Aqueous Equilibria

The __________________________ 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

A buffer solution is a solution of:

1. and

2.

Both must be present! (and should NOT neutralize EACH OTHER!!!!)

A buffer solution has the ability to ________________ upon the addition of small amounts of either acid or base.

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)

(b)

(c)

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)

Mixture of weak acid and conjugate base!

OR…… Use the Henderson-Hasselbalch 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-Hasselbalch equation

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)

Common ion effect pH = pKa + log [HCOO-][HCOOH]

Mixture of weak acid and conjugate base!

HCl H+ + Cl-

HCl + CH3COO- CH3COOH + Cl-

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

[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

What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?

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

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+]

Equilibrium Calculation NH4+ (aq) H+ (aq) + NH3 (aq)

= 9.20

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?

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

Chemistry In Action: Maintaining the pH of Blood

TitrationsIn a __________________ 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.

________________ – 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)

Titration Curve

A titration curve is a plot of _________________ ____________________.

Typically the titrant is a strong (completely) dissociated acid or base.

Such curves are useful for determining endpoints and dissociation constants of weak acids or bases.

Features of the Strong Acid-Strong Base Titration Curve

1. The pH starts out _________, reflecting the high [H3O+] of the strong acid and increases gradually as acid is neutralized by the added base.

2. Suddenly the pH ________________. This occurs in the immediate vicinity of the ____________________. For this type of titration the pH is _____ at the equivalence point.

3. Beyond this steep portion, the pH ______________ as more base is added.

Sample Calculation: Strong Acid-Strong Base Titration Curve

Consider the titration of 40.0 mL of 0.100 M HCl with 0.100 M NaOH.

Region 1. Before the equivalence point, after adding 20.0 mL of 0.100 M NaOH. (Half way to the equivalence point.)

Initial moles of H3O+ = 0.0400 L x 0.100 M = 0.00400 mol H3O+

- Moles of OH- added = 0.0200 L x 0.100 M =0.00200 mol OH-

base added of volume acid of volumeoriginal

remaining OH of (mol)amount ]O[H 3

3

Sample Calculation: Strong Acid-Strong Base Titration Curve (Cont. I)

Region 2. At the equivalence point, after adding 40.0 mL of 0.100 M NaOH.

Initial moles of H3O+ = 0.0400 L x 0.100 M = 0.00400 mol H3O+

- Moles of OH- added = 0.0400 L x 0.100 M =0.00400 mol OH-

base added of volume acid of volumeoriginal

remaining OH of (mol)amount ]O[H 3

3

An equal number of moles of NaOH and HCl have reacted, leaving only a solution of their salt (NaCl). The pH=7.00 because the cation of the strong base and the anion of the strong acid have no effect on pH.

Sample Calculation: Strong Acid-Strong Base Titration Curve (cont. II)

Region 3. After the equivalence point, after adding 50.0 mL of 0.100 M NaOH. (Now calculate excess OH-)

Total moles of OH- = 0.0500 L x 0.100 M = 0.00500 mol OH- -Moles of H3O+ consumed = 0.0400 L x 0.100 M =0.00400 mol

base added of volume acid of volumeoriginal

remaining OH of (mol)amount ][OH

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):

The four Major Differences Between a Strong Acid-Strong Base Titration Curve and a Weak

Acid-Strong Base Titration Curve

1. The initial pH is ___________.

2. A gradually rising portion of the curve, called the ____________, appears before the steep rise to the equivalence point.

3. The pH at the equivalence point is _____ ___________________

4. The steep rise interval is less pronounced.

Strong Acid-Weak Base Titrations

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

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

At equivalence point (pH < 7):

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

The four Major Differences Between a Weak Acid-Strong Base Titration Curve and a Weak

Base-Strong Acid Titration Curve

1. The initial pH is ______________.

2. A gradually decreasing portion of the curve, called the buffer region, appears before a steep fall to the equivalence point.

3. The pH at the equivalence point is _______ ____________.

4. Thereafter, the pH _______________ as excess strong acid is added.

Features of the Titration of a Polyprotic Acid with a Strong Base

1. The loss of each mole of H+ shows up as separate equivalence point (but only if the two pKas are separated by more than 3 pK units).

2. The pH at the midpoint of the buffer region is equal to the _______ of that acid species.

3. The same volume of added base is required to remove each mole of H+.

Acid-Base Indicators

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

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)

Equilibrium Calculation 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-]

Solubility Equilibria

• Looking at dissolution or precipitation reactions of ionic compounds

• Ksp (solubility product) is the equilibrium constant for the equilibrium reaction between a solid and its saturated ions– Indicates

• ________, _______ and ________ do not appear in the equilibrium equation (just like before)

Solubility and Ksp

• Solubility is how much actually dissolves while Ksp is the equilibrium constant.– ____, __________, and the presence of ________

________________in solution affect solubility– Only ________________ affects Ksp

• Can use Ksp to calculate solubility but need to be careful…not always correct!

Factors that Affect Solubility(4)

1. Common Ion Effect• presence of other ions ________________• presence of 2nd solute that gives a common ion

___________________

2. pH• dissolving increases when pH is made __________(more of the

basic ion X- will be made to react with extra H+ ions)

• the more basic the anion, the more its solubility is ________________________

3. Formation of a Complex Ion4. Amphoterism

Complex Ion Equilibria and Solubility

A complex ion is an ion containing a central ____________ bonded to ________________________________.

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

Co(H2O)62+ CoCl4

2-

Where they “smush “ it all together

•Metal ions can act as _________

•Can react with water (a Lewis base)

•Can react with other Lewis bases (must be able interact more than water can)

•Keq becomes Kf (formation constant)

•Kf’s size shows stability of complex ion in solution

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.

Factors that Affect Solubility (cont.)

4. Amphoterism• Some insoluble, metal hydroxides will dissolve in

strong acids or strong bases- will dissolve in acid because of __________________- Will dissolve in base because a __________________

• Solubility varies with the metal bonded

Precipitation & Separation of Ions

Q Ksp get precipitation until Q = Ksp

Q Ksp at equilibrium

Q Ksp solid dissolves until Q = Ksp

• Selective Precipitation of ions– Separate ions in an aqueous solution using a

reagent that forms a precipitate 1 or few ions in solution