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Previously in Chem104:
• more acid/base reactions:
• weak / weak• strong / strong• strong / weak
• calculations
• Polyprotic acids
Today in Chem104:
•Titrations
•Buffers
• calculations
Titrations: a summary
1) strong acid + strong base titrations
2) weak acid or base titrations (by strong base or acid)
• Have pH 7 at equivalence pt• have flat slopes at beginning and end
• have pH at equivalence pt determine by conjugateweak acid titrations have basic pH at eq. pt.weak base titrations have acidic pH at eq. pt.
• have more pronounced slope at beginning• have pH = pKa at ½ volume to equivalence point • have buffer region where [AH] ~ [A], i.e., where conjugate species have about the same concentrations
Buffers: a summary
1. Resist change in pH
2. Made from conjugates in ~equal concentrations
• Acid form [AH] reacts with added base• Base form [A] reacts with added acid
• pH = pKa + log [A] / [AH]But you don’t need to memorize this:
you can derive it! Fast!
4. Buffer pH determined from theHenderson-Hasselbalch equation
3. An acid or base may have multiple buffer regions
• Give me 2 examples
Buffers: one new point
Buffer capacity: how much acid or base can it “absorb”, or compensate for before pH changes
Consider these two buffer solutions and answer,“Which has higher buffer capacity?”
• 0.100 M Acetic acid + 0.100 M sodium acetate
• 0.001 M Acetic acid + 0.001 M sodium acetate
Buffers: how would you make one?
My research students have that very problem in research lab. Let’s do it and I can report to them that my Gen Chem students can help them out!
How would you make 1 L of a 0.100 M phosphate buffer at pH 7?”
1st: find the Ka’s for the acid/base system
2nd: determine the conjugate pair appropriate for the pH
3rd: use the HH equation (or derive it) or the Ka expression to find the relative proportions of conjugates
Phosphoric acid, H3PO4 …which conjugate pair to use at pH 7?
Step 1. H3PO4 + H2O
H2PO4 - + H3O+
Ka1 = 7.6 x 10-3
Step 2. H2PO4- +
H2OHPO4
2- + H3O+
Ka2 = 6.2 x 10-8
Step 3. HPO42-
+ H2O
PO4 3- + H3O+
Ka3 = 2.12 x 10-13
Let’s do it!
Step 2. H2PO4- +
H2OHPO4
2- + H3O+
Ka2 = 6.2 x 10-8
pH = pKa + log [A] / [AH]
7.00 = 7.21 + log [A] / [AH]
-0.21 = log [A] / [AH]
0.62 = [A] / [AH] Or 0.62 = mol A / mol AH
For 1L of 0.100 M: mol A + mol AH = 0.100mol
So: (0.62 mol AH) + mol AH = 0.100mol
So 0.62 mol AH = 1.00 mol A
1.62 mol AH = 0.100mol
mol AH = 0.100 / 1.62 mol = 0.0617 mol AH
mol A = 0.62 x 0.0617 mol AH = 0.0383 mol A
Let’s do it!
Step 2. H2PO4- +
H2OHPO4
2- + H3O+
Ka2 = 6.2 x 10-8
To make the buffer solution:
0.0617 mol AH = 0.0617 mol NaH2PO4 0.0617 mol NaH2PO4 x 119.98 g/mol = 7.40 g NaH2PO4
0.0383 mol A = 0.0383 mol Na2HPO4 0.0383 mol Na2HPO4 x 141.96 g/mol = 5.44 g Na2HPO4
Dissolved in 1 L water
All Definitions of Acid and Base use Donor /Acceptor
Bronsted Acid/Base: proton H+ donor/acceptor
Remember this reaction?
Lewis Acid/Base: electron pair donor/acceptor
CuCl2(H2O)2 (s) + 3H2O [CuCl(H2O)5]+ + Cl-
Cu
H2O
H2O OH2
OH2
Cl
OH2Cu2+:OH2
e- acceptor :e- donorLewis Acid :Lewis Base
All ionic solids dissolve using Lewis A/B interactions
NaCl(s) + 6H2O [Na(H2O)6]+ + Cl-
Na+:OH2
e- acceptor :e- donorLewis Acid :Lewis Base
Mn+
H2O
H2O OH2
OH2
H2O
OH2
Written simply:
This is typical expression for solubility equilibriumGiven by the Solubility Product Ksp
All ionic solids dissolve using Lewis A/B interactions
AgCl(s) + 2H2O [Ag(H2O)2]+ + Cl-
Ksp = 1.8 x10-10 Ksp = [Ag+][Cl-]1.8 x10-10 = [Ag+][Cl-]1.3 x10-5 M = [Ag+] = [Cl-]
AgCl(s) Ag+ + Cl-
Very low solubility due to weak Lewis A/B interactions which does not compensate for large lattice energy
1.3 x10-5 M = [Ag+] = [Cl-] This is the molar solubility of AgCl
Ionic solids which completely dissolve are highly soluble and cannot be described with a Ksp
NaCl(s) + 6H2O [Na(H2O)6]+ + Cl-
Mn+
H2O
H2O OH2
OH2
H2O
OH2
Solubility obeys
AgCl(s) Ag+ + Cl- + excess Cl-
Ksp = 1.8 x10-10 Solubility =1.3 x10-5 M = [Ag+] = [Cl-]
AgCl(s) Ag+ + Cl-
If more chloride is added the equilbirum shifts left,and Solubility Product Ksp requires that less AgCl dissolves
Ksp = 1.8 x10-10 Solubility, [Ag+] <1.3 x10-5 M
otherwise called The Common Ion Effect obeys
AgCl(s) Ag+ + Cl- + excess Cl-
Ksp = 1.8 x10-10 Solubility =1.3 x10-5 M = [Ag+] = [Cl-]
AgCl(s) Ag+ + Cl-
If more chloride is added the equilbirum shifts left,and Solubility Product Ksp requires less AgCl dissolves
Ksp = 1.8 x10-10 butSolubility, [Ag+] <1.3 x10-5 M because [Cl- ] >>1.3 x10-5 M
The pH Effect obeys
Ksp = 3.7 x10-9
Ca(CO3)(s) Ca2+ + CO32-
If pH is lowered by adding acid,more CaCO3 dissolves
…. and cleans the dishwasher:
+ AH
HCO3-
H2CO3
+ AHH2O + CO2
+ AH
The Chelate Effect obeys
Ksp = 3.7 x10-9
Ca(CO3)(s) Ca2+ + CO32-
If Ca2+ is removed by adding a ligand,
more CaCO3 dissolves
…. and also cleans the dishwasher:
+ citric acid
Ca(citrate)
COOHHOC
COOHHOO
-OCCO-
-OC
OH
O
O
OCa2+
Making better (stronger) Lewis A/B interactionscan improve solubility and clean, too
We have seen:AgCl(s) + 2H2O [Ag(H2O)2]+ + Cl-
Ksp = 1.8 x10-10
AgCl can be completely dissolved!
Very low solubility due to weak Lewis A/B interactions which do not compensate for large lattice energy
But if ammonia is Lewis base:AgCl(s) + 2 NH3 [Ag(NH3)2]+ + Cl-