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Acids & Bases (Topics To be covered, Chang 15)

1. Lewis / Brønsted acids and bases 2. The acid/base properties of water 3. Strong acids and bases - fully ionized 4. Weak acids AND Weak bases 5. Equilibrium 6. Equilibrium constant expression 7. Titrations e.g., Formic acid

H

O

O

HThis hydrogen isn't acidic

This hydrogen is weakly acidic, why?

Acids & Bases

- reactions in water (H3O+, OH-) - equilibria (not rates)

Strong Acids

HA + H2O H3O+ + A- (aq) i.e., solution, not heterogeneous equilibria ---------

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--------------- What are strong acids? Compounds for which the mixture of HO+H2 + A- is much more

stable than HA + H2O in solution Typical examples of strong acids: HCl HBr HI NOT HF

H2SO4 HNO3 in water, all of these compounds completely dissolve AND completely dissociate. e.g. HCl (g)+ H2O (l) H3O+ (aq) + Cl- (aq) [H3O+] at equilibrium = [HCl] that was put into water Kc = [H3O+]] [Cl-] /[HCl][H2O] equilibrium constant for dissociation (%

dissociation = 100%)

Strong Bases

---------

-----------------

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What about water?

H2O + H2O H3O+ + OH-

Kc = [H3O+] [OH-] / [H2O] [H2O] Kc [H2O]2 = [H3O+] [OH-] = 1 x 10-14 (by conductivity measurements @ 25 oC)

[H2O] ~ 55M

1 L = 1000 g / 18g /mol = ∼ 55 M Because the concentration of water essentially doesn’t change (55M - 10-14 M ≈ 55 M), it

is more convenient to replace the equilibrium constant Kc with Kw Kc [H2O]2 = Kw = 1 x 10-14 = [H3O+] [OH-] i.e. Water is a very weak acid Take negative log of both sides -log (Kw) = -log (1 x 10-14) = -log([H3O+]) - log([OH-])

DEFINE pH

----------------

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-------------------------- What about solutions of acids in water If makes solution of 0.1 mol HCl in 1L H2O

HCl + H2O H3O+ + Cl- t=0 0.1M 55 M - - t = infinity - 55 M 0.1M 0.1 M essentially no

change

[H3O+] = 0.1M pH = -log (0.1) =1 If one had 0.3 mol of HCl in 1 L H2O, the pH = -log (0.3) = 0.52 If one had 0.001 of HCl in 1L H2O, the pH is 3 Note: Only OH- + H+ contribute to pH, pOH, not Cl-, Na+ etc. (most counterions aren’t

important) What about strong bases?

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----------------- 1) What about weak acids? Formic acid

H OH

O

+ OH2 + H3O+H O

O

(Note, not )C

OH

O

WHY? if 0.1 M in H2O, how much is dissociated?

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H OH

O

+ OH2 + H3O+H O

O

t = 0 0.1 55 _ _

t = infinity 0.1-x x x

Kc = H O

H

O

OH2

H3O+

H O

O

Kc [H2O] = Ka = H O

H

O

H3O+

H O

O

Ka = acid dissociation constant (a reaction where H+ is transferred from an acid to H2O to

give H3O+) Ka = x2 / 0.1 - x or assume x << 0.1 (5%) x2 /0.1 - x = 1.8 x 10-4 (this comes from table on Page 607) x2/0.1 = 1.8 x 10-4 x2 = 1.8 x 10-5

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= 4.2 x 10-3

(x at 4.2 x 10-3 is less than 5%, therefore this approximation is OK). Otherwise, you have

to solve the quadratic. for quadratic ax2 + bx + c = 0 x2 + 1.8 x 10-4x - 1.8 x 10-5 = 0

x = − ± −b b ac

a

2 42

= − ±− −18 10 7 2 10

2

4 5. .x x

x = 4.3 x 10-3, that is, you get the same answer (within < 5%)

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A weak base example ------------- What is the OH- concentratioon in a 0.10 M aqueous solution of methylamine, CH3NH2?

What is the percent dissociation of methylamine?

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CH3NH2 + H2O

HO- + CH3N+H3

t=0 0.1M 55M - - t = infinity 0.1-x - x x essentially no change

Keq = [CH3N+H3] [OH-] / [CH3NH2] [H2O] Kb = Keq [H2O] = [CH3NH3

+] [OH-] / [CH3NH2] Kb = base equilibrium constant. A reaction where a molecule accepts H+ from water to

produce HO- (x)(x) / (0.10 - x) = 3.9 x 10-4 = Kb for methylamine x2 ∼ (3.9 x 10-4) (0.10) - assume x << 0.10 then check x ∼ 6.2 x 10-3 mol L-1

% dissociation = 6.2 x 10-3 / 0.10 x 100 ≈ 6% Here we can not make approximation (we are over 5%)!! Therefore solve quadratic x2 / ( 0.10 - x) = 3.9 x 10-4 x2 + (3.9 x 10-4)x - 3.9 x 10-5 = 0

x = − ± −b b ac

a

2 42

= − ±− −18 10 7 2 10

2

4 5. .x x

x = .0060 % dissociation = 6.0 x 10-3 / 0.1 = 6%

------------------ 2) Weak bases in water NH3 + H2O NH4

+ + OH- Keq = [NH4

+] [OH-] / [H2O][NH3] Kb = [NH4

+] [OH-] / [NH3] = 1.8 x 10-5 (from the table) if 0.02 M solution NH3

NH3 + H2O HO- + NH4+

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t=0 0.2M 55M - - t = infinity 0.2-x - x x essentially no change

Kb = x2 / 0.02 - x = 1.8 x 10-5 assume x << 0.02 (5%) x2 = 1.8 x 10-5 x 0.02 = 3.6 x 10-7 x= 6.0 x 10-4 [OH-] = 6 x 10-4 pOH = 3.22 pH = 14 - 3.22 = 10.7 Check that x << 0.02 x=.0006 =3% Weak acid in water What about solution of NH4

+ Cl in water

NH4+Cl- + H2O

H3O+ + NH3

t=0 0.2M 55M - - t = infinity 0.2-x - x x essentially no change

But we only know Kb Kb = 1.8 x 10-5 [OH-] [NH4

+] / [NH3] We need Ka for NH4

+ - from where? RECALL: HA and H2O A- + H3O+ Ka = [H3O+] [A-] / [HA] A- + H2O HO- + HA Kb= [OH-] [HA] / [A-] watch this

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Ka x Kb = [H3O+] [A-] / [HA] x [OH-] [HA] / [A-] Cancel italicized terms and Ka x Kb = [H3O+] [OH-] = Kw i.e. For any conjugate acid/base pair Ka x Kb = Kw pKa + pKb = 14 = pKw ie. If Kb = 1.8 x 10-5 pKb = 4.74 pKa = 14 - 4.74 = 9.26 Ka = 5.50 x 10-10 NOW WE CAN SOLVE THE PROBLEM OF NH4

+ in water. (DO THIS ON YOUR OWN)

Leveling effect in different solvents Can we measure the relative strengths of strong acids? All of them, e.g. HCl, HNO3,

H2SO4, HClO4, are almost completely dissociated in water. We must give them a more difficult test to allow us to discriminate between them. [Levelling effect of water.] Therefore, examine the equilibrium in a less basic medium than water (e.g., acetic acid). In this case, the equilibrium will not lie as far to products.

HNO3 + CH3-CO2H CH3C+(OH)2 + -NO3 Such experiments show that: HClO4 > H2SO4 > HNO3 > HCl Acid strength of oxyacids depends on the number of oxygens. (We can write more

canonical forms for the anion, and so delocalize the -ve charge) Why is this order observed?

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More stable anions are those that have the charge best dispersed over space (it’s thermodynamically favourable to spread charge on bigger atoms than smaller atoms and on several atoms if possible. For example,

O Cl

O

O

O

O

Cl O

O

O OCl

O

O

O

O

ClO

O

O

The perchlorate anion has 4 resonance forms. ------

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----------------------

Buffer Solutions Chang Chap. 16

Recall use of “Buffer zone”, Buffers contain weak acid + conjugate base salt of weak acid or weak base + conjugate acid salt of weak base Must use salts that are fully ionized in water Two types of Buffers i) weak base + conjugate acid of weak base e.g.,

NaHCO3 + H3O+ H2CO3 + H2O Base Conj acid

or ii) Weak acid + conjugate base fo the weak acid

e.g. H4N+Cl- + OH- H2O + NH3

Acid Conj base

Let’s do an example of a buffer solution, which has 100 mL of a solution that is 0.4 M PhCO2

- Na+ AND 0.5 M PhCO2H (C6H5COOH) What is pH of this solution? 1. PhCO2

- Na+ + H2O NaOH + PhCO2H (aq) 2. PhCO2H + H2O H3O+ + PhCO2

-

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Ka = [H3O+][PhCO2-] / [PhCO2H] = 6.0 x 10-5 (from table)

PhCO2H + H2O

H3O+ + PhCO2-

t=0 0.5M 55M - 0.4M t = infinity 0.5-x - x 0.4 + x essentially no change

Ka = 6.0 x 10-5 = x (0.4 + x) / (0.5 - x) We can assume that x is small compared to 0.4 or

0.5 (<5%) and neglect it for the same reasons as before. x = 6.0 x 10-5 x 0.5/0.4 = 7.5 x 10-5 = [H3O+] pH = -log (7.5 x 10-5) = 4.12 i.e. [H3O+] = Ka [PhCO2H] / [PhCO2

_ ] negative log each side - log [H3O+] = -log Ka - log [PhCO2

- ] / [PhCO2H] pH = pKa + log [PhCO2

- ] / PhCO2H] pH = log 6.0 x 10-5 + log (0.4/0.5) = 4.22 - 0.097 = 4.12 pH = pKa + log [A-] / [HA] Henderson - Hasselbalch equation. This is an equation YOU SHOULD REMEMBER It applies if Ka of acid or Kb of base < 10-3 IT CAN ONLY BE USED IF THE CONCENTRATIONS OF A- and HA are similar. THE

RATIO MUST BE: 0.1 < [A-]/[HA] < 10

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Buffering action If add 10 mL of 0.15M HCl to the above solution (100 mL) solution Adding a strong acid will remove some PhCO2

- PhCO2

- + H3O+ PhCO2H + H2O

Added HCl 10 mL x 10-3 L/mL x 0.15 mol/L = 1.5 x 10-3 mol ie removed 1.5 x 10-3 mol PhCO2

_ produced 1.5 x 10-3 mol PhCO2H New volume = 100 mL + 10 mL = 110 mL

PhCO2H + H2O H3O+ + PhCO2-

New Initial 50 x 10-3 mol 40 x 10-3 mol

+ 1.5 x 10-3 mol - 1.5 x 10-3 mol 110 x 10-3L 110 x 10-3L

At Equi 51.5 x 10-3 - x 38.5 x 10-3 + x

110 x 10-3L 110 x 10-3L Substituting Ka = 6.0 x 10-5 = x (38.5/110 + x) / (51/110 - x) (again assume x << 38.5/110 M) x= 8.02(6) x 10-5 = [H3O+] pH=4.10 (compare 4.12)

i.e., we added some strong acid to the solution, but the pH only changed a small amount -

the solution was buffered. The easier way to calculate this is to use the Henderson Hasselbalch Equation pH = pKa + log [PhCO2

-] / [PhCO2H] [PhCO2H]= 0.100 L x 0.5 mol/L + 0.01L x 0.15 mol/L / 0.110 L = 0.05 + 0.0015 mol / 0.11 L = 0.0515 mol / 0.11 L = 0.468 M [PhCO2

-]= 0.100 L x 0.4 mol/L + 0.01L x 0.15 mol/L / 0.110 L = 0.04 + 0.0015 mol / 0.11 L = 0.0385 mol / 0.11 L = 0.35M [PhCO2

-] / [PhCO2H] = 1.338

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log = 1.13 pH = pKa - 0.13 = 4.22 - 0.13 = 4.09 compare with 4.12 [WE MAY DO A Demo on Buffering ----------------

---------------

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Now addition of base to same solution

take 100 mL of buffer 0.5 M PhCO2H

0.4 M PhCO2- + 10 mL of 0.15 M KOH 1) KOH + PhCO2H PhCO2

_ + H2O 2) PhCO2H + H2O H3O+ + PhCO2- as before pH = 4.14 If 10 mL 0.15M KOH in buffer PhCO2H + H2O H3O+ + PhCO2

- AS ABOVE pH = pKa + log (41.5 x 10-3 / 48.5 x 10-3) pH = 4.17

Indicators of pH Changes Chang p. 662

A compound that changes colour when it picks up a proton (becomes positively charged)

and becomes charged or more likely loses a proton to become negatively charged In = indicator in the following pages HIn + H2O H3O+ + In- e.g., Anthocyanidins flowers, sweet cherries

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O

OH

OH

OH

OHOH

OH acid

base

O+

OH

OH

OH

OHOH

OH

Acid form, REDBasic Form, BLUE_GREEN

[WE MAY DO A Demo on Indicators ----------------

--------------- In order to really be sure of the colour, need at least a 10 fold excess of one form i.e. Ratio [In-]/[HIn] ≈ 10/1 to see the blue-green colour for an indicator with a pKa = 5 pH = pKa + log [In-] / [HIn]

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For [In-] / [Hin] = 1/1 pH= pKa + 0 = 5 (log 1) but to see red (ie, to be sure you are not looking at orange) in acid need at least a 10 fold

excess of acid Ratio [In-] / [HIn] = 1/10 pH = 5 + log (1/10) = 5 - 1 = 4 To see blue-green in base Ratio [In-] / [HIn] = 10/1 pH = 5 + log (10/1) = 5 + 1 = 6 This is why each pH indicator works over about a 2 pH unit range (2 order of magnitude in

concentration)

Summary

Titration of weak acid with strong base or vice versa. A small amount of indicator (much much less than the compound you are titrating (neutralizing) is added. Once the compound you are titrating is neutralized, a small amount of acid (base) can titrate the indicator, changing its colour. Once its colour has changed, you know the stuff you are also interested in has changed..

Indicator pH range Colour Change pKa Methyl violet 0 - 3 yellow - violet 1.6 Methyl orange 3.3 - 4.6 red - yellow 4.2 Methyl red 4.2 - 6.2 red - yellow 5.2 Bromthymol blue 6.0-7.8 Yellow - blue 7.2 Thymol blue 7.9 - 9.4 Yellow - blue 8.2 Phenolphthalein 8.3 - 10.0 Colourless - red 9.5 alizarine yellow 10.1 - 12.1 yellow - red 11.0

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For a all titrations, choose an indicator whose pKa is similar to the pH at the equivalence

point. For a strong acid/strong base titration, the equivalence point is at pH7. Therefore, while bromthymol blue is the best (range = 6.0-7.8) because of the height of the curve, methyl red and phenolphthalein will also suffice

Titrations

Strong Acids with Strong Bases

ii) Neutralization point = pH system 7 Strong acid + Strong Base water + salt e.g., take 50.0 mL of 0.10 M HCl titrateed with 0.10 M HCl

base added (mL) pH 0 1 10 1.18 20 1.37 40 1.95 45 2.28 48 2.69 49 3.0 50 7.0 51 11.0 55 11.68 60 11.96

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80 12.36 100 12.52 large excess 13

ii) Weak acid + strong base

at the beginning 100 mL Acetic acid CH3CO2H 0.1M (Ka = 1.8 x 10-5) pKa = 4.74 Titrate with: 100 mL 0.1 M KOH a) at beginning 0.1M HOAc (THIS IS A Ka calculation) CH3CO2H + H2O H3O+ + CH3CO2- x2/(0.1-x) = 1.8 x 10-5 x = 1.34 x 10-3 pH = 2.87 b) at the equivalence point (where there are equimolar amounts of initial acid and

added base) note that at the equivalence point, the volume of the original solution has changed - we

have dumped not dry powder, but a solution into it. Therefore, also keep track of the new volume - all calculations are done in molarity

Started 100 mL 0.1 M CH3COOH. At the equivalence point, we have added 100 mL x 0.1

M NaOH. This produces MOLES 0.1L x 0.1 mol/L / 0.1L (Starting solution) + 0.1L (added NaOH) = 0.01 moles / 0.2 L @ equivalence point = 0.05 M NaOAc THIS IS A Kb calculation CH3CO2Na + H2O OH- + H3CCO2H 0.05 - x x x x2 / 0.05 = Kb = 1 x 10-14 / 1.8 x 10-5 (Kw/Ka)

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x2=2.7 x 10-11 x= 5.3 x 10-6 pOH 5.27 pH = 8.72 FOR THIS SOLUTION, WITH EQUIVALENCE POINT 8.72, WE NEED A

DIFFERENT INDICATOR. PHENOLPHTHALEIN (RANGE 8.3-10) IS THE BEST. The pKa of phenolphthalein is close to that of the pH of the final solution.

c) when the solution is half titrated (at 50 mL addition of KOH) pH = pKa + log [A-]/[HA] pH = 4.74 + log [0.005/150 mL]/[0.005/150 mL] = 4.74 d) at 25 mL addition pH = pKa + log [A-]/[HA] pH = 4.74 + log [0.0025/125 mL]/[0.075/125 mL] = 4.74 + log [0.0025]/[0.075]

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=4.26 THESE ARE A PAIR OF OTHER EXAMPLES. WHICH INDICATOR SHOULD YOU

USE?

SUMMARY OF TITRATIONS

A titration curve has 3 parts. I) The beginning. Here we have a weak acid (Ka calculation, Fig. 18.7) or a weak base (Kb calcuation, Fig. 18.8). ii) From 10%-90% of the titration, we have a buffer (where 0.1 < [A-]/[HA] < 10) iii) Finally, we have the equivalence point, where the added moles of strong base = initial moles of weak acid (Fig. 18.7) or added moles of strong acid = initial moles of weak base (Fig. 18.8). At this point, the weak acid is converted to conjugate base and the pH is calculated using a Kb calculation (Fig. 18.7). OR the weak base is converted to the conjugate acid, and the pH is calculated doing a Ka calculation (Fig. 18.8).

These 3 points are illuminated in the following graph

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Note which indicator works best here (Methyl red)

The Periodic Table: Acid and Basic Solutions

Oxides of metals (green part of the table) typically hydrolyse to give basic solutions

(containing –OH). K2O + H2O → 2 KOH → 2K+ + 2 HO-

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MgO + H2O → Mg(OH)2 → Mg2+ + 2 HO- On the other hand, main group oxides (blue part of the table) typically hydrolyse to give

acidic solutions. Note that not all these compounds dissolve well: SiO2 is glass (quartz). N2O3 + H2O → 2 HNO3 → 2 H3O+ + 2 NO3

-

P4O10 + 6 H2O → 4 H3PO4 → H3O+ + H2PO4

- SO3 + H2O → H2SO4 → H3O+ + HSO4

-

I2O7 + H2O → 2HIO4 → H3O+ + IO4

-

FINALLY, there are some metal compounds that are amphoteric (orange part of the table,

go both ways!). That is, under acidic conditions it reacts one way and under basic conditions another.

Al2O3 + HCl → AlCl3 + 3 H2O (Acts as base, removes acid) Al2O3 + NaOH → NaAl(OH)4 (acts as acid, removes base) Aluminum hydroxide reacts with both acids and bases Al(OH)3 (s) 3H+ (aq) → Al3+ (aq) + 3 H2O (l) Al(OH)3 (s) -OH (aq) -Al(OH)4 (aq)

Lewis Acids

We have talked almost exclusively about Bronsted acids (which donate H+ to water to give H3O+). There are other kinds of acids.

(H+ is a special example of a Lewis acid AND a Bronsted acid). Lewis acids are molecules that are unable to make an octet with their available valence

electrons AND which are somewhat electronegative. Thus, compounds in Group 13 have only 3 valence electrons. They can achieve an octet

by forming 3 normal bonds and, IN THE EMPTY ORBITAL a fourth bond by accepting/sharing a lone pair of electrons – acceptors are Lewis acis.

The donor of the lone pair must be an element that is willing to share the electrons – that is

not too electronegative. It is the Lewis base.

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FB

F

F

ONLY 6 electrons

NHH

H

6 electrons in covalent bonds2 available to share

B NH

H

H

F

FF

Boric acid is a Lewis acid with water (the Lewis base)

BOHOH

OH

H2OB

OH

OH

OH

O+ H3O

+

Generally, good Lewis acids come from Group 13 or Group 2. Good Lewis bases come

from Group 15 or Group 16. Metal ions may also be good Lewis acids (e.g., Ag+), but are beyond the discussion in this course.