1

Acids, Bases and the Common Ion Effect

HF + H2O H3O+ + F-

H+

+Cl-

Consider the following acid equilibrium of a weak acid:Ka = [H3O+][F-]

[HF]

HCl(aq)

What happens when we add some strong acid to the mixture?

HCl completely dissociates, adding free H3O+ to solution. This is a common ion to the weak acid equilibrium.

By LeChatelier’sprinciple, we predict the HF dissociation should be driven left, suppressing the dissociation.

More quantitativea) Determine [F-] in a solution of 0.500M HF. b) Determine [F-] in a solution of 0.500M HF and 0.10M HCl.KHF = 6.6x10-4.

[ ] [ ]H F K K K CHF HF HF HF+ −= =− + +2 4

2

a) A solution of a weak acid. Let’s use the quadratic.

[F-] = 0.0178M

2

Continued

Initialweak acid 0.50 0 0strong acid 0 0.10 0

Change -x + x + xEquilib 0.50-x 0.10 +x x

HF + H2O H3O+ + F-

b) Use ICE table. Let HF and HCl dissociate separately.

Ka = [H3O+][F-] [HF]

Ka = (0.10 +x)(x) / (0.50-x)

Assumption: ionization of HF is suppressed. Assume x<<0.1; x<<0.5

Ka ~ (0.10)(x) / (0.50)

ContinuedRearrange: x = [F-] = (0.50M)Ka/ 0.10M = 0.0033 M

Comparison a) no common ion, [F-] = 0.0178 Mb) common ion, [F-] = 0.0033 M

5 X less!

This is one example of the common ion effect.

When a strong acid supplies the common ion, H3O+, the ionization of a weak acid is suppressed.

When a strong base supplies the common ion, OH-, the ionization of a weak base is suppressed.

3

Buffering and titrations!16_12

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12pH

5 10 15 20 25 30 35 40 45 50Volume of NaOH added (mL)

Volume NaOHadded (mL)

012345

1015202122232425262728293035404550

2.923.473.793.984.134.254.675.035.455.575.725.916.238.78

11.2911.5911.7511.8711.9612.2212.3612.4612.52

pH

Colorless

Phenolphthalein

Equivalence point

0

Yellow

Pink

Bromcresol green

Blue

Example: any chemical system in equilibrium

aA + bB + cC dD + eE + fF

K [D] [E] [F][A] [B] [C]eq

d e f

a b c=

aA + + cC dD + eE + fF

K [D] [E] [F][A] [B] [C]eq

d e f

a b c=

aA + bB + cC dD + eE + fF

K [D] [E] [F][A] [B] [C]eq

d e f

a b c=

Acid/Base buffer – a system that resists changes to pH caused by addition of excess base or acid.

Chemical buffer – a chemical system that resists change

4

Acid Base Buffer SystemsA mixture of a weak acid and it’s conjugate base

O

O

H O

O

Benzoic acid benzoate

CH3

OHO

acetic acid acetate

CH3

O O

A mixture of a weak base and it’s conjugate acidNCH3

CH3 CH3 NH

+

CH3

CH3 CH3

trimethylamine trimethylammonium

Add as sodium salts for example

Add as chloride salt for example

An acid/base buffer system

K [H ][A ][HA]a =+ −

What happens if we add acid and conjugate base in equal concentrations? i.e. equal moles of HA and NaA

HA + H2O H3O+ + A-

Initialweak acid 0.10 0 0conj base 0 0 0.10

Change -x +x + xEquilib 0.10-x +x 0.10+x

K [H ][0.10 + x][0.10- x]a =+

But due to presence of A-, ionization of HA is suppressed more than usual. x<<0.10

[H+] = Ka

pH = pKa!

5

Henderson-Hasselbalch Equation

K [H ][A ][HA]a =+ −

Solve for [H+]

-log [H+] = -log Ka –log [HA] + log [A-]

[H ] K [HA][A ]a+

−=

Take –log of both sides

-log [H+] = -log Ka + log [A-]/[HA]

pH pK log [A ][HA]a

-= + pH pK log

CCa

AHA

-= +

From previous ICE table, we see: [HA] = CHA – x[A-] = CA

- + x But in general, CHA = CA

-. If x<<CHA

Buffer ratiosFor correct buffering, there should be significant amounts of HA and A-.Just as importantly, the ratio should be such that:

0.10 CC

10.0HA

A≤ ≤

−

pH pK 1buffer a= ±

Substituting this constraint into the H-H equation

6

How to prepare a buffer1. What pHbuffer do you require?

2. Find an acid/conj base system from tables that has pKa within 1 unit of the required pH.

3. Use the H-H equation to determine the exact ratio of the acid and the base you will need.

4. What buffer capacity do you need…0.1M, 0.01M, 0.001M? Answer depends on how much acid or base could potentially impact the system that you are trying to buffer.

5. Prepare solutions taking into account 3) and 4).

ExamplePrepare 1L of buffer with pH = 3.5, with a total concentration, [A-] + [HA] = 0.01M.

2). Pick buffer system.

Acid pKa

Citric Acid 3.13Benzoic acid 4.20Acetic acid 4.77Carbonic acid 6.36Ammonium ion 9.25Phenol 9.89

7

Continued

pH pK logCC

logCC

pH pKa

CC

[A ][HA]

= 10

CC

10 2.34

aAHA

AHA

AHA

-(pH pKa)

AHA

(3.5 3.13)

-

-

-

-

= +

= −

≈

= =

−

−

3) Determine ratio of [A-]/[HA] needed. Use either H-H equation or equilibrium expression.

Cont’d5) Prepare solution (i.e. What masses or volumes are needed?) CA-/CHA = 2.34 (1)AlsoCA- + CHA = 0.01M (2)

from (1) CA- = 2.34 CHA ,substitute this into (2)

2.34 CHA + CHA = 0.01M CHA = 0.003 M

CA- = 2.34 CHA CA- = 0.007M

What substances, and quantities, are needed to make these solutions?

8

cont’dCitric acid, C6H8O7 MW = 192.1g/molemonosodium citrate NaC6H7O7 MW = 214.1g/mole

Since we are preparing 1L, we need to add 0.003 moles of citric acid and 0.007 moles of monosodium citrate.

massHA = 0.003 mol x 192.1 g/mol = 0.576g

massNaA = 0.007 mol x 214.1 g/mol = 1.50g

Dissolve to make 1L of solution.OR

Dissolve 0.01mol x 192.1g/mol = 1.92g of citric acid in about 1L of water, and neutralize with NaOH until pH = 3.5! (Total volume must be 1L.)

Examples of Buffers in NaturepH of blood is 7.40+0.05. pH is maintained by a series of buffer systems, H2CO3/HCO3

-, phosphate and proteins. pH regulation in the body is critical since functioning of enzymes is highly pH dependent.

Alkalosis – raising of pH results from hyperventilation, or exposure to high elevations (altitude sickness).

Acidosis – lowering of pH in blood by organ failure, diabetes or long term protein diet.

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What causes alkalosis at high elevation?…a

multiple equilibrium hypothesis.

The bicarbonatebuffer is essential

for controllingblood pH

CO2(aq) + H2O H2CO3(aq)

H2CO3(aq) + H2O H3O+ + HCO3-

CO2(g)

[HCO ][H CO ]

~ 1032 3

−

pKa = 6.4+

OH-

2 H2O

pH increases !

in the lungs

in the blood

At high elevation,P drops,

hyperventilation!

Acid Base TitrationsAcid base titrations are examples of volumetric techniques used to analyze the quantity of acid or base in an unknown sample.

Acid + Base H2O + salt

This is done by detecting the point at which we have added an equal number of equivalents of base to the acid. This is the equivalence point. For neutralization of an unknown monoprotic acid (A) with a base (B), we have at equivalence:

moles of acid = moles of base CAVA = CBVB

We detect the equivalence point with a pH meter or by identifying the end-point with an acid base indicator.

A

equivBA V

VCC =

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Acid Base Titrations - some terminologyequivalence point: the point at which

moles of acid = moles of baseend-point: the experimental approximation of

the equivalence pointtitration error: the difference between the

equivalence point and the end-pointtitrant: the solution that is added in a

measured quantitystandard solution: a solution of known

concentration

IndicatorsAcid-base indicators are highly coloured weak acids or

bases.HIndic Indic- + H+

colour 1 colour 2Colour transition occurs for 0.1< [Indic-]/[HIndic]<10

They may have more than one colour transition. Example. Thymol blue

Red Yellow BlueOne of the forms may be colourless - phenolphthalein

(colourless to pink)

1pK[HIn]

][InlogpKpH inda

indatransition ±=+=

−

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IndicatorsSelection of an indicator.

The colour transition range for an indicator is pKa + 1. Choose an indicator that has a pKa close to the pH at the equivalence point.

phenolphthalein

Indicator Examples

bromthymol blue

methyl red

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Acid Base Titrations16_11

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0 5 10 15 20 25 30 35 40 45 50

Volume of NaOH added (mL)

Volume NaOHadded (mL)

05

1015202122232425262728293035404550

1.001.181.371.601.952.062.202.382.697.00

11.2911.5911.7511.8711.9612.2212.3612.4612.52

pH

Pink

ColorlessPhenolphthalein

Equivalence point

Bromcresol green

Blue

Yellow

Titration of strong acid with strong base

Low initial pH, large pH change at equivalence point

ExampleWhat is the pH at 0mL, 10mL, 25.0mL and 35mL in titration of 25mL of 0.100 M strong acid (HCl) with 0.100M NaOH?

0 mL [H+] = CA = 0.100M pH = -log (0.1) pH = 1.00

10mL (all calculated in moles)HCl(aq) + NaOH (aq) H2O + NaCl(aq)

Initialstrong acid 0.0025 0 0 0strong base 0 0.001Change -0.0010 -0.001 0.001 0.001Equilib 0.0015 ~0 0.001 0.001

[H+] = mol H+ /VT = 0.0015 /0.035L = 0.0429 pH = 1.37

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Cont’dIn general, before the equivalence point:

[H+] = (CAVA – CBVB) / (VA+VB)

25.0 mL moles acid = moles base(i.e. this is the equivalence point)

all acid is neutralized… no excess baseWe are left with pure NaCl(aq) solution, pH = 7.00

35.0mL pH is determined by amount of excess base

[OH-] = (CBVB – CAVA) / (VA+VB) = {(.1M)(.035L)-(.1M)(.025L)}/{.025+.035L} = 0.0167M

pH = 14 – pOH = 14 – (-log [OH-]) = 14 -1.78 pH = 12.22

Titration of Weak Acid with Strong Base16_12

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pH

5 10 15 20 25 30 35 40 45 50Volume of NaOH added (mL)

Volume NaOHadded (mL)

012345

1015202122232425262728293035404550

2.923.473.793.984.134.254.675.035.455.575.725.916.238.78

11.2911.5911.7511.8711.9612.2212.3612.4612.52

pH

Colorless

Phenolphthalein

Equivalence point

0

Yellow

Pink

Bromcresol green

Blue

moderate initial pH, buffer region, basic equivalence point, change in pH not as great as with strong acid strong base

1

43

2

14

Titration of Weak Acid with Strong Base1. Initial Point – calculate pH of solution of a weak acid.

(Use approximation for weak acid or solve quadratic.)

2. Buffer Region - Use H-H equation to determine pH.moles HA = initial moles HA – moles OH- addedmoles A- = moles OH- added

3. Equivalence Point- a pure solution of a weak base, A-

moles A- = moles HA we started with(Use approximation for weak base or solve quadratic.)

4. Excess base (region 4) – pH is determine solely by the amount of excess strong base added.(See Strong Acid/Strong Base example.)

ExampleWhat is the pH at the equivalence point in the titration of 50 mL of a 0.1M solution of acetic acid with 0.1M NaOH?Ka(acetic acid) = 1.8 x10-5

What is the volume of NaOH at the equivalence point?

For a monoprotic acid, at the equivalence point, moles of acid = moles of base

CAVA = CNaOHVNaOH

Vequiv=VNaOH=CAVA/CNaOH=(0.100M)(0.050L)/(0.100M) = 50mL

At equivalence point, [conj. base]= CAVA/VTOTAL = [A-]

[A-], = (0.100M)(0.050L)/(0.050+0.050L) = 0.05M

15

Cont’dSo, at the equivalence point, what is the pH of a 0.05M solution of CH3COO- Na+?

[OH ] C K C KK

(0.05M) (1x10 )1.8x10

5.27x10B BB w

A

-14

-56− −= = = =

pH = 14 – pOH = 14 – (-log (5.27 x 10-6)) pH = 8.7215_8

0

yellow violetMethyl violet

Thymol blue (acidic range)

Bromphenol blue

Methyl orange

Bromcresol green

Methyl red

Bromthymol blue

Thymol blue (basic range)

Phenolphthalein

Alizarin yellow R

red yellow

yellow blue

red yellow

yellow blue

red yellow

yellow blue

yellow blue

colorless pink

yellow red

2 4 6pH range for color changeIndicator name

8 10 12 Which indicator would we use?

pKa(indicator) = pHequivalence + 1