Chapter 17: Acid-Base Reactions

Key to solving acid-base problems is:

- Writing the correct chemistry

- Determining which species acts as a base and which as an acid

- Knowing what remains after the reaction has occurred

Look for the limiting reagent

- Also look at what remains and what has been formed

Consider 3 options:

1. Strong Base + Strong Acid

2. Strong Base + Weak Acid

3. Weak Base + Strong Acid

[ see notebook]

Strong acid weak conjugate baes

Ka x Kb = Kw = 10-14

When Ka = 1, strong acid.. Kb = 10-14

as Ka decrease and becomes weaker, Kb gets larger and becomes stronger

17.4 Acid-Base Titrations - limiting reactant questions: either acid or base is always in excess, except at the equivalence

point (mol acid = mol base for 1:1 acid:base titration)

- In 3 titrations diff species are formed as pH changes:

- Strong Acid + Strong Base

- Strong acid neutral salt strong base

- Weak Acid + Strong Base

- Weak acid buffer basic salt strong base

o (Beaker of weak acid slowly dropping strong base into it )

o Equiv point will come when we form basic salt

- Weak base + Strong Acid

- Weak base buffer acidic salt strong acid

o Beaker of weak base slowly dropping strong acid into it

Strong Acid – Strong Base Titration (Lab 1, 6)

- Slowly adding NaOH to HCl, measure pH as you add NaOH

- Start at low pH of HCl since it’s very acidic, as we progress up the line we’re adding strong base

- pH stays around 1 since it doesn’t take much strong acid to keep pH low

- relatively close to equivalence point pH shoots up

Practice Calculations: Strong Acid – Strong Base

Titrate 25.0 mL 0.100 M HCl (aq) with 0.100 M NaOH (aq)

Calculate pH at 4 points: 0 mL, 12.5 mL, 25.0 mL, 37.5 mL of NaOH

0 mL NaOH = only strong acid present HCl is a strong acid, so [H3O+] = [HCl]

12.5 mL NaOH = ½ equiv point; ½ HCl consumed HCl (aq) + NaOH (aq) NaCl(aq) + H2O (l)

B 2.5 x 10-3 1.25 x 10-3 0

A 1.25 x 10-3 0 1.25 x 10-3

- NaCl doesn’t affect pH

Some HCl remains – in a new TOTAL VOLUME

pH = -log [H3O+] = - log (1.25 x 10-3/0.0375L ) = 1.47

25.0 mL NaOH = equivalence point All HCl consumed

HCl (aq) + NaOH (aq) NaCl(aq) + H2O (l)

B 2.5 x 10-3 2.5 x 10-3 0

A 0 0 2.5 x 10-3

pH = 7

37.5 mL NaOH = only strong base present All HCl consumed; excess NaOH

HCl (aq) + NaOH (aq) NaCl(aq) + H2O (l)

B 2.5 x 10-3 3.75 x 10-3 0

A 0 1.25 x 10-3 2.5 x 10-3

pOH = -log [OH] = - log ( 1.25 x 10-3mol / 0.0625L) = 1.699

pH = 14 – pOH = 12.301

Weak Acid- Strong Base Titration - not as much of a dramatic steep increase

- more than one inflection point, couple changes happening

- where pH = pKa around 12.5 … half-equivalence point

- between 0 and half equivalence point – buffer region

Titrate 25.0 mL 0.100 M CH3COOH with 0.100 M NaOH

0mL NaOH = only weak acid present Solve pH of weak acid, use Ka

CH3COOH + H2O CH3COO - + H3O +

I 0.100 M - 0 0

C - x + x M + x M

E 0.100 – x M x M x M

Ka = [ CH3COO-][H3O+]/[CH3COOH] = 1.8 x 10-5, soleve as usual…

X = [H3O+] = 1.3 x 10-=3

pH = -log{1.3 x 10-3) = 2.87

12.5 mL NaOH = ½ equivalence point ½ CH3COOH consumed

CH3COOH (aq) + NaOH (aq) CH3COONa(aq) + H2O(l)

Bef 2.5 x 10-3 1.25 x 10-3 0 -

Aft 1.25 x 10 -3 0 1.25 x 10-3

CH3COONA is a basic salt and DOES affect pH; some CH3COOH remains.. this is a buffer

Practice Calculations: Weak Acid – Strong Base

Solve for pH of a buffer with use of an ICE table or the Henderson-Hasselbalch equation below

= 4.74 + log (1.25 x 10-3 mol / 0.0375 L) / (1.25 x 10-3 mol / 0.0375 L ) = 4.74

pH = pKa at ½ equivalence point of a weak/strong titration

BECAUE we’ve consumed half the original acid and converted it to its conjugate base

Half the weak acid remains, half has been converted to weak conjugate base

Weak base + strong acid – relationship still holds pH will be pKa of the weak acid….???

pOH = pKb ?

Ka = [A-][H3O+] / [HA]

25.0 mL NaOH = equivalence point All CH3COOH consumed

Mol CH3COOH (aq) + NaOH (aq) CH3COOHNa (aq) + H2O(l)

Bef 2.5 x 10-3 2.5 x 10-3 0 -

Aft 0 0 2.5 x 10-3

CH3COONa is a basical salt, and does affect pH

Solve for the pH of a weak base

[CH3COO-] = 2.50 x 10-3 / 0.0500L = 0.0500 M ( taking into account the total volume to find [ x] )

CH3COO- (aq) + H2O (l) CH3COOH + OH

I 0.050 M - - 0

C -x + x + x

E 0.050 – x x x

Kb = [CH3COOH][OH-]/ [CH3COO-] = x2/0.05 – x = Kw / Ka = 5.5 x 10-10

X = [OH-] = 5.27 x 10 -6 , pOH = -log [OH-] = 5.28

pH = 14 – pOH = 8.72

37.5 mL naOH = only strong base present All CH3COOH consumed; excess naOH

Mol CH3COOH + NaOH CH3COONa (aq) + H2O (l)

Bef 2.5 x 10-3 3.75 x 10-3 0 -

Aft 0 1.25 x 10-3 2.5 x 10-3

Excess strong base and weak base : strong base determines pH

pOH = -log[OH-] = -log (1.25 x 10-3 / 0.0625L ) = 1.699

pH = 14 – 1.699 = 12.301 weak acid - strong base titration

Features different from strong

acid – strong base:

Start at a higher pH

Buffer region

½ equivalence point, pH = pKa

equivalence point pH > 7

Iclicker 1: B

Weak base strong acid titration

1) All weak base

2) Red line – Buffer region… pH=pKa +/- 1 pH unit

3) ½ equivalence , pH = pKa of the conjugate acid

4) Equvalence pt, acidic salt, typically pH<7

5) Excess strong acid

Titration – Key Concepts Equivalence point: mol acid = mol base ( if 1:1 reaction)

Strong acid- strong baes titration:

- pH – 7 at equivalence point

- sharp pH change at equivalence

Weak acid – strong base/ strong acid – weak base titrations

- buffer region

- pH = pKa at ½ equivalence

- pH doesn’t equal 7 at equivalence ( pH depends on the salt present – acidic or basic )

17.3 Acid-Base Indicators - equivalence point: theoretical point where mol acid = mol base (1:1 titration)

- With an indicator, we observe the endpoint (colour change)

- Colour depends on pH

- Indicators are weak acids/bases.. just like acetic acid etc (Hln, ln-):

- Hln(aq) + H2O(l) H3O+ (aq) + ln-(aq)

Acid colour base colour

- KHln = [ln-][H3O+]/[Hln] or pH = pKHln + log [ln-]/[Hln]

- Eq 17.4 ^ relates solution pH to pKa of indicator

- Doesn’t interfere with titration stoichiometry – only a few drops

- Want the indicator’s pKa to be as close as possible to equivalence point pH of titration solution

- Need a bit of excess of base / acid to see the colour – works out to about 10/1 excess to see

basic colour or 1/10 excess to see acid colour [ln-]/[Hln]

- We see:

- Acid colour when pH < (pKHln – 1 ) (10-fold excess of Hln)

- Base colour when pH > (pKHln + 1) (10-fold excess of ln-)

- Colour change over ~ 2 pH units

- Want the colour change ot be ator close to the equivalence point

- How to choose a suitable indicator?

o Select an indicator with pKa (pKHln) close to expected equivalence point pH

- Bromthymol blue:

- Blue (base colour), pH = pKa + 1

- ‘green’ (colour change mid-point), pH = pKa

- Yellow (acid colour), pH = pKa – 1

- Icicker 2: HCl + NaOH .. indicator? C) phenol red (6.3 – 8.0 pH range)

- Iclicker 3: NH3 + HCl .. indicator? B) bromocresol green (4.0-5.8 pH range)

Indicators – Key Concepts - Indicators are weak acids / bases

- Colour change from acid to base form

- Colour change occurs over ~ 2 pH units

- Choose indicator with pKa close to pH of titration equivalence point

Buffers and Titrations - Where do we see the importance of buffers?

o Patients’ blood pH (eg. Acidosis)

o Biochemical assays

o Lakes and streams

- How do we control buffer pH?

o With addition of appropriate acid or base

The Chemistry of a Buffer - Buffer chemistry: equilibrium reaction of either ewak species with water (eg. NH3/NH4+)

NH3(aq) + H2O(l) NH4+ (aq) + OH-(aq)

-and-

NH4+ (aq) + H2O (l) NH3 (aq) + H3O+(aq)

- Looks like weak base or weak acid ionization in water, but now weak base and weak acid

concentrations are both significant

- Both conjugates are present in large amounts in a buffer

- Concentrations of them have to be similar and in reasonable quantity

17-1: Common Ion Effect 0.10 M C6H5COOH has pH 2.60 and is 2.5% ionized (slide 13)

C6H5COOH (aq) + H2O (l) Ka C6H5COO- (aq) + H3O+ (aq)

Concept check: what would be the effect on the pH and on the % ionization of 0.10 M benzoic acid

solution with the addition of C6H5COONa(aq)?

Add C6H5COO-:

- ionization is happening left to right, but increase in CH6H5COO- supresses ionization by shifting

equilibrium and causing backwards reaction to happen, moves towards the left

- equilibrium shifts left

- ionization decrease

- pH increases (we added a weak base)

Adding a `common`ion reduces the extent of ionization

Adding C6H5COO-(aq) to C6H5COOH(aq) produces a buffer solution – an equilibrium mixture of a

weak conjugate acid-base pair

17-2 Buffers: Applying the Common Ion Effect - a buffer solution is:

o a solution of a weak acid-base conjugate pair

o both concentrations > 100 x Ka; concentrations need to be similar

- buffer capacity:

- a buffer is effective when:

0.1 < [weak base]/[weak acid] < 10 (eq. 17.3)

- best when [weak acid] & [weak base] are large and equal

Buffer Capacity Ph= phenyl

- Buffer capcity: maximum at equal [OhCOOH] and [PhCOO-]

- Effective range pH = pKa +/- 1 ( shown by green lines)

- PhCOOH = benzoic acid

- PhCOO- = bezoiate ion

- Buffers minimize / resist changes in pH

How to make a buffer: 3 ways 1) Mix weak conjugate acid and base together

Eg. HCOOH and HCOONa (formic acid and formate)

NH3 and NH4Cl

2) Titrate a weak acid with a limiting amount of STRONG BASE (to produce the conjugate weak

base)

Eg. HCOOH + NaOH

NH4Cl + NaOH

3) Titrate a weak base with a limiting amount of STRONG ACID (to produce the conjugate weak

acid)

eg. HCOONa + HCl

NH3 + HCl

Iclicker 4: which combination of 1M solutions will give a buffer? C) 25 mL NaOH + 50mL C6H5COOH

- A) 50 mL HCl + 50 mL NaOH NO b/c SA/SB

- B) HCl + C6H5COOH NO b/c 2 acids

- C) 25 mL NaOH + 50mL C6H5 COOH – YES b/c limiting amt of strong base (1/2 equiv pt)

- D) 25mL NaOH + 50mL C6H5COONa NO b/c 2 bases

- E) 50mL NaOH + 50mL C6H5COOH NO b/c too much strong acid it wouldn’t be a buffer.. equiv pt

Calculate pH of buffer from Ka or Kb - Calculate pH and % ionization for 250 mL of a solution of:

0.011M C6H5COOH(aq) and 0.0850 M C6H5COONa(aq)

Write down the chemistry (start with weak acid or base)

- The conjugate weak acid-base pair do NOT react with each other; one of them reacts with H2O

- When both reactants are weak equilibrium arrow

- Mol tables when you have two reactants consuming / using each other up.. these two are

coexistent without consuming one another

C5H5COOH (aq) + H2O(l) Ka C6H5COO-(aq) + H3O+(aq)

I 0.100 M - 0.0850M 0M

C - x M + x M + x M

E 0.100 – x 0.0850 + x x

Ka (benzoic acid) = 6.3 x 10-5

Ka = [6H5COO-][H3O+]/[C6H5COOH] = 6.3 x 10-5

6.3 x 10-5 = (0.0850 + x)(x) / (0.100 – x)

The common ion suppresses acid ionization, so we can ignore the +x and –x terms:

[Ha]/Ka = 0.100/6.3 x 10-5 = 1.6 x 103 , which is > 100

6.3 x 10-5 = (0.085)(x)/0.100

X = 7.41 x 10-5 = [H3O+]

pH = -log (7.41 x 10-5) = 4.13

pH increased compared to solution of C6H5COOH only, where pH = 2.60 (slide 13)

% ionization: [H3O+]/[HAinitial] x 100% = 7.41 x 10-5 / 0.100 x 100% = 0.0741% (negligible)

Buffer ionization decreased vs C6H5COOH only, where ionization was 2.5%.

Therefore, we can ignore ionization in a buffer (ie. The small x approximation is always valid in a buffer)

Alternative calculation of Buffer pH: Henderson – Hasselbalch Equatoin Ka = [C6H5COO-][H3O+]/[C6H5COOH]

Rearrange to get:

[H3O+] = Ka[C6H5COOH]/C6H5COO-]

Take –log of all terms

-log[H3O+] = -log (Ka) – log ( [C6H5COOH]/C6H5COO-] )

pH = pKa + log ( [C6H5COO-]/[C6H5COOH] )

Or more generally..

pH = pKa + log ( [A-]/[HA] Henderson-Hasselbalch equation (eq 17.2)

- Only applies when you have a buffer

In our example..

pH = 4.20 + log {0.0850 / 0.100) = 4.13

What does a buffer do? - Resist changes in pH, to the limit of buffer capacity

- Demonstration: addition of acid and base to water vs pH 7 buffer

Effect of Adding strong acid or base?

CH3CO2- + H3O+ CH3CO2H + H2O CH3CO2H + OH- CH3CO2- + H2O

WB + SA WA + SB

pH decreases pH increases

Middle:

- equal capacity to add acid or base.. if you add acid, responds by sending base.. weak base is

converted into conjugate acid

- ½ equivalence point: conjugate acid and base are present in equal amounts and pH = pKa + log

wk base / weak acid concentration (which are equal so its 1 so log 1 = 0) so pH = pKa as well

Left side:

- Strong acid, converted into weak acid because we have buffer component there

o pH will change but only a little bit; buffer is doing its job

right side:

- take buffer with equal resevoirs, add strong base acid part of buffer will step forward and

react with strong base, convert it into weak base

- strong base hydroxide converted to weak base acetate ion .. pH increases a little

- what if: adding so much OH- you depleted CH3CO2H so much that you exceeded 1:10 ratio…

nowhere near midpoint of buffer range – little capacity to accept more base but tremendous

capacity now to accept more base to back titrate towards the middle again

one way arrows because: we have a strong reagent so therefore full dissociation, one way arrow

iclicker 5: C6H5COOH + H2O Ka C6H5COO- + H3O+, add HCl (aq) to buffer which is true:

i) equilibrium shifts left

ii) C6H5COOH ionization decreases

iii) pH increases

answer: e) I, ii because:

- increasing HCl, a strong acid, into solution causes more H3O+ to be produced so due to le

chatelier’s principle equilibrium therefore shifts to the left (i)

- if it’s shifting to the left, the reverse reaction is favoured so less of forward reaction is

favoured/taking place therefore C6H5COOH isn’t being ionized as much

- ( Ka = [A-+*H3O++/*HA+ … amount of A- goes down and amount of H3O+ increases a lot.. the

reason why it ionizes is because of Ka.. more benzoic acid doesn’t mean its going to

automatically ionize since we still have a lack of?.. need to still meet Ka value )

Example: Buffer + Strong Acid Add 20 mL of 0.150 M HCl (aq) to our buffer. What is pHÉ (Buffer = 250 mL of 0.1M C6H5COOH and

0.0850 M Nac6H5COO (aq)

- The strong acid will react with the basic component of this buffer

- Mol table since we need to diagnose / assess change

Mol (cxv) HCl + C6H5OO-(aq) C6H5COOH (aq) + Cl- (aq)

Bef 3.0 x 10-3 2.12 x 10-2 2.5 x 10-2 0

Aft 0 1.82 x 10-2 2.8 x 10-2 3.0 x 10-3

- HCl is limiting… Cl produced has no basic properties, doesn`t matter

- Still have our buffer, concentrations are fairly similar of weak conjugate acid base pair

- We add a limiting amount of strong acid

- What remains is a buffer 0.1 < [weak base]/[weak acid] < 10

Concept check: Does Cl- (aq) impact pH? No – so we will not include it in calculations (spectator)

New concentrations: divide moles by TOTAL volume (250 + 20 = 270 mL)

C6H5COOH (aq) + H2O Ka C6H5COO- (aq) + H3O+

I 0.104 M - 0.0676 M 0 M

C - x M + x M + x M

E 0.104 – x M 0.0676 + x M x M

Ka = 6.3 x 10-5

6.3 x 10-5 = (0.0676 + x)(x) / (0.104 – x )

Assume x is small.

6.3 x 10-5 = (0.0676)(x)/(0.104)

x = 9.69 x 10-5 = [H3O+]

pH = -log[H3O+] = 4.01

Hasselbalch equation: pH = pka + log [base]/[acid]

pH = -log{6.3 x 10-5) + log(0.0676/0.104) = 4.01

you can ignore total volume and just put moles in rather than concentrations… only for hasselbalch

pH decreased slightly (from 4.13) upon addition of a limiting amount of strong acid – as expected

BUFFER CHANGE:

Concept check: what is the effect on pH of our buffer if:

Add small amount of NahOH? pH increases slightly – some benzoic acid converted to benzoiate ion

Add large amount of NaOH? pH increases a lot - ALL of the benzoic acid is converted to bezoiate ion,

and possible excess naOH remains; buffer is destroyed.

add small amount of water? No change in pH - moles of bezoic acid and benzoate ion are unchanged;

and their concentrations are still sufficiently high; equally diluting both weak acid and base by a little

Iclicker 6: what acid would be the best choice to make a buffer of pH 8.0?

a) HClO2 pKa = 1.96

b) CH3COOH pKa = 4.74

c) HOCl pka = 7.54

d) NH4Cl pKa = 9.26

e) (C2H5)2NH2Cl

pKa of weak acid as close as possible to pH

Iclicker 7: what reagent could be used to reach the correct pH?

a) NaOH

b) HCl

Iclicker 7b: target pH is 7.0, which reagent could be used?

a) NaOH

- (KHln -1 .. so its acutally 6.54… so to get to 7.0 you still add base due to the range )

- Pay attention to where you’re STARTING from, weak acid.. you still have to add NaOH to

increase pH at all

- Looking at pKa to choose the appropriate acid to make the buffer of pH certain #

Buffer: Key Concepts - Buffer effective range: pKa +/-1

- +x / -x is always negligible

- A buffer solution contains a weak conjugate acid-baes pair with both concentrations >100 x Ka

- Buffer capacity: a buffer is effective when 0.1 < [weak base]/[weak acid] < 10

- Best when [weak acid] & [weak base] are large and equal (max capacity) to neturalize acid or

base

- If it has more acid present, more capacity to add base; more base present, more capacity to add

acid

- Capcity to neutralize acid / base changes over its effective range

Make a Buffer:

- - directly mix conjugate weak acid-base pair together

- React weak acid with limiting amount of strong base

- React ewak base with limiting amount of strong acid

Calculate buffer pH from Ka expression (or Henderson-Hasselbalch equation):

pH = pKa + log([A-]/[HA]

- Buffer resists changes in pH by converting strong acid into weak acid, or strong base into weak

base

Truama, Critical Care, and pH Monitoring - Blood pH (7.4) is a highly regulated parameter of physiology

- Acidosis of blood (pH < 7.35) can be triggered in cases of acute infection (sepsis), cardiac arrest

or pulmonary dysfunction (hypoxia)

- Blood pH < 6.8 or >7.8 is fatal

- Blood pH regulated primarily by volatile CO2 and non-volatile acids, such as bicarbonate,

phosphate, lactate, protein, etc.

Blood pH - Equations:

- Two-step dissolution and ionization equilibria

- CO2 (g) + H2O (l) K H2CO3 (aq)

- H2CO3 (aq) + H2O (l) Ka1 HCO3 – (aq) + H3O+ (aq)

- Net Reaction: CO2 (g) + 2H2O (l) K * Ka1 HCO3- (aq) + H3O+ (aq)

K = 3.4 x 10-2 (25 C)

Ka1 = 4.5 x 10-7 (25 C)

Persistent cases of acidosis treated with bicarbonate infusion (NaHCO3) to prevent total organ failure

titration! Acid = CH3COOH, Base = NaOH; H+ (aq) + OH (aq) H2O (l)

Persistent Acidosis: Treatment Net Reaction

CO2 + 2H2O K * Ka HCO3- (aq) + H3O+ (aq)

K*Ka1 = [HCO3-][H3O+] / PCO2

Multiply by –log

pK = pKa1 = -log [HCO3-

]/PCO2 + pH