Applications of aqueous Applications of aqueous equilibriaequilibria

NeutralizationNeutralizationCommon-Ion effectCommon-Ion effect

BuffersBuffersTitration curvesTitration curves

Solubility and KSolubility and Kspsp

Learning objectivesLearning objectives

Determine extent of neutralizationDetermine extent of neutralization Construct buffer solutionsConstruct buffer solutions Derive Henderson-Hasselbalch equationDerive Henderson-Hasselbalch equation Apply HH to calculate pH of buffer Apply HH to calculate pH of buffer

solutionssolutions Calculate pH titration curvesCalculate pH titration curves Write solubility product expressionsWrite solubility product expressions Identify factors that affect solubilityIdentify factors that affect solubility

NeutralizationNeutralization

Acid + Base Acid + Base → Salt + Water→ Salt + Water Extent Extent → → depends on type of acid and basedepends on type of acid and base What are the major species left when base What are the major species left when base

neutralizes an acid?neutralizes an acid? Four combinations:Four combinations:

Strong-strongStrong-strong Strong-weakStrong-weak Weak-strongWeak-strong Weak-weakWeak-weak

Strong-strongStrong-strong

Net ionic equationNet ionic equation

KKnn = 1/K = 1/Kww = 10 = 101414

Neutralization goes to completionNeutralization goes to completion NaNa++, Cl, Cl-- and H and H22OO

)()()( 2 aqNaClOHaqNaOHaqHCl

OHaqOHaqOH 23 2)()(

Manipulating equilibria expressionsManipulating equilibria expressions

If reaction A can be written as the If reaction A can be written as the sumsum of of reactions B + C + D +...+ N (reactions B + C + D +...+ N (ΣΣRRii))

Then K is Then K is productproduct over over ΠΠKKii

This approach is used routinely in solution This approach is used routinely in solution equilibria problemsequilibria problems

...A B C D NK K xK xK x xK

)()()( 2 aqAcOHaqOHaqHAc

Weak-strongWeak-strong Acetic acid is not completely ionizedAcetic acid is not completely ionized Net ionic equation:Net ionic equation:

What is K? Write KWhat is K? Write K in terms equilbria we know:in terms equilbria we know:

KK = 1.8x10= 1.8x10-5-5x1.0x10x1.0x101414 = 1.8x10 = 1.8x1099

Goes to completion – attraction of OHGoes to completion – attraction of OH-- for protons for protons NaNa++, Ac, Ac--, H, H22O (v. small amount OHO (v. small amount OH--))

532 108.1)...()()()(.1 xKaqAcaqOHaqOHaqHAc a

1423 100.1/1.....2)()(.2 xKOHaqOHaqOH w

9

3

3 108.1]][[

]][][[1 xOHAc

OHOHHAcKxKKw

a

Strong-weakStrong-weak Net ionic equation for neutralization of ammonia with HClNet ionic equation for neutralization of ammonia with HCl

What is K? Write in terms of known equilibria:What is K? Write in terms of known equilibria:

Add 1 and 2:Add 1 and 2:

KK1+21+2 = K = K11KK22 = = 1.8x101.8x10-5-5x1.0x10x1.0x101414 = 1.8x10 = 1.8x1099

KK1+21+2>>1 - Neutralization complete>>1 - Neutralization complete NHNH44

++, Cl, Cl--, H, H22O and v. small amount of HO and v. small amount of H33OO++

)()()( 4233 aqNHOHaqNHaqOH

5432 108.1)...()()()(.1 xKaqNHaqOHaqNHaqOH b

1423 100.1/1.....2)()(.2 xKOHaqOHaqOH w

42323 2.21 NHOHOHNHOHOHOH

Weak-weakWeak-weak

Net ionic equation for neutralization of acetic acid by ammonia Net ionic equation for neutralization of acetic acid by ammonia – neither ionized– neither ionized

Obtain KObtain Kn n from equilibria we knowfrom equilibria we know

KK1+2+31+2+3 = K = K11KK22KK33

= = 1.8x101.8x10-5-5x1.8x10x1.8x10-5-5x1.0x10x1.0x101414 = 3.2x10 = 3.2x1044

NHNH44++, Ac, Ac--, H, H22O (v. small amounts of HAc and NHO (v. small amounts of HAc and NH33))

)()()()( 43 aqAcaqNHaqNHaqHAc

532 108.1)...()()(.1 xKaqAcOHaqOHaqHAc a

5432 108.1)...()()()(.2 xKaqNHaqOHaqNHaqOH b

1423 100.1/1.....2)()(.3 xKOHaqOHaqOH w

Common ion effect - bufferingCommon ion effect - buffering

Solutions of Solutions of weak acid weak acid andand conjugate conjugate bases bases have important applications for have important applications for “buffering” pH – resisting change to pH “buffering” pH – resisting change to pH from added acid or base. (weak base and from added acid or base. (weak base and conjugate acid perform the same function)conjugate acid perform the same function)

The pH of operation will depend on the The pH of operation will depend on the dissociation constants for the particular dissociation constants for the particular acid (base).acid (base).

Use weak acid strategy for Use weak acid strategy for calculating pH in buffercalculating pH in buffer

Acetic acid and sodium acetate 0.1 M.Acetic acid and sodium acetate 0.1 M. Initial species are HAc, NaInitial species are HAc, Na++, , AcAc-- and H and H22OO

Two proton exchange reactions, but one Two proton exchange reactions, but one does not alter concentrationsdoes not alter concentrations

532 108.1)...()()( xKaqAcOHaqOHaqHAc a

1)...()()()( KaqHAcaqAcaqAcaqHAc

The Big Table of concentrationsThe Big Table of concentrations Determination of final concentrations in terms of initial Determination of final concentrations in terms of initial

concentrations concentrations Note difference between acid case and buffer case:Note difference between acid case and buffer case:

[Ac[Ac--]]ii is 0 with HAc only is 0 with HAc only

[Ac[Ac--]]ii is > 0 in the buffer is > 0 in the buffer

Principal Principal processprocess

HAc(aq) + HHAc(aq) + H22O = HO = H33OO++(aq) + Ac(aq) + Ac--(aq)(aq)

Initial concInitial conc 0.100.10 00 0.100.10

ChangeChange -x-x xx xx

Equilibrium Equilibrium concconc

0.10 - x0.10 - x xx 0.100.10 + x + x

Solving for xSolving for x

Put concentrations into expression for KPut concentrations into expression for Kaa

0.10 – x 0.10 – x ≈ 0.10 ≈ 0.10 + x≈ 0.10 ≈ 0.10 + x x = Kx = Kaa = 1.8 x10 = 1.8 x10-5 -5 M (life is good)M (life is good) pH = 4.74pH = 4.74 Note: when [HAc] = [AcNote: when [HAc] = [Ac--], pH = pKa], pH = pKa In In allall buffers pH = pK buffers pH = pKaa when [HB] = [B when [HB] = [B--]]

53 108.1)10.0(

)10.0(

][

]][[

xx

xx

HAc

AcOHKa

X << 0.1

Common-ion effect – Le Chatelier Common-ion effect – Le Chatelier in actionin action

Without added acetate ion the pH of 0.10 Without added acetate ion the pH of 0.10 M acetic acid is 2.89.M acetic acid is 2.89.

Addition of AcAddition of Ac-- causes [H causes [H33OO++] to decrease] to decrease

Consequence of Le Chatelier: increasing Consequence of Le Chatelier: increasing [product] equilibrium goes to reactants[product] equilibrium goes to reactants

pH changes in different bufferspH changes in different buffers

)()()( 32 aqAcOHaqOHaqHAc

Modulating pH in acetic acid by Modulating pH in acetic acid by adding acetate ionadding acetate ion

Do buffers really work?Do buffers really work?

Compare effect of adding OHCompare effect of adding OH-- to solution to solution of given pHof given pH

1.1. Buffer solutionBuffer solution

2.2. Strong acid at same initial pHStrong acid at same initial pH

Adding base to bufferAdding base to buffer

Analytical proof in HAc/AcAnalytical proof in HAc/Ac-- system system

Initial pH = 4.74Initial pH = 4.74 What happens when 0.01 mol of NaOH is What happens when 0.01 mol of NaOH is

added to I Ladded to I L

Principal Principal processprocess

HAc(aq) + HHAc(aq) + H22O = HO = H33OO++(aq) + Ac(aq) + Ac--(aq)(aq)

Equilibrium Equilibrium conc/Mconc/M

0.10 - x0.10 - x xx 0.10 + x0.10 + x

What happens when 0.01 mol of What happens when 0.01 mol of NaOH is added?NaOH is added?

Addition of OHAddition of OH- - causes conversion of HAc into causes conversion of HAc into AcAc--

Neutralization Neutralization processprocess

HAc(aq) + OHHAc(aq) + OH-- = H = H22OO (aq) + Ac(aq) + Ac--(aq)(aq)

Initial concInitial conc 0.100.10 0.010.01 0.100.10

ChangeChange -0.01-0.01 -0.01-0.01 0.01

Equilibrium Equilibrium concconc

0.090.09 0 0.110.11

After addition of OHAfter addition of OH--, recompute [H, recompute [H33OO++]]

[H[H33OO++] = 1.8x10] = 1.8x10-5-5x0.09/0.11 = 1.5x10x0.09/0.11 = 1.5x10-5-5 M M

pH = 4.82 – a change of + 0.08 pH unitspH = 4.82 – a change of + 0.08 pH units

][

][][

][

]][[

3

3

Ac

HAcKOH

HAc

AcOHK

a

a

Same deal when adding acidSame deal when adding acid

Almost all the added HAlmost all the added H++ is converted into is converted into HAcHAc

In same way as with base, new [HAc] = In same way as with base, new [HAc] = 0.11 M and new [Ac0.11 M and new [Ac--] = 0.09 M] = 0.09 M

[H[H33OO++] = 1.8x10] = 1.8x10-5-5x0.11/0.09 = 2.2 x 10x0.11/0.09 = 2.2 x 10-5 -5 MM

pH = 4.66 – a change of -0.08 pH unitspH = 4.66 – a change of -0.08 pH units

)()()( 23 aqHAcOHaqOHaqAc

Adding OHAdding OH-- to strong acid at pH = to strong acid at pH = 4.744.74

What is change in pH if added to a What is change in pH if added to a solution of HCl at pH 4.74?solution of HCl at pH 4.74?

[H[H++] = 1.8 x 10] = 1.8 x 10-5-5 M M HCl removes 1.8x10HCl removes 1.8x10-5-5 mol of OH mol of OH--. . Excess OHExcess OH- - = 0.01 - 1.8x10= 0.01 - 1.8x10-5-5 mol of OH mol of OH- - ≈ ≈

0.01 mol.0.01 mol. New pH = 12 – a change of 7 pH unitsNew pH = 12 – a change of 7 pH units

In buffer at same initial pH, change was only In buffer at same initial pH, change was only 0.08 units0.08 units

Buffer capacityBuffer capacity

A measure of the amount of acid or base A measure of the amount of acid or base that a buffer solution can absorb before a that a buffer solution can absorb before a significant change in pH occurs.significant change in pH occurs.

Or – the amount of acid or base required Or – the amount of acid or base required to yield a given change in pH.to yield a given change in pH.

Messin’ with buffersMessin’ with buffers

][

][log

][

][loglog][log

][

][][

310

3

Acid

BasepKpH

Base

AcidKOHpH

Base

AcidKOH

a

a

a

Henderson-Hasselbalch:Henderson-Hasselbalch:Send the pain belowSend the pain below

Buffer calculation short-cutBuffer calculation short-cut DerivationDerivation

Assumes that [HAssumes that [H++] << [HA]] << [HA]

)(

)(

][

]][[ 3

xA

xBx

A

BOHKa

ApplicationsApplications

Able to predict percent dissociation of an Able to predict percent dissociation of an acid from the difference between pH and acid from the difference between pH and pKpKaa

Calculate relative quantities of acid and Calculate relative quantities of acid and conjugate base required to achieve a conjugate base required to achieve a given pH (pKgiven pH (pKaa must be reasonably close to must be reasonably close to

pH)pH)

][

][log

Acid

BasepKpH a

The glass half-neuteredThe glass half-neutered

Instead of using a source of weak acid and Instead of using a source of weak acid and a conjugate base, take a weak acid and a conjugate base, take a weak acid and neutralize with strong base to make the neutralize with strong base to make the conjugate base. (HAc and NaOH).conjugate base. (HAc and NaOH).

When When halfhalf the HAc is neutralized [HAc] = the HAc is neutralized [HAc] = [Ac[Ac--]]

pKa = pHpKa = pH Useful buffers can be made in pH range of Useful buffers can be made in pH range of

pKa pKa ± 1 pH units± 1 pH units