 9 moles acid initialmoles acid rxted moles base added moles acid total

Titration Curves: Weak acid – Strong base Higher initial pH (weak acid, lower K a ) Buffer region –gradual pH rise –Midpoint: ½ initial HA reacted Equivalence point: pH > 7.00 Slow pH increase 10  [HA] = [A - ]  pH = pK a

Calculating pH during titration Original solution of weak HA –ICE table Buffer Region At the equivalence point: After the equivalence point 11 Excess moles of OH - added  x = [H 3 O + ]

Titration Curves: Strong acid – Weak base Initial pH > 7.00 (weak base) Buffer region –gradual pH decrease Equivalence point: pH < 7.00 Slow pH decrease 12 Less common than strong base-weak acid (fewer appropriate indicators)

Titration Curves: Polyprotic Acids 13

Salts “slightly soluble” Equilibrium between solid and dissolved ions PbSO 4 (s) Pb 2+ (aq) + SO 4 2- (aq) H2OH2O NaCl (s)Na + (aq) + Cl - (aq) H2OH2O soluble Solubility product Ion-product expression (at saturation) Solubility-Product Constant larger K sp : more dissolution at equil. (saturation) Smaller K sp : less dissolution at equil. (saturation)

Insoluble Metal Sulfides MnS (s)Mn 2+ (aq) + S 2- (aq) H2OH2O S 2- (aq) + H 2 O (l)HS - (aq) + OH - (aq) MnS (s)Mn 2+ (aq) + + H 2 O (l)HS - (aq) + OH - (aq)

3-minute Practice Write K sp expression for each of the following: Silver bromide in H 2 O Silver sulfide in H 2 O AgBr (s)Ag + (aq) + Br - (aq) H2OH2O + H 2 O (l) Ag 2 S (s)2Ag + (aq) + HS - (aq) + OH - (aq)

Higher K sp = greater solubility ? Yes, for compounds with same total number of ions CompoundK sp Solubility Ca(OH) x x PbSO x x MgCO x x BaF x x minute Practice

What else affects solubility? Presence of a common ion: PbSO 4 (s)Pb 2+ (aq) + SO 4 2- (aq) H2OH2O Add Na 2 SO 4 ? Decreases solubility pH: CaCO 3 (s)Ca 2+ (aq) + CO 3 2- (aq) H2OH2O ↑ [H 3 O + ] ↑ solubility if compound contains anion of weak acid CO 3 2- (aq) + H 3 O + (aq)H 2 O (l) + HCO 3 - (aq)

Homework problems Chap 19: #9, 13, 19, 29, 50, 63, 70, 76, 78, 90 Due Tuesday, 4/28 More lecture notes will be added next week! Stay tuned. 19

Precipitation Will it occur? Q sp = K sp : Q sp > K sp : Q sp < K sp : Selective precipitation –Way to separate ions –Form slightly soluble compounds with different K sp Saturated solution Precipitation occurs Unsaturated solution

Selective Precipitation 21 Mix 0.2 M Zn(NO 3 ) 2 and 0.4 M Mn(NO 3 ) 2. Precipitate? Add NaOH…Zn(OH) 2 and Mn(OH) 2 K sp Zn(OH) 2 = 3.0 x K sp Mn(OH) 2 = 1.6 x minute Practice Which product is more soluble? What [OH-] would need to make a saturated solution of the more soluble product? Hint: use K sp expression! Products?

Complex Ions Central metal ion + ligands Ionic ligands: OH-, CN-, halides Molecular ligands:H 2 O, NH 3 Lewis acid Lewis base M(H 2 O) 4 2+ (aq) + 4 NH 3 (aq)M(NH 3 ) 4 2+ (aq) + 4 H 2 O (l) Formation constant:

Effects of ligands A slightly soluble compound becomes more soluble when its cation forms a complex ion. AgBr (s)Ag + (aq) + Br - (aq) Add Na 2 S 2 O 3 : Ag + (aq) + S 2 O 3 2- (aq)Ag(S 2 O 3 ) 2 3- (aq)2 Amphoteric Hydroxides: Very slightly soluble in water More soluble in acidic or basic solutions Al(OH) 3 (s)Al 3+ (aq) + 6 H 2 O (l)+ 3H 3 O + Al(H 2 O) 6 (s)+ 4 OH - Al(H 2 O) 2 (OH) 4 - (aq) + 4 H 2 O (l)