ACIDS AND BASES

REVISE Brønsted-Lowry acids and bases Amphoteric substances Conjugate acid base pairs Neutralisation Neutral: pH = 7 ([H+] = [OH-]) Acidic: pH < 7 ([H+] > [OH-]) Basic: pH > 7 ([H+] < [OH-]) pH = -log [H+] pOH = -log [OH-] pH + pOH = 14 at 25oC Kw = [OH-][H+] Kw = 14.0 at 25oC Kw = KaKb

STRONG ACIDS AND BASES Strong acids and bases  react nearly “completely” to produce H+ and OH-  equilibrium constants are large e.g.: HCl H+ + Cl- Complete dissociation: large small Common strong acids: HCl, HBr, HI, H2SO4, HNO3, HClO4 (Why is HF not a strong acid?) Common strong bases: LiOH, NaOH, KOH, RbOH, CsOH, R4NOH

LiOH  Li+ + OH- Example: Calculate the pH of 0.1M LiOH. 0.1M 0.1M Strong base  Dissociates completely LiOH  Li+ + OH- 0.1M Start: 0.1M 0.1M Complete rxn: pOH = -log [OH-] = -log (0.1) = 1  pH = 14 - 1 = 13

IMPOSSIBLE!!!!!! Problem: What is the pH of 1x10-8 M KOH? As before: pOH = -log (1x10-8) = 8 pH = 14 – 8 = 6 BUT pH 6  acidic conditions and KOH is a strong base IMPOSSIBLE!!!!!!

We do this by systematic treatment of equilibrium. Since the concentration of KOH is so low (1x10-8 M), we need to take the ionisation of water into account. In pure water [OH-] = 1x10-7 M, which is greater than the concentration of OH- from KOH. We do this by systematic treatment of equilibrium. Charge balance: [K+] + [H+] = [OH-] Mass balance: [K+] = 1x10-8 M Equilibria: [H+][OH-] = Kw = 1x10-14 M 3 equations + 3 unknowns  solve simultaneously Find: pH = 7.02 Hint: You end up with a quadratic equation which you solve using the formula.

Also note that: Only pure water produces 1x10-7 M H+ and OH-. If there is say 1x10-4 M HBr in solution, pH = 4 and [OH-] = 1x10-10 M But the only source of OH- is from the dissociation of water.  if water produces 1x10-10 M OH-  it can only produce 1x10-10 M H+ due to the dissociation of water.  pH in this case is due mainly to the dissociation of HBr and not the dissociation of water. It is thus important to look at the concentration of acid and bases present.

Some guidelines regarding the concentrations of acids and bases: When conc > 1x10-6 M  calculate pH as usual When conc < 1x10-8 M  pH = 7 (there is not enough acid or base to affect the pH of water) When conc  1x10-8 - 1x10-6 M  Effect of water ionisation and added acid and bases are comparable, thus: use the systematic treatment of equilibrium approach.

WEAK ACIDS AND BASES Weak acids and bases  react only “partially” to produce H+ and OH-  equilibrium constants are small HA H+ + A- Partial dissociation small large Acid dissociation constant Common weak acids: carboxylic acids (e.g. acetic acid = CH3COOH) ammonium ions (e.g. RNH3+, R2NH2+, R3NH+) Common weak bases: carboxylate anions (e.g. acetate = CH3COO-) amines (e.g. RNH2, R2NH, R3N)

base hydrolysis constant/ base “dissociation” constant B + H2O BH+ + OH- Weak base partial dissociation Kb small base hydrolysis constant/ base “dissociation” constant NOTE: pKa = -log Ka pKb = -log Kb As K increases, its p-function decreases and vice versa.

Problem: Find the pH of a solution of formic acid given that the formal concentration is 2 M and Ka = 1.80x10-4. HCOOH H+ + HCOO- H2O H+ + OH- Systematic treatment of equilibria: Charge balance: [H+] = [HCOO-] + [OH-] Mass balance: 2 M = [HCOOH] + [HCOO-] Equilibria: 4 equations  4 unknowns  difficult to solve

 [HCOO-] >> [OH-] Make an assumption: [H+] due to acid dissociation  [H+] due to water dissociation Produces HCOO- Produces OH- [HCOO-] large [OH-] small  [HCOO-] >> [OH-]  Charge balance: [H+]  [HCOO-]

Charge balance: [H+]  [HCOO-] Mass balance: 2 M = [HCOOH] + [H+] Equilibria: Let [H+] = [HCOO-] = x Or x = -0.019  No negative conc’s [H+] = [HCOO-] = 0.019 M  pH = 1.7

OR since [HCOOH] > 1x10-6, we can calculate pH as usual Weak acid equilibrium conditions HCOOH H+ + HCOO- 2M Start: 2-x x x Equilibrium: Solve as before

FRACTION OF DISSOCIATION,  Fraction of acid in the form A- For the above problem:  = [HCOO-] F = 0.019 M 2 M = 0.0095  Acid is 0.95% dissociated at 2 M formal concentration Weak electrolytes dissociate more as they are diluted.

FRACTION OF ASSOCIATION WEAK BASE EQUILIBRIA B + H2O BH+ + OH- Charge balance: [BH+] = [OH-] Mass balance: F = [B] + [BH+] Equilibria: Let [BH+] = [OH-] = x FRACTION OF ASSOCIATION

CONJUGATE ACIDS AND BASES Relationship between Ka and Kb for a conjugate acid- base pair: Ka.Kb = Kw = 1x10-14 at 25oC If Ka is very large (strong acid) Then Kb must be very small (weak conjugate base) And vice versa Base so weak it is not a base at all in water If Ka is very small, say 1x10-6 (weak acid) Then Kb must be small, 1x10-8 (weak conjugate base) Greater acid strength, weaker conjugate base strength, and vice versa.

Calculate the pH of 0.1 M NH3, given that pKa = 9.244 for ammonia. Problem: Calculate the pH of 0.1 M NH3, given that pKa = 9.244 for ammonia. Kb NH3 + H2O NH4+ + OH- base acid pKa = -log Ka Ka = 5.70x10-10 Kb = Kw Ka = 1x10-14 5 .70x10-10 = 1.75 x10-5

Solve for x using the quadratic equation Problem: Calculate the pH of 0.1 M NH3, given that pKa = 9.244 for ammonia. Kb = x2 F - x Where x = [OH-] = [NH+] = x2 0.1 - x = 1.75 x10-5 Solve for x using the quadratic equation Negative value discarded Find x = 1.31x10-3 M = [OH-] pOH = -log [OH-] = 2.88 pH = 14 – 2.88 = 11.12

BUFFERS Mixture of an acid and its conjugate base. Buffer solution  resists change in pH when acids or bases are added or when dilution occurs. Mix: A moles of weak acid + B moles of conjugate base Find: moles of acid remains close to A, and moles of base remains close to B  Very little reaction HA H+ + A- Le Chatelier’s principle

HENDERSON-HASSELBALCH EQUATION For acids: When [A-] = [HA], pH = pKa For bases: pKa applies to this acid Kb  B + H2O BH+ + OH-  Ka base acid acid base

Derivation: HA H+ + A-

The acid or base is consumed by A- or HA respectively Why does a buffer resist change in pH when small amounts of strong acid or bases is added? ? The acid or base is consumed by A- or HA respectively A buffer has a maximum capacity to resist change to pH. Buffer capacity, :  Measure of how well solution resists change in pH when strong acid/base is added. Larger   more resistance to pH change

A buffer is most effective in resisting changes in pH when: pH = pKa i.e.: [HA] = [A-]  Choose buffer whose pKa is as close as possible to the desired pH. pKa  1 pH unit

Problem: Calculate the pH of a solution containing 0.200 M NH3 and 0.300 M NH4Cl given that the acid dissociation constant for NH4+ is 5.7x10-10. NH3 + H2O NH4+ + OH- acid base Ka pKa = 9.244 pKa applies to this acid pH = 9.244 + log (0.200) (0.300) pH = 9.07

POLYPROTIC ACIDS AND BASES Can donate or accept more than one proton. In general: Diprotic acid: H2L HL- + H+ Ka1  K1 HL- L2- + H+ Ka2  K2 Diprotic base: L2- + H2O HL- + OH- Kb1 HL- + H2O H2L + OH- Kb2 Relationships between Ka’s and Kb’s: Ka1. Kb2 = Kw Ka2. Kb1 = Kw

Using pKa values and mass balance equations, the fraction of each species can be determined at a given pH.

ACID-BASE TITRATIONS We will construct graphs to see how pH changes as titrant is added. Start by: writing chemical reaction between titrant and analyte using the reaction to calculate the composition and pH after each addition of titrant

TITRATION OF STRONG BASE WITH STRONG ACID Example: Titrate 50.00 ml of 0.02000 M KOH with 0.1000 M HBr. HBr + KOH KBr + H2O What is of interest to us in an acid-base titration: H+ + OH- H2O Mix strong acid and strong base  reaction goes to completion H+ + OH-  H2O

* Calculate volume of HBr needed to reach the equivalence point, Veq: Example: Titrate 50.00 ml of 0.02000 M KOH with 0.1000 M HBr. * Calculate volume of HBr needed to reach the equivalence point, Veq: C1V1 C2V2 n1 n2 = But n1 = n2 = 1  CHBrVeq = CKOHVKOH (0.1000 M)Veq = (0.02000 M)(50.00 ml) Veq = 10.00 ml

There are 3 parts to the titration curve: 1 Before reaching the equivalence point  excess OH- present At the equivalence point  [H+] = [OH-] 2 3 After reaching the equivalence point  excess H+ present

Before reaching the equivalence point  excess OH- present HBr + KOH KBr + H2O Say 2.00 ml HBr has been added. Starting nOH- = (0.02 M)(0.050 L) = 1x10-3 mol COH- = nunreacted Vtotal nH+ added = (0.1 M)(0.002 L) = 2x10-4 mol COH- = 0.01538 M Vtotal = 50 + 2 mL = 52 mL = 0.052 L nOH- unreacted = 8x10-4 mol Kw = [H+][OH-] 1x10-14 = [H+](0.01538 M) [H+] = 6.500x10-13 M  pH = 12.19

At the equivalence point  nH+ = nOH- pH is determined by dissociation of H2O: H2O H+ + OH- x x Kw = [H+][OH-] 1x10-14 = x2 x = 1x10-7 M  [H+] = 1x10-7 M  pH = 7 pH = 7 at the equivalence point ONLY for strong acid – strong base titrations!!

After reaching the equivalence point  excess H+ present HBr + KOH KBr + H2O Say 10.10 ml HBr has been added. Starting nOH- = 1x10-3 mol CH+ = nexcess Vtotal nH+ added = (0.1 M)(0.0101 L) = 1.010x10-3 mol CH+ = 1.664x10-4 M pH = 3.78 nH+ excess = 1x10-5 mol Vtotal = 50 + 10.1 mL = 60.1 mL = 0.0601 L

Note: A rapid change in pH near the equivalence point occurs. Equivalence point where: slope is greatest second derivative is zero (point of inflection)

Calculate titration curve by calculating pH values after a number of additions of HBr.

TITRATION OF WEAK ACID WITH STRONG BASE Example: Titrate 50.00 ml of 0.02000 M formic acid with 0.1000 M NaOH. HCO2H + NaOH HCO2Na + H2O OR HCO2H + OH- HCO2- + H2O HA A- pKa = 3.745 Equilibrium constant so large  reaction “goes to completion” after each addition of OH- Ka = 1.80x10-4 Kb = 5.56x10-11 Strong and weak react completely

* Calculate volume of NaOH needed to reach the equivalence point, Veq: Example: Titrate 50.00 ml of 0.02000 M formic acid with 0.1000 M NaOH. HCO2H + OH- HCO2- + H2O * Calculate volume of NaOH needed to reach the equivalence point, Veq: C1V1 C2V2 n1 n2 = But n1 = n2 = 1  CNaOHVeq = CFAVFA (0.1000 M)Veq = (0.02000 M)(50.00 ml) Veq = 10.00 ml

There are 4 parts to the titration curve: Before base is added  HA and H2O present. HA weak acid,  pH determined by equilibrium: HA H+ + A- Ka From first addition of NaOH to immediately before equivalence point  mixture of unreacted HA and A- HA + OH- A- + H2O BUFFER!!  use Henderson-Hasselbalch eqn for pH 2 1

At the equivalence point  all HA converted to A-. A- is a weak base whose pH is determined by reaction: A- + H2O HA + OH- Kb 4 3 4) Beyond the equivalence point  excess OH- added to A-. Good approx: pH determined by strong base (neglect small effect from A-)

Before base is added  HA and H2O present. HA = weak acid. Ka = 1.80x10-4 HA H+ + A- F- x x 1.80x10-4 = x2 0.02 - x x2 + 1.80x10-4x – 3.60x10-6 = 0 x = 1.81x10-3  [H+] = 1.81x10-3  pH = 2.47

Say 2.00 ml NaOH has been added. From first addition of NaOH to immediately before equivalence point  mixture of unreacted HA and A- . BUFFER!! HCO2H + OH- HCO2- + H2O Say 2.00 ml NaOH has been added. Starting nHA = (0.02 M)(0.05 L) = 1x10-3 mol nOH- added = (0.1 M)(0.002 L) = 2x10-4 mol HA + OH- A- + H2O 1x10-3 2x10-4 8x10-4 Start - End pH = 3.745 + log 2x10-4 8x10-4  pH = 3.14

But VA = VHA = VTot

Special condition: When volume of titrant = ½ Veq pH = pKa Since: nHA = nA-

At the equivalence point  all HA converted to A-. A- = weak base. (nHA = nNaOH) Starting nHA = 1x10-3 mol HA + OH- A- + H2O 1x10-3 Start - End  nOH- = 1x10-3 mol Solution contains just A-  a solution of weak base

A- + H2O HA + OH- Kb = 5.56x10-11 F- x x x FA- = nA- V = 1x10-3 mol Vtotal = 50 + 10 mL = 60 mL = 0.060 L = 0.0167 M 5.56x10-11 = x2 0.0167 - x [OH-] = 9.63x10-7 M x2 + 5.56x10-11x – 9.27x10-13 = 0 pOH = 6.02 x = 9.63x10-7 pH = 7.98 pH is slightly basic at equivalence point for strong base-weak acid titrations

CALCULATED TITRATION CURVE

Titration curve depends on Ka of HA. As HA becomes a weaker acid the inflection near the equivalence point decreases until the equivalence point becomes too shallow to detect  not practical to titrate an acid or base that is too weak.

Titration curve depends on extent of dilution of HA. As HA becomes a more dilute the inflection near the equivalence point decreases until the equivalence point becomes too shallow to detect  not practical to titrate a very dilute acid or base.

TITRATION OF WEAK BASE WITH STRONG ACID This is the reverse of the titration of weak base with strong acid. The titration reaction is: B + H+ BH+ Recall: Strong and weak react completely

There are 4 parts to the titration curve: Before acid is added  B and H2O present. B weak base  pH determined by equilibrium: B + H2O BH+ + OH- Kb F-x x

From first addition of acid to immediately before equivalence point  mixture of unreacted B and BH+ B + H+ BH+ BUFFER!!  use Henderson-Hasselbalch equation for pH pKa applies to this acid

At the equivalence point  all B converted to BH+. BH+ is a weak acid  determined pH by reaction: BH+ B + H+ Ka F-x x FBH+ = n Vtotal Take dilution into account pH is slightly acidic (pH below 7) for strong acid-weak base titrations

4) Beyond the equivalence point  excess H+ added to BH+. Good approx: pH determined by strong acid (neglect small effect from BH+) Example: 50.00 ml of 0.05 M NaCN is titrated with 0.1 M HCl. Ka for NaCN = 6.20x1010 Draw the titration curve by calculating pH at various volumes of HCl.

TITRATION CURVE OF WEAK BASE WITH STRONG ACID

TITRATIONS IN DIPROTIC SYSTEMS Example - a base that is dibasic: B + H+  BH+ BH+ + H+  BH22+ pKb1 = 4.00 pKb2 = 9.00 With corresponding reactions: Two end points will be observed.

FINDING END POINTS WITH A pH ELECTRODE After each small addition of titrant the pH is recorded and a titration curve is plotted. 2 ways of determining end points from this: using derivatives using a Gran plot

Setup

But there are autotitrators! Titrando from Metrohm

USING DERIVATIVES End point is taken where the slope is greatest Or where the 2nd derivative is zero

USING A GRAN PLOT A problem with using derivatives  titration data is most to obtain near the end point Example – titration of a weak acid, HA Ka = [H+]H+[A-] A- [HA] HA HA H+ + A- NOTE: pH electrode responds to hydrogen ion ACTIVITY, not concentration

Say we titrated Va ml of HA (formal conc = Fa) with Vb ml of NaOH (formal conc = Fb) : HA + OH- A- + H2O [A-] = nOH(titrated) VTotal = VbFb Va + Vb [HA] = nHA(initial) – nOH(titrated) VTotal = VaFa- VbFb Va + Vb Substitute into the equilibrium constant: Ka = [H+]H+[A-] A- [HA] HA

Rearrange: VaFa Fb - Vb [H+]H+ = 10-pH = Ve - Vb

Gran plot equation: Gran plot  Graph of Vb10-pH vs Vb If is constant, then: A- HA Slope = -Ka A- HA and x-intercept = Ve Use data taken before end point to find end point Can determine Ka from slope

Use only linear portion of graph Extrapolate graph to get Ve

FINDING END POINTS WITH INDICATORS Acid-base indicator  acid or base itself Various protonated species have different colours HIn H+ + In-

Choose indicator whose colour change is as close as possible to the pH of the end point Indicators transition range overlaps the steepest part of the titration curve

Indicator error: difference between the observed end point (colour change) and the true equivalence point. Systematic error Random error Visual uncertainty associated with distinguishing the colour of the indicator reproducibly Why do we only add a few drops of indicator? Indicator is an acid/base itself  will react with analyte/titrant Few drops  neglible relative to amount of analyte

Good luck with the test and don’t panic!

HOW LONG DOES IT TAKE YOU TO FIND THE MAN’S HEAD?