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John A. SchreifelsChemistry 212

Chapter 17-1

8–1

Chapter 17

Acid-Base Equilibria


Chapter 17-2

8–2

Overview

• Solutions of a Weak Acid or Base– Acid ionization equilibria– Polyprotic acids– Base ionization equilbria– Acid-Base properties of Salts

• Solutions of a Weak Acid or Base with Another Solute– Common Ion Effect– Buffers– Acid-Base Titration Curves


Chapter 17-3

8–3

Acid –Ionization Equilibria

• Weak acids and weak bases only partially dissociate; their strengths are experimentally determined in the same way as strong acids and bases by determining the electrical conductivity.

• The reaction of a weak acid (or base) with water is the same as discussed in previous section.

• Consider the reaction:

– Hydronium ion concentration must be determined from the equilibrium expression. • Relative strengths of weak acids can be determined from the value of the

equilibrium constant. – Large equilibrium constant means strong acid– Small equilibrium constant means weak acid

E.g. determine which acid is the strongest and which the weakest.Acid KaHCN 4.9x1010

HCOOH 1.8x104

CH3COOH 1.8x105

HF 3.5x104

]HA[]A][OH[

K )aq(A)aq(OH)l(OH)aq(HA 3a32


Chapter 17-4

8–4

Determining K from pH

• Ka determined if pH and CHA known.– Use the equilibrium expression for the acid.E.g. Determine the equilibrium constant of acetic acid if the pH of a

0.260 M solution was 2.68. Determine [H3O+]; [HA]; and [A].

– Strategy• Calculate the [H3O+] from pH; this is x in the table above. • The rest of the quantities are obtained from the bottom row.

HA(aq) + H2O(l) H3O+(aq) + A (aq) Initial conc 0.100 M 0 0 to Equil. x +x +x

Equil. 0.100 M x + x +x


Chapter 17-5

8–5

Calculating Equilibrium Concentrations in Weak–acid Solutions

• pH determined if Ka and Ca known; for the dissociation of acetic acid:

– [H3O+]total = [H3O+]CH3COOH + [H3O+]H2O.

– [H3O+]total [H3O+]CH3COOH.

• The total hydronium ion concentration is often equal to the contribution from the weak acid which is usually a lot stronger acid than water.

• The total hydronium ion concentration is needed for the equilibrium calculation.

)()( lOHaqCOOHCH 23

)aq(COOCH)aq(OH 33

53

33a

10x8.1

]COOHCH[]COOCH][OH[

K

)l(OH2 2 )aq(OH)aq(OH3

143w

10x00.1

]OH][OH[K


Chapter 17-6

8–6

pH from Ka and Ca

E.g. Calculate the pH of 0.100M acetic acid. Given pKa = 4.76

Method I:• Substitute into equilibrium equation to get • x2 + 1.75x105x 1.75x106 = 0. • Solve using quadratic equation (see book).Method 2• Assume x << CHA. Then x = (KaCHA)1/2. • Check (confirm assumption to be correct)

– Analytical concentration should be: Ca = 100x[H3O+] Method 3 method of successive approximations.• As in Method 2; then • x = (Ka(CHA x1))1/2; repeat if necessary.

E.g. Calculate pH of 0.0200M lactic acid if its Ka = 8.4x104M.

CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO (aq) Initial conc 0.100 M 0 0 to Equil. x +x +x

Equil. 0.100 M x + x +x

x100.0x10x75.1

25


Chapter 17-7

8–7

% Dissociated (also called % Ionized) Weak Acids

• % ionization – a useful way of expressing the strength of an acid or base. – 100% ionized a strong acid. – Only partial ionization occurs with weak acids.

E.g. determine the % ionization for 0.100 M, 0.0100 M, 0.00100M HCN if Ka = 4.9x1010.

– Solution: determine x for each and sub into definition above. Check assumptions.

• Notice % ionization increases with dilution.

HA(aq) + H2O(l) H3O+(aq) + A (aq) Initial conc CHA 0 0 to Equil. x +x +x

Equil. CHA x +x +x

%100C

xIonization %HA

%100CK

%100C

KCIonization %

2/1

HA

a

HA

2/1aHA


Chapter 17-8

8–8

Polyprotic Acids

• Some acids can donate more than one proton to the solution. Thus a diprotic acid has two protons such as H2S and H2SO4, while a common triprotic acid has three acidic protons that can be donated (H3PO4).

• First proton easily removed; others much more difficult.

Treat Polyprotic acids as if they were monoprotic acids; Use Ka1.

•The equilibrium constant for removal of each successive proton is about 105 times the equilibrium constant for removal of the preceeding proton. •E.g. determine the pH of 0.100 M H2SO3. Then determine .

Formula Ka1 Ka2 Ka3 H3PO4 7.5x10 3 6.2x10 8 4.8x10 11 H2SO3 1.5x10 2 6.3x10 8

]SO[ 23


Chapter 17-9

8–9

Equilibria:Weak bases (WB) (proton acceptor)

• Treat bases just like we did the weak acid; except you are calculating [OH].

• The general equation that describes the behavior of a base in solution is:

• Set up the equilibrium table as before for the acids and substitute values for all the quanitities in the equilibrium expression.

• Since usually CB is supplied, we have one unknown which we can evaluate using standard equil. equation for weak base.

• Remember that x = [OH] and not [H3O+]. • E.g. Calculate the pH of 0.10M NH3(aq). • Hint: Expect pH > 7 when with weak base.

]B[]OH][BH[K )aq(OH )aq(BH)l(OH)aq(B b2

B(aq) + H2O(l) BH+(aq) + OH (aq) Initial conc. CB 0 0 to Equil. x +x +x

Equil. CB x +x + x

xCx

]B[]OH][BH[K

B

2

b


Chapter 17-10

8–10

Equilibria:Weak bases Structure

• Many nitrogen containing compounds are basic –the amine most important.

• Most of the amines have a lone pair of electrons that are available for bonding with an acidic proton (Brønsted-Lowry base).

• Amines usually have a carbon residue in place of a hydrogen.

N

H

R 1 R 2

Amine Structure


Chapter 17-11

8–11

Relation between Ka and Kb

• Ka and Kb are always inversely related to each other in aqueous solutions.

• Inverse relationship explains why conjugate base of very weak acid is relatively strong.E.g. given the Ka’s of the following acid list their conjugate bases in terms of relative strength.

Acid KaHF 3.5x104

HCOOH1.8x104

HOCl 3.5x108

HCN 4.9x1010

HA(aq) + H2O(l) H3O+(aq) + A (aq)

]HA[]OH][A[

K 3a

Add A (aq) + H2O(l) HA(aq) +OH (aq) ]A[

]OH][HA[Kb

2H2O(l) H3O+(aq) +OH (aq) Kw = Ka•Kb


Chapter 17-12

8–12

Salts of WA and WB

• Salt: an ionic substance formed as a result of an acid–base neutralization reaction.– Salt of an acid(base) obtained by its neutralization with acid if it is a

base and base if it is an acid. E.g. NaCl is a salt from the reaction of HCl with NaOH. – The properties of the salt will depend upon the strengths of the acid

and base that formed the salt.E.g.1: determine the acid–base reaction that would produce CH3COONa, NaCN, NH4Cl, (NH4)2CO3.

• Salts are usually soluble in water because of their ionic character.

• When they dissolve, they affect the pH of the solution. Depends upon relative strengths of the conjugate acid and base.


Chapter 17-13

8–13

Salt of Strong Acid and Strong Base

• Neutral solution results if the salt is from the reaction of a SA + SB.E.g. NaCl

• Other cations and anions producing neutral solutions: Li+, Na+, K+, Ca2+, Sr2+, Ba2+ and Cl, Br, I, , ). E.g. what is the approximate pH of the following. NaCl, KCl, LiClO4, etc.?

• Salt of WA + SB (basic) and Salt of WB + SA (acidic).• Ignore cation (or anion) from SA (base).• Conjugate of WA is WB basic solution.• Conjugate of WB is WA acidic solution.

3NO

4ClO

SA + SB Neutral (very WA & WB)SA + WB Acidic (WA)WA + SB Basic (WB)whereSA = Strong Acid; SB = Strong BaseWA = Weak Acid; WB = Weak Base


Chapter 17-14

8–14

Calculating the pH of Salt of WA or WB (other ion from SA(SB))

• Salt of WA: Use Kb of the conjugate base and treat it as a weak base:

A(aq) + H2O(l) HA(aq) + OH(aq)

E.g. determine the pH of 0.100M NaCH3COO. Ka (CH3COOH) = 1.75x105.

E.g. determine the pH of 0.200 M NaCN. Ka(HCN) = 4.9x1010.

• Salt of WB: Use Ka of conjugate acid and treat as a weak acid: E.g. determine pH of 0.250M NH4Cl. Kb = 1.8x105.

E.g. determine pH of 0.100 M N2H5Br. Kb = 1.1x108.


Chapter 17-15

8–15

Salt of WA + WB

• Determine Ka and Kb of acidic and basic portions of salt.

• Largest K dominates to make solution either acidic or basic.

• E.g. determine if 0.100 M NH4CN is acidic or basic. E.g. 2 predict if 0.100 M C6H5NH3F is acidic or basic.


Chapter 17-16

8–16

The Common Ion Effect

• Common–Ion Effect: the change in the equilibrium that results from the addition of an ion that is involved in the equilibrium. E.g. NaOCl is added to 0.100 M HOCl; is added to NH3.

• Setting up the standard equilibrium table can show the effect.• E.g. determine the pH of a solution prepared by mixing 50.0 mL of 0.100 M

HOCl with 50.0 mL of 0.100 M NaOCl (Ka = 3.5x108). • Set up equilibrium table after calculating the concentrations of each in the

final mixture. • Initial concentrations change slightly as a result of a change reaction.

• Solve using either approximations or quadratic equation.• Shifts equilibrium towards the basic side.

ClNH4

HOCl + H2O H3O+ + OCl 0.0500M 0 0.0500M

x +x +x 0.0500 x +x 0.0500 + x


Chapter 17-17

8–17

Buffers

• Buffer solution: a mixture of conjugate acid and base that resists pH changes.

– Significant buffering capacity occurs when [acid] = [base], pH = pKa.– An example of the common ion effect.

E.g. Calculate pH of solution containing 0.040M Na2HPO4 and 0.080M KH2PO4. pKa2=7.20.

– Set up equilibrium table.– Ignore the value of x compared to the concentrations of the common ion. – pH in buffering region related to the relative amount of conjugate acid and

base.• Let then the equilibrium equation is:

]acid[]base[r

rK

]OH[

r]OH[K

a3

3a


Chapter 17-18

8–18

Addition of Acid or Base to a Buffer

• Upon addition of a SB to the buffer we have:– Addition of either acid or base changes ratio of acidic and basic

forms. – Big changes in pH occur only when nearly all of one species is

consumed.

E.g. determine r after addition of 5.00 mL of 0.100 M NaOH to 10.00 mL of 0.100 M HOCl. Determine pH if Ka = 3.5x108.

E.g. Determine pH of 50.00 mL of phosphate buffer containing equilmolar concentrations (0.200M) of acid/base forms, after 10.00 mL 0.100 M NaOH or 10.00 mL of HCl. pKa2 =7.20

• Changes in volume don't affect pH.

bbaa

bbVCVC

VCr


Chapter 17-19

8–19

Henderson-Hasselbalch Equation

• The effect of r (=[A]/[HA]) on pH is better understood by taking log of both sides of equation between K and conc. To give

• Called Henderson-Hasselbach equation.• Allows us to predict pH when HA/A mixed.• When [A] /[HA] = 1 (i.e. [HA]=[A]), pH = pKa

E.g. Calculate pH of solution containing 0.040M Na2HPO4 and 0.080M KH2PO4. pKa2=7.20.

E.g.2 determine the ratio of the concentration of the conjugate acid to concentration of the conjugate base for a weak acid in which the pH was 5.45 and pKa was 5.75.

E.g. determine the pH of a solution consisting of 0.100 M NH3 and 0.150 M NH4Cl.

HA

Aa

a

n

nlog+pK =

[HA]

]A[log+pK=pH


Chapter 17-20

8–20

Neutralization Reactions

• Neutralization Reaction: the reaction of an acid with a base to produce water.

• Extent of reaction nearly quantitative (except if both acid and base are weak.

• SA–SB:– E.g. HNO3 + NaOH NaNO3 + H2O

• SA produces: H3O+

• SB produces: OH

• Overall reaction:• WA–SB: thought of as two step reaction.

– E.g. HOCl + NaOH NaOCl + H2OK = ?

• Large equilibrium constant means reaction nearly quantitative.

H3O+ + OH 2H2O 1wK = 1/Kw = 1.00x1014

HOCl H+ + OCl Ka = 3.5x10 8 H+ + OH H2O 1

wK = 1.00x10+14 HOCl + OH H2O + OCl K = Ka 1

wK = 3.5x106


Chapter 17-21

8–21

Neutralization Reactions – WB + SA and WA + WB

• WB + SA– SA produces H3O+ ions; use base as is.– E.g. NH3 + HCl + Cl or

• Conclusion: Quantitatively generate product (nearly).• WA + WB: initially undissociated species dominates.

• Conclusion: Reaction will sometimes, but not always, be quantitative.

E.g. determine the extent of reaction when di methyl amine (Kb = 5.4x104) reacts with either HF (Ka = 3.5x104) or HOCl (Ka = 3.5x108).

NH3 + H2O 4NH + OH Ka = 1.8x10 5

H3O+ + OH 2H2O 1wK = 1.00x10+14

NH3 + H3O+ 4NH + H2O K = Ka 1

wK = 1.8x109

HOCl + H2O H3O+ + OCl Ka = 3.5x10 8 NH3 + H2O

4NH + OH Kb = 1.8x10 5

H3O+ + OH 2H2O 1wK = 1.00x10+14

NH3 + HOCl 4NH + OCl K = KaKb 1

wK = 63

4NH


Chapter 17-22

8–22

pH Titration Curves

• Titration curve: plot of pH of the solution as a function of the volume of base (acid) added to an acid (base).

• Sharp rise in curve is equivalence point.

• pH at equivalence point is 7.0 for SA but higher for WA.

Titration of 0.100 M HA with 0.100 M NaOH

0

2

4

68

10

12

14

0 10 20 30 40

Volume Base Added, mL

pH WASA

• Equivalence point can be used to determine the concentration of the titrant.E.g. the equivalence point for 15.00 mL of an acid occurred when 25.00 mL of 0.075 M NaOH was added. What was the molarity of the acid?


Chapter 17-23

8–23

SA–SB Titrations

• Base removes some acid and pH increases.• Let nb = moles of base added

na,r = moles of acid remainingna,r = na nb = CaVa CbVb

• Moles of hydronium ion same as moles of acid remaining. nH3O+ = na,r;

• Valid until very close to equivalence point. • Equivalence point(EP): pH = 7.00• Beyond EP: pH due only to base added (i.e. excess base). Use

total volume.• E.g. Determine pH of 10.0 mL of 0.100M HCl after addition of 5.00,

10.0 and 15.0mL of 0.100M NaOH.

V + V

VCVCV + V

n]OH[

ba

bbaa

ba

OH3

+3


Chapter 17-24

8–24

Titration of SB with SA

• Acid removes some of the base and pH is changed by amount of base removed.Let na = moles of acid added nb,r = moles of base remaining

nb,r = CbVb CaVa

• Moles of hydroxide ion same as moles of base remaining. nOH = nb,r;

– Valid until EP. • EP: pH = 7.00• Beyond EP: pH due only to excess acid. Use total volume.

E.g. Determine pH of 10.0 mL of 0.100M NaOH after addition of 5.00, 10.0 and 15.0mL of 0.100M HCl.

V + VVCVC

V + V

n]OH[

Baaabb

BaOH


Chapter 17-25

8–25

WA with SB Titration

• As above base removes some of the acid and pH is changed by amount of acid removed. Let nb = moles of base added

nHA = moles of acid remainingnHA = CHAVHA CbVb

nA = nb = CbVb

• Up to equivalence point moles of hydronium ions must be determined from equilibrium expression.

• Equivalence point: pH = pH of salt of WA• Beyond Equivalence point: Use amount of excess base to determine pH.• E.g. determine pH of 10.0 mL of 0.100M HA after addition of 5.00, 10.0

and 15.0mL of 0.100M NaOH. Ka = 1.75x105.

bbaa

bba

HA

Aa

VCVCVC

logpK

nnlogpKpH


Chapter 17-26

8–26

WB–SA Titrations

• Acid removes some of the base and decreases the pH. Let na = moles of acid added nb,r = moles of base remainingnb,r = CbVb CaVa nBH+ = na = CaVa

• Moles of hydroxide ions must be determined from equilibrium expression. Valid until EP.

• EP: pH = pH of salt of weak base.• Beyond EP: pH due only to presence of acid added after endpoint (i.e.

excess acid) as seen for strong base. Volume correction needed as above (total volume).E.g. Determine pH of 10.0 mL of 0.100M B after addition of 5.00, 10.0 and 15.0mL of 0.100M HCl. Kb = 1.75x105.

aabb

aab

B

BHb

VCVCVC

logpK

n

nlogpKpOH

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