CO(g) + H2O(g) CO2(g) + H2(g)

CO(g) + H2O(g) CO2(g) + H2(g)

CO(g) + H2O(g) CO2(g) + H2(g)

K >>1 Forward rxn dominates (rxn lies to the right). Mostly products at equilibrium, [products] >> [reactants]

Cl2(g) + 2 NO(g) 2 NOCl(g) K = 6.2 x 104

K <<1 Reverse rxn dominates (rxn lies to

the left). Mostly reactants at equilibrium,

[products] << [reactants]

COCl2(g) CO(g) + Cl2(g) K = 2.0 x 10-10

H2(g) + F2(g) 2 HF(g) K = 10

K 1 Forward and reverse rxn occur to roughly

the same extent, [products] [reactants]

2.0 moles of NH3 gas are introduced into a previously

evacuated 1.0 L container. At a certain temperature the

NH3 partially dissociates by the following equation.

At equilibrium 1.0 mol of NH3 remains. Calculate the

equilibrium constant for this reaction.

2 NH3(g) N2(g) + 3 H2(g)

Reaction Quotient (Q)

Q = K: The rxn is at equilibrium. No shift.

Q < K: The rxn shifts right to produce

products to increase Q.

Q > K: The rxn shifts left to produce

reactants to decrease Q.

If a change (stress) is imposed on a system

at equilibrium, the position of the equilibrium

will shift in a direction that tends to reduce

that change (stress).

Le Chatelier’s Principle

Le Chatelier’s Principle

1. SO2(g) is removed.

2. O2(g) is added.

3. SO3(g) is added.4. The volume of the reaction container is

halved.5. An inert gas like Ar is added.6. A catalyst is added.7. Temperature is increased.

2 SO2(g) + O2(g) 2 SO3(g) H = 198 kJ

Le Chatelier’s Principle

1. CO2(g) is added.

2. CaCO3(s) is added.

3. The volume is increased.

4. The temperature is decreased.

CaCO3(s) CaO(s) + CO2(g) H = 556 kJ

N2O4(g) 2 NO2(g)

1.0 mol of N2O4(g) is placed in a 10.0 L

vessel and then reacts to reach equilibrium.

Calculate the equilibrium concentrations of

N2O4 and NO2. K = 4.0 x 10-7

Brønsted-Lowry Model

Acids – are proton donors

Bases – are proton acceptors

HC2H3O2 is a stronger acid then HCN which

Has the stronger conjugate base?

Comments on the Conjugates of Acids and Bases.

• The weaker the acid the stronger its conjugate base.

• The weaker the base the stronger its conjugate acid.

• The conjugate base of a weak acid is a WEAK base.

• The conjugate base of a strong acid is worthless.

• The conjugate acid of a weak base is a WEAK acid.

Acid Ka Realative acid

strength

Conjugate base

Kb Relative base

strength

HCl ~106

HF 7.2 x 10-4

HC2H3O2 1.8 x 10-5

HOCl 3.5 x 10-8

NH4+ 5.6 x 10-10

Acid Ka Realative acid

strength

Conjugate base

Kb Relative base

strength

HCl ~106 Cl- ~10-20

HF 7.2 x 10-4 F- 1.4 x 10-11

HC2H3O2 1.8 x 10-5 C2H3O2- 5.6 x 10-10

HOCl 3.5 x 10-8 OCl- 2.9 x 10-7

NH4+ 5.6 x 10-10 NH3 1.8 x 10-5

Stuff you should now know.

1. Ka value is directly related to acid strength.

2. Weak acids vs. strong acids (Ka’s and % dissociation.3. Conjugate acid-base pairs.

4. KaKb=Kw

5. Kb value is directly related to base strength.

6. How to write out Ka and Kb rxns and expressions.7. The weaker the acid the stronger the conjugate base

(and vice versa).8. Conjugate bases of strong acids have no basic

properties whatsoever! (Kb << Kw)

Calculate the pH of a 0.10 M HBr solution.

Calculate the pH of a 0.10 M HOCl solution.

KaHOCl = 3.5 x 10-8

Calculate the pH of a 0.10 M NaF.

KaHF = 7.2 x 10-4

Calculate the pH of a 0.10 M Ca(OH)2

solution.

Calculate the pH of a solution containing

0.10 M HOCl and 0.02 M NaOCl.

KaHOCl = 3.5 x 10-8

Calculate the pOH of 0.05 M Ba(OH)2.

Calculate the pOH of 0.50 M KOCl.

KaHOCl = 3.5 x 10-8

Calculate the pOH of 1.00 M HI.

Calculate the pOH of 0.25 M NH4Cl.

Ka NH4+ = 5.6 x 10-10

Calculate the pOH of a solution containing

0.25 M NH4Cl and 0.10 M NH3.

Ka NH4+ = 5.6 x 10-10

Calculate the pH of 1.6 x 10-13 M HNO3.

A solution of 8.00 M HCOOH is 0.47%

Ionized. What is the Ka for the acid? pH?

Acid Ka

HF 7.2 x 10-4

C6H5NH3+ 2.6 x 10-5

HC2H3O2 1.8 x 10-5

HCN 6.2 x 10-10

NH4+

Acidic, Basic, or Neutral?

1. NaCN2. NH4NO3

3. KI4. LiC2H3O2

5. C6H5NH3Cl6. KF7. NaNO3

8. HClO4

9. Ca(OH)2

10. NH4CN11. NH4C2H3O12. CaO13. SO3

Acidic, Basic, or Neutral?

1. NaCN Na+ - worthless, CN- - weak base, basic2. NH4NO3 NO3

- - worthless, NH4+ - weak acid, acidic

3. KI K+ - worthless, I- - worthless, neutral4. LiC2H3O2 Li+ - worthless, C2H3O2

- - weak base, basic5. C6H5NH3Cl Cl- - worthless, C6H5NH3

+ - weak acid, acidic6. KF K+ - worthless, F- - weak base, basic7. NaNO3 Na+ - worthless, NO3

- - worthless, neutral8. HClO4 HClO4 – strong acid, acidic9. Ca(OH)2 Ca(OH)2 – strong base, basic10. NH4CN KaNH4 < KbCN- - basic11. NH4C2H3O KaNH4 = KbC2H3O- - neutral12. CaO metal oxide - basic13. SO3 nonmetal oxide - acidic

Buffers

Buffer – A solution where a weak acid and

its conjugate base are both present

in solution.

• Buffers resist changes in pH

Good Buffers

• Good buffers will have the following:– EQUAL concentrations of the weak acid and

its conjugate base.

– LARGE concentrations of the weak acid and its conjugate base.

– pKa = pH of desired pH.

Examples of Buffers

• HCN/CN-

• NH4+/NH3

• H2PO4-/HPO4

2- - intracellular fluid buffer

• H2CO3/HCO3- - blood buffer

Calculate the pH of a solution that is 1.00 M

HNO2 and 1.00 M NaNO2.

KaHNO2 = 4.0 x 10-4

Calculate the pH when 0.10 mol of HCl is

Added to a 1.00 L solution containing 1.00 M

HNO2 and 1.00 M NaNO2.

KaHNO2 = 4.0 x 10-4

Calculate the pH when 0.10 mol of NaOH

are added to a 1.0 L solution containing

1.00 M HNO2 and 1.00 M NaNO2.

KaHNO2 = 4.0 x 10-4

Calculate the pH of a solution formed by

Mixing 500.0 mL of 0.100 M NH3 and 500.0

mL of 0.0500 M HCl. KbNH3 = 1.8 x 10-5

You want to prepare a HOCl buffer of pH

8.00. You want to make a 500. mL solution

and use all of the 0.75 mol of HOCl you

have on hand. How many mol of KOCl must

you add? KaHOCl = 3.5 x 10-8

Calculate the pH of a solution formed by

mixing 500. mL of 1.50 M HCN with 250. mL

of 1.00 M NaOH. KaHCN = 6.2 x 10-10

• Total Points in course: 800

• Points to be decided next week: ~415

Proposed Study Plan

• Thursday: HE III Material (finish Lon Capa)• Friday: HE I Material• Saturday: HE II Material• Sunday: HE III Material• Monday: HE III Material• Tuesday: He III Material• Wednesday: HE I, II Material• Thursday: HE I, II, III Material

A 100. mL solution of 0.10 M HF is titrated by 0.10 M NaOH. Calculate the pH when 0.0, 25.0, 50.0, 100.0, and 125.0 mL of NaOH have been added. KaHF = 7.4 x 10-4