Acid Base Equilibria 2

Acid-Base Equilibria

Chapter 16

The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.

The presence of a common ion suppresses the ionization of a weak acid or a weak base.

Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).

CH3COONa (s) Na+ (aq) + CH3COO- (aq)

CH3COOH (aq) H+ (aq) + CH3COO- (aq)

common ion

16.2

A buffer solution is a solution of:

1. A weak acid or a weak base and

2. The salt of the weak acid or weak base

Both must be present!

A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.

16.3

CH3COOH (aq) H+ (aq) + CH3COO- (aq)

Consider an equal molar mixture of CH3COOH and CH3COONa

Adding more acid creates a shift left IF enough acetate ions are present

Which of the following are buffer systems? (a) KF/HF (b) KCl/HCl, (c) Na2CO3/NaHCO3

(a) KF is a weak acid and F- is its conjugate basebuffer solution

(b) HCl is a strong acidnot a buffer solution

(c) CO32- is a weak base and HCO3

- is it conjugate acidbuffer solution

16.3

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?

HCOOH (aq) H+ (aq) + HCOO- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.30 0.00

-x +x

0.30 - x

0.52

+x

x 0.52 + x

16.2

Mixture of weak acid and conjugate base!

KKaa for HCOOH = 1.8 x 10 for HCOOH = 1.8 x 10 -4-4

[H+] [HCOO-]Ka = [HCOOH]

x = 1.038 X 10 -4

pH = 3.98

OR…… Use the Henderson-Hasselbach equationConsider mixture of salt NaA and weak acid HA.

HA (aq) H+ (aq) + A- (aq)

NaA (s) Na+ (aq) + A- (aq)Ka =

[H+][A-][HA]

[H+] =Ka [HA]

[A-]

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

[A-]

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

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

pKa = -log Ka

Henderson-Hasselbach equation

16.2

pH = pKa + log[conjugate base]

[acid]

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?

HCOOH (aq) H+ (aq) + HCOO- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.30 0.00

-x +x

0.30 - x

0.52

+x

x 0.52 + x

Common ion effect

0.30 – x 0.30

0.52 + x 0.52

pH = pKa + log [HCOO-][HCOOH]

HCOOH pKa = 3.77

pH = 3.77 + log[0.52][0.30]

= 4.01

16.2

Mixture of weak acid and conjugate base!

HCl H+ + Cl-

HCl + CH3COO- CH3COOH + Cl-

16.3

pH= 9.18

Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?

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

Initial

End

0.30 0.36 0

0.30 - x 0.36 + x x

16.3

[NH4+] [OH-]

[NH3]Kb =

Change - x + x + x

= 1.8 X 10-5

(.36 + x)(x)

(.30 – x)

1.8 X 10-5 =

1.8 X 10-5 0.36x

0.30

x = 1.5 X 10-5 pOH = 4.82

pH = 9.20


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

start (M)

end (M)

0.29 0.01 0.24

0.28 0.0 0.25

final volume = 80.0 mL + 20.0 mL = 100 mL

16.3

NH4+ 0.36 M x 0.080 L = 0.029 mol / .1 L = 0.29 M

OH- 0.050 x 0.020 L = 0.001 mol / .1 L = 0.01M

NH3 0.30 M x 0.080 = 0.024 mol / .1 L = 0.24M

Ka= = 5.6 X 10-10[H+] [NH3]

[NH4+]

= 5.6 X 10-10[H+] 0.25

0.28

[H+] = 6.27 X 10 -10

= 9.20


NH4+ (aq) H+ (aq) + NH3 (aq)

pH = pKa + log[NH3][NH4

+]pKa = 9.25 pH = 9.25 + log

[0.30][0.36]

= 9.17

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

start (M)

end (M)

0.29 0.01 0.24

0.28 0.0 0.25

pH = 9.25 + log[0.25][0.28]

final volume = 80.0 mL + 20.0 mL = 100 mL

16.3

Chemistry In Action: Maintaining the pH of Blood

16.3

TitrationsIn a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

Equivalence point – the point at which the reaction is complete

Indicator – substance that changes color at the endpoint (hopefully close to the equivalence point)

Slowly add baseto unknown acid

UNTIL

The indicatorchanges color

(pink) 4.7

Strong Acid-Strong Base Titrations

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

OH- (aq) + H+ (aq) H2O (l)

16.4

100% ionization!

No equilibrium

Weak Acid-Strong Base Titrations

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

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

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

At equivalence point (pH > 7):

16.4

Strong Acid-Weak Base Titrations

HCl (aq) + NH3 (aq) NH4Cl (aq)

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

At equivalence point (pH < 7):

16.4

H+ (aq) + NH3 (aq) NH4Cl (aq)

Acid-Base Indicators

16.5

pH

16.5

The titration curve of a strong acid with a strong base.

16.5

Which indicator(s) would you use for a titration of HNO2 with KOH ?

Weak acid titrated with strong base.

At equivalence point, will have conjugate base of weak acid.

At equivalence point, pH > 7

Use cresol red or phenolphthalein

16.5

Finding the Equivalence Point(calculation method)

• Strong Acid vs. Strong Base– 100 % ionized! pH = 7 No equilibrium!

• Weak Acid vs. Strong Base– Acid is neutralized; Need Kb for conjugate base

equilibrium• Strong Acid vs. Weak Base

– Base is neutralized; Need Ka for conjugate acid equilibrium

• Weak Acid vs. Weak Base– Depends on the strength of both; could be

conjugate acid, conjugate base, or pH 7

Exactly 100 mL of 0.10 M HNO2 are titrated with 100 mL of a 0.10 M NaOH solution. What is the pH at the equivalence point ?

HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l)

start (moles)

end (moles)

0.01 0.01

0.0 0.0 0.01

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.05 0.00

-x +x

0.05 - x

0.00

+x

x x

[NO2-] =

0.010.200 = 0.05 MFinal volume = 200 mL

Kb =[OH-][HNO2]

[NO2-]

=x2

0.05-x= 2.2 x 10-11

0.05 – x 0.05 x 1.05 x 10-6 = [OH-]

pOH = 5.98

pH = 14 – pOH = 8.02

Complex Ion Equilibria and Solubility

A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions.

Co2+ (aq) + 4Cl- (aq) CoCl4 (aq)2-

Co(H2O)62+ CoCl4

2-

16.10

16.10

Complex Ion Formation

• These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions).

• As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH3)4

2+ (ammonia is used as a test for Cu2+ ions), and Ag(NH3)2

+.• Memorize the common ligands.

Common LigandsLigands Names used in the ion

H2O aqua

NH3 ammine

OH- hydroxy

Cl- chloro

Br- bromo

CN- cyano

SCN- thiocyanato (bonded through sulphur) isothiocyanato (bonded through nitrogen)

Names• Names: ligand first, then cation

Examples:– tetraamminecopper(II) ion: Cu(NH3)4

2+

– diamminesilver(I) ion: Ag(NH3)2+.

– tetrahydroxyzinc(II) ion: Zn(OH)4 2-

• The charge is the sum of the parts (2+) + 4(-1)= -2.

When Complexes Form• Aluminum also forms complex ions as do some post transitions

metals. Ex: Al(H2O)63+

• Transitional metals, such as Iron, Zinc and Chromium, can form complex ions.

• The odd complex ion, FeSCN2+, shows up once in a while• Acid-base reactions may change NH3 into NH4

+ (or vice versa) which will alter its ability to act as a ligand.

• Visually, a precipitate may go back into solution as a complex ion is formed. For example, Cu2+ + a little NH4OH will form the light blue precipitate, Cu(OH)2. With excess ammonia, the complex, Cu(NH3)4

2+, forms.• Keywords such as "excess" and "concentrated" of any solution

may indicate complex ions. AgNO3 + HCl forms the white precipitate, AgCl. With excess, concentrated HCl, the complex ion, AgCl2-, forms and the solution clears.

Coordination Number

• Total number of bonds from the ligands to the metal atom.

• Coordination numbers generally range between 2 and 12, with 4 (tetracoordinate) and 6 (hexacoordinate) being the most common.

Some Coordination Complexes

molecular formula

Lewis base/ligand

Lewis acid donor atom

coordination number

Ag(NH3)2+ NH3 Ag+ N 2

[Zn(CN)4]2- CN- Zn2+ C 4

[Ni(CN)4]2- CN- Ni2+ C 4

[PtCl6] 2- Cl- Pt4+ Cl 6

[Ni(NH3)6]2+ NH3 Ni2+ N 6

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Acid Base Equilibria 2