Section IIB Acid Base & Complexation Chemistry & Titrations About 4 lectures Chapter 10 Monoprotic A-B Equilibria Coverage All Sections Problems All except the few involving activity or spreadsheets as needed Chapter 11 Polyprotic A-B Equilibria Coverage Sections 11-1 to 11-5 (11-3 & 5 only lightly) Problems 3-6, 12, 13, 15-19, 22, 24, 26, 27 Chapter 12 A-B Titrations Coverage - 12-1 to 12-4 & 12-6 (note I may skip from Chapter 10 to parts of 12) Problems 1-32, 36-44 Chapter 13 Compleximetric Titrations EDTA Coverage All Sections Problems 1, 5-8, 13, 22-24, 26-32 More on Monofunctional Acids and Bases Strong Acids and Bases e.g., HCl H+ + Cl (as a strong acid the HCl tends to completely ionize and the concentration Of H+ is the formal concentration of HCl) Question: Do you know the difference between formal and analytical concentrations? Problem: K a 10 4 so determine the percent dissociation of a 0.1000 M HCl solution ([H+] [Cl-]) / [HCl] x 2 /(0.1000 x) 10 4 x 0.09999 so 99.99 % ionized ph 1.00 Complete ionization of strong bases also occurs and thus 0.1000 M NaOH is 0.1000 M OH- and since ph + poh 14, ph 13 How do you know if you are dealing with a strong acid or base Just look and a table of ionization constants if not there it is strong! 1
More on Monofunctional Acids and Bases Weak Acid Calculations HA H + + A - x x F (or C A ) formal concentration [HA] added to solution [ H ][ A [ HA] α ] x F x + 2 [ A [ A ] ] + [ HA ] F K a α fraction of dissociation x More on Monofunctional Acids and Bases Weak Base Calculations B + H 2 O BH + + OH - x x F (or C B ) formal concentration [B] added to solution [ BH α ][ OH [ B] x F + 2 ] α fraction of dissociation [ BH [ BH + + ] + ] x x [ B] F K b 2
More on Monofunctional Acids and Bases Buffers A buffer is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid. A buffered solution resists changes in ph when a solution is diluted or when acids and bases are added to the solution (It stabilizes the ph!). Henderson Hasselbalch equation: Be able to derive these K b [OH-] ([BH+]/[B]) -logk b -log[oh-] - log [BH+]/[B] ph poh pk pk b a [ A ] + log [ HA ] [ BH + ] + log [ B] 3
Buffers ph pk a [ A ] + log [ HA ] Add a strong acid and this happens H+ + A- HA ; the ratio decreases, but since it s a log term ph does not change that much Add a strong base and this happens B + HA A- + BH+ ; the ratio changes, but since it s a log term ph does not change that much However, when A- or HA is nearly consumed (buffer s capacity exceeded), the ratio does change rapidly and ph is appreciably altered Buffers ph pk a [ A ] + log [ HA ] Dilute and both [A-] and [HA] decrease equally so ratio does not change Intuitively, extensive dilution should cause system to move Toward neutral ph so what s going on?? Acid rxn increases A- HA H+ + A- & The rigorous H-H Equation is as follows C A - + [H+] - [OH-] ph pka + log C HA - [H+] + [OH-] Hydrolysis rxn increases HA A- + H 2 O HA + OH- In most cases the formal concentrations C A&HA are much larger than [H+] & [OH-], that is until excessive dilution 4
Buffer Capacity β dcb dph dca dph ph pk a [ A ] + log [ HA ] How do you increase β? Increase the buffer concentration. Buffers Practical Preparation 1. Add the amount of either the acid or base form of your buffer compound to make the desired concentration (mol/l). 2. Dissolve the compound in a volume of water ~70-90% of your final total volume. 3. Adjust to the desired the ph by either adding NaOH or HCl using a ph meter. (Other strong acids or strong bases can do the job!) 4. Quantitatively transfer to a volumetric flask and dilute to the mark. ph pk a [ A ] + log [ HA ] You should note that a buffer is created when performing a titration of A weak acid or weak base! 5
Buffers Acid-Base Titrations Calculations (monofunctional A-B) 1. Write out the reaction! 2. Determine the volume at the equivalence point. 3. Determine what chemical species are present at different parts of the titration. Perform appropriate calculations to determine ph as the titration proceeds a) Initial conditions b) Before the equivalence point c) At the equivalence point d) After the equivalence point 6
Strong Acids Strong Bases Determine what chemical species are present at different parts of the titration. Perform appropriate calculations to determine ph as the titration proceeds a) Initial conditions determined by Ca or Cb b) Before the equivalence point determined by how much has not be titrated yet (but consider dilution) c) At the equivalence point about ph 7 (no hydrolysis takes place because the conjugates are extremely weak) d) After the equivalence point determined by amount of titrant in excess (but consider dilution) Titration of a Weak Acid or Base with a Strong Base or Acid Determine what chemical species are present at different parts of the titration. Perform appropriate calculations to determine ph as the titration proceeds a) Initial conditions determined using the K a or K b equilibrium expression b) Before the equivalence point in the buffer region so use H-H equation c) At the equivalence point must consider the hydrolysis of the conjugate d) After the equivalence point determined by the excess of the strong reagent 7
General Shapes of Titration Curves Strong Base with Strong Acid Weak Acid with Strong Base General Shapes of Titration Curves Effect of pka Effect of initial concentration 8
Example Titration Problem a) At the start 6.2 x 10-10 [H+][CN-] / [HCN] [H+] 2 /(0.200 [H+]) [H+] {0.200 (6.2 x 10-10 )} 1/2 1.11 x10-5 ) ph 4.95 b) At 20 ml note : 40% of the way to the equiv pt ph pka + log [CN-]/[HCN] 9.21 + log 40/60 9.03 note: ph pk a at 50% to equiv pt c) At equivalence point note : consider CN- hydrolysis CN- + H 2 O HCN + OH- K b K w / K a [HCN] [OH-] / [CN-] 1.61 x 10-5 [OH-] 2 / 0.200 (25/75) [OH-] 1.28 x 10-3 poh 2.89 & ph 11.1 d) At 55 ml note : consider excess NaOH [OH-] 5/80 (0.100 M ) 6.25 x 10-3 M poh 2.20 ph 11.8 Consider the titration of 25.0 ml of 0.200 M HCN (K a 6.2 x 10-10 ) We will work a base titration With 0.100 M NaOH. e.g., titration of 50.0 ml of 0.100 M ammonia with 0.100 M HCl Equiv Pt is obviously at 50.0 ml ASIDE Nanostuctured Microcantilvers as Biosensors Actual silicon microcantilevers (MCs) in small array with surface modifications Nanostructures Cavitand receptors Thin films Bioaffinity phases Focused Ion Beam milling can alter the thickness and shape of Cantilevers and form nanolevers 9
ASIDE Amino Acids, Chirality, and Antibodies R + H 3 N C CO 2 - H AA have weak acid & weak base functionality Isoelectric point Chirality (handedness) IgG Proteins MW ~ 160,000 Binding constants K > 10 8 Lock & Key Fit (selectivity) ASIDE Cantilever-Based Chemi-Mechanical Transducer z R Demonstrated applications: gas & liq sensors metal ions VOCs ph DNAPLs Proteins DNA Chiral Stoney Response Equation: z 3l 2 (1-ν) σ / Yt 2 σ - differential stress Y- Young s modulus ν - Poisson s ratio l & t length & thickness 10
ASIDE Microcantilever Deflection Measurements Readout accuracy - 0.25 nm Mechanical noise ~ 10 nm (3 mn / m) Cantilever mounted in a 100µL flow cell Flow rate: 0.15 ml/min (liq.) to 3 ml/min (gases) ASIDE Chiral Discrimination Using Cantilever Deflection 250 Deflection (nm) 200 150 100 50 0 D /L-AA 1 2 3 4 Nanostructured cantilevers have anti-d or -L amino acid antibodies chemically attached 28-50 5-100 6-150 0 5 10 15 20 25 30 35 Time (minute) 1 50 mg/l L-tryptophan (anti-l-aa) 2 50 mg/l L-phenylalanine (anti-l-aa) 3 50 mg/l D-tryptophan (anti-l-aa) 4 50 mg/l D-phenylalanine (anti-l-aa) 5 & 6 show no enantioselectivity W/O Ab 5 50 mg/l L-phenylalanine (IgG) 3 & 4 show no response to wrong AA 6 50 mg/l D-phenylalanine (IgG) Maximum Deflection (nm) Slope of the response curve (nm/minute 24 20 16 12 8 4 0 Slope (nm/minute) 5 4.5 4 3.5 3 100 fold excess D-Phe 1000 fold excess D-Phe 0 2 4 6 Concentration of L-Phe (mg/l) 0 20 40 60 80 100 120 Concentration of L-Phenylalanine (mg/l) 11
Indicators An indicator changes color near the equivalence point of the titration (endpoint). For an acid-base titration the indicator is an acid or base that changes color (visible absorbance) when the ph changes. The acid-base indicator is an acid/base. If you add it at high concentration, you will titrate it too!! (titration error, indicator error) The ph change for the indicator should take place in the steep portion of the titration curve near the equivalence point. If you can perform titration calculations for an acid-base titration, you can use the information to find an appropriate indicator! Indicators ph pk in + log [In-] / [HIn] the ratio changes (& color as well) when the ph changes from 1 below to 1 above pk in 12
Calculate the ph throughout the titration of 100.0 ml of a solution that is 0.050 M HCl and 0.10 M nitrophenol (Ka 4.0 x 10-9 ) with 0.20 M NaOH. At start: ph if only HCl -log 0.05 1.3 ph if only HNP 4.0 x 10-9 x 2 /0.10 x x [H+] 2.0 x 10-5 (ph 4.7) Of mixture ph 1.3 (ionization of HNP is minute, especially in HCl) At 1 st equivalence pt: Note: 100 ml x 0.05 5 mmol of HCl so 25 ml (5 mmol of NaOH needed) ph if only HCl ph ~ 7.0 Of mixture 4.0 x 10-9 x 2 / 0.100 [100/125] x x [H+] 1.8 x 10-5 (ph 4.7) Note: there are 10 mmol of the weak acid so another 50 ml of NaOH is needed for 2 nd eq. pt. After total of 50.0 ml NaOH After total of 75.0 ml of NaOH Pick indicator(s) (at Board) Polyfunctional Acids and Bases Let s consider a difunctional base first: B + H 2 0 BH+ + OH- K b1 BH+ + H 2 O BH 2 2+ + OH- K b2 And for the conjugates BH 2 2+ BH+ + H+ K a1 K w /K b2 BH+ B + H+ K a2 13
Polyfunctional Acids and Bases Let s consider phosphoric acid now: H 3 PO 4 H 2 PO - 4 + H+ K a1 7.1 x 10-3 H 2 PO - 4 HPO 2-4 + H+ K a2 6.3 x 10-8 HPO 2-4 PO 3-4 + H+ K a3 7.1 x 10-13 If the pkas of a polyprotic acid differ by about 4 or more than distinctive breaks are observed in a titration curve. Thus, in principle three breaks will be observed above. However the last acid step is so weak that it is difficult to observe. Polyfunctional Acids and Bases Suppose we titrate 50 ml of 0.10 M H 3 PO 4 with 0.10 M NaOH. Calculate the ph at a) 25 ml b) 50 ml c) 75 ml d) 100 ml e) 125 ml Note: pka1 2.2 pka2 7.2 pka3 12.2 Solution will only be approximate a H 3 PO 4 H 2 PO 4 - b c H 2 PO 4 - HPO 4 a) 50% to first equiv. pt. ph ~ pka1 ph 2.2 d e HPO 4 2- PO 4 3-25 50 75 100 125 ml b) Many species to account for but if pkas differ enough there is an approximation ph at 1 st equiv. pt. ~ (pka1 + pka2)/2 ph 4.7 c) 50% to second equiv. pt. ph ~ pka2 ph 7.2 d) Just like in (b) (pka2 + pka3)/2 ph 9.7 e) Well you should know the routine by now!! 14
The predominate form & α-values Phosphoric acid is like the case in this Chp 11 Figure For a monoprotic acid HA α HA [HA]/([HA] + [A-]) Divide num & denon on right by [HA] α HA 1/(1 + K a /[H+]) [H+] / ([H+] + K a ) In a similar fashion one can derive α A- K a / ([H+] + K a ) The predominate form & α-values For a diprotic acid like fumaric acid we have the following α-plot a) ph pka + log [A-]/[HA] -1 log R ; R 0.1 so HA dominates b) At ph 8 the ratio is 10 so A- dominates Instead consider what happens when 50 ml of 0.10 M Na 3 PO 4 is mixed with 50 ml of 0.20 M NaH 2 PO 4 15
Metal Complexation Complexation a metal forms a coordinate covalent bond M n+ + L (M L) n+ K f Examples 1. Cd 2+ + Cl - CdCl + + Cl - CdCl 2 + etc. 2. Fe (1,10 Phen) 3+ 3 3. Cu-(Oxime) 2 1. Stepwise formation of CdCl 4 2- ;monodentated ligand; Coordination # of 4) 2. Bidentated ligand; Coordination # of 6 3. Bidentated ligand; Coordination # of 4 Metal Complexation Ideally one would like to Have a one-to-one ligandto- metal stoichiometry In comes EDTA as a chelating ligand 16
Other EDTA-like Derivatives EDTA is also a Polyprotic Acid Notice the difference between the following reactions (charges largely ignored) H4Y + M MY + 4 H+ Y 4- + M MY 17
EDTA s Conditional Formation Constant Relevant EDTA Tables 18
EDTA Titrations Region 1 - No equilibria - Just consider how much metal has yet to be titrated - Don t ignore volume change Region 2 - K f equilibria involved - C T [M] Region 3 - K f equilibria involved - C T excess EDTA EDTA Titrations 19
EDTA Titrations EDTA Titrations; ph Effects ph can be adjusted to target Metals (differential titrations) 20
EDTA Titrations; Indicators EDTA Titrations; Indicators (in water hardness titrations using EBT indicator some Mg is needed Even if one needs to add it) Ca has bigger EDTA formation constant than Mg but smaller Indicator formation constant 21
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