Chem 222

#6 Ch 6Jan 27, 2005

Announcement

♦ Quiz 2 will be held on next Thursday (2/3).

We plan to give• One question from Ch3-5• One question from Ch6 (+ Ch7)

Please memorize all the equations highlighted by red and solve home work in this section.

♦ Submit your report for Exp 2 on next M/T.

Lab Report• Lab reports will have the

following information in them:

1.Cover sheet with the final results2.Analysis section3.Photocopy of the experiment

description you placed in your notebook before the start of the experiment.

4. Photocopy (or carbon) copies of the signed data pages you recorded in lab. Do not recopy these for neatness.

Lab Report1. Cover sheet

• the final results• a statement of what you measured (i.e., Nicotine in unknown tobacco sample #_____) unknown number• average value of your determinations • both absolute and relative standard deviations • Write numbers neatly. If we cannot read your numbers, you gain no points.

2. Analysis section (2-4 pages in double space). • Clearly explain how you calculated the results filled in the cover sheet. • Show the equation used and example calculations at least for one measurement. • Also, show graphs etc if they are required for analysis. • Explain why you cannot obtain three measurements.• Type this section for clarity.

1.32 M Unit & Significant Figures

Analysis Example

X: Moles of CaC2O4·H2OX = {Weight of CaC2O4·H2O}/

{Mol weight of CaC2O4·H2O}

For measurement 1,X1 = 0.502 g/ (120.2 g/mol)*

= 2.30 mmolThe weights for measurement 2 and 3 are

0.530 g and 0.350 g. Using these values, X2 = 2.05 mmol & X3 = 1.80 mmol.(* Note: the numbers are incorrect)

So molarity of Ca2+ (y) is y = [Moles of CaC2O4·H2O]/[ Q1 ]

Using this, for measurement 1y1 = 1.32 M Similarly, y2 = 1.25 M & y3 = 2.45 M.

Equation used

Sample calculation

Exp 5-7 Acid Base Titration

• Ch 6 Chemical Equilibrium• Ch 7 Titration• Ch 10 Monoprotic Acid-Base

Equilibria• Ch 11 Polyprotic Acid-Base

Equilibria• Ch12 Acid-Base Titration

Ch6. Chemical Equilibrium

• 6-1 The equilibrium constantv1

aA + bB cC + dD [6-1]v2

v1 = k1[A]a[B]b & v2 = k2[C]c[D]D,where [A] stands for the “concentration” of A.

If the reaction reaches equilibrium, v1 = v2 k1[A]a[B]b = k2[C]c[D]d

Namely, K ≡ k1/k2 = [6-2]←The left side in [6-1] in the denominator

In evaluating an equilibrium constant, 1. Concentrations of solutes should be

expressed as M (moles/L).2. Concentration of gases should be

expressed in bars3. Concentration of pure solids, pure

liquids, solvents are omitted because they are unity.

←→

ba

dc

BADC][][][][

6-2 Manipulating Equilibrium Constants

Q1.• H2O H+ + OH-

KW =[H+][OH-]; KW = 1.0 × 10-14 at 25 °C.When no other chemicals are added, show[H+] = 1.0 × 10-7.

←→

HA H+ + A- K1 = [H+][A-]/[HA]←→

If the direction of the reaction is reversed, the new equilibrium constant is simply the reciprocal of the original value K

H+ + A- HA K1‘ = [HA] /[H+][A-] = 1/K1Combining reactionsIf the two reactions are addedHA H+ + A- K1H+ + C CH+ K2_________________________HA +C A- + CH+ K3

The equilibrium const is the product of K1 and K2.

←→

Reverse Reaction

213 ]][[][

][]][[

]][[]][[ KK

CHCH

HAAH

CHACHAK === +

+−++−

←→

←→

←→

Ex. (p101)• The equilibrium constant for the reactionH2O H+ + OH- (1)is called Kw ([H+][OH-] = 10-14) at 25 °C. • Liquid ammonia reacts H2O as

NH3(aq) + H2O NH4+ + OH- (2)KNH3 = 1.8 × 10-5

Q. Find the equilibrium constant (K3) for the reaction

NH4+ NH3 + H+ K3

Step 1: Make the equation using adding (1) & (2)

H2O H+ + OH- KwNH4+ + OH- NH3 + H2O [Q1]__________________________NH4+ NH3 + H+ K3 = [Q2]

←→

←→

←→

= 1.0×10-14/[Q3] = 5.6 ×10-10

←→

←→

←→

6-2 Equilibrium and Thermodynamics

EnthalpyThe enthalpy change, ∆H, for a reaction is the

heat absorbed or released when the reaction takes place.

HCl(g) H+(aq) + Cl-(aq) ∆H° = -74.85 kJ/molH1 H2 ∆H = H2 – H1

Heat is released by 74.85 kJ/mol____

H1 ↑∆H↓ ____ H2 If ∆H is positive, heat is

absorbed.

∆H° denotes ∆H at 25 °C

←→

Entropy• The measure of “disorder” for a

system: SKCl(s) K+(aq) + Cl-(aq)S1 S2

∆S ≡ S2 – S1 = 76.4J/(K·mol)∆S is the change is entropy. ∆S > 0 More disordered∆S < 0 Less disordered

←→

Free Energy∆Chemical reaction is favored when ∆ H < 0 or ∆ S >0. ∆G combines the two effects

∆G = ∆ H - T ∆ S (6-5)

When ∆G < 0, the reaction is favored.Problem 6-7 (b) A favorable enthalpy change occurs when ∆H

Free energy and Equilibrium

• ∆G is important in chemical reactions, since the equilibrium constant K depends on ∆G as

K = exp(-∆G/RT),where R is the gas constant [= 8.31 J/K·mol]

For exampleHCl(g) H+(aq) + Cl-(aq) ∆H = -74.85 kJ/mol & ∆S = -130.4 J/K ·mol

∆G = -74.85 ×103 J/mol + (298.15 K) (-130.4 J/K ·mol) = -35.97 kJ/mol

K = exp(-∆G/RT) = 2.00×106 = [H+(aq)][Cl-(aq)]/[HCl]

←→

Equilibrium constant is affected by T

K = exp(-∆G/RT) = exp{-(∆H-T∆S)/RT}= exp(-∆H/RT + ∆S/R)= exp(-∆H/RT)exp(∆S/R)

Ex. Kw = [H+][OH-]

At 10 °C 2.97×10-15

At 25 °C 1.01×10-14

At 35 °C 2.07×10-14

Le Châtelier’s Principle (p103)

• If a system is changed from an equilibrium state, the system moves to a direction that partially cancels the change.

Q1. Suppose the above reaction is at equilibrium.When the concentration of Cr2O72- is changed from 0.10 to 0.20 M, in what direction will the reaction proceed to reach the equilibrium?

Q ≡ [Br-][Cr2O72-][H+]8/{[BrO3-][Cr3+]2} = 2K > K (Q reaction quotient) [Br-][Cr2O72-][H+]8 will be reduced.

Q2. When the temperature is raised from equilibrium, in what direction will the reaction proceed to reach the equilibrium?

6-3 Solubility Product & 6-4 Common Ion Effects

• The solubility product is the equilibrium constant for the reaction in which solid salt dissolves to give its constituent ions.

Hg2Cl2(s) Hg22+ + 2Cl-

Ksp = [Hg22+][Cl-]2 = 1.2 × 10-18←→

x (2x)2 = 1.2 × 10-18

4x3 = 1.2 × 10-18 x = 6.7 × 10-7 M

What happens if NaCl is already dissolved at 0.030 M?

6-7 Protic Acids and Bases (p111)

• In aqueous chemistry, an acid is a substance that increases [H3O+] (hydronium ions) when added to water.

• Conversely, a base decreases [H3O+].

• Because Kw =[H3O+][OH-] = const., a decease of [H3O+] always increases [OH-].

a base increases [OH-].

♦ The word protic refers chemistry involving transfer of H+ from one molecules to another. ♦ H+ is also called a proton. H+ and H3O+ are often interchangeably used.

pH

6-8 pH (p113)

• Autoprotolysis of water

H2O H+ + OH-←→

The autoprotolysis constant for water has the special symbol Kw.

Kw = [H+][OH-]

At 25 °C, Kw = 1.01 ×10-14

Ex. How much is [OH-] if [H+] is 1.0 × 10-3 M at 25 °C?

pH = -log[H+]γH+ ~ -log[H+] (γH+ ~ 1)(True definition will be given in (8-10)

Ex2. How much is [OH-] if pH is 5 at 25 °C?

pH & pOH• pOH = -log[OH-]• pH + pOH = -log[H+][OH-]

= -logKW = 14.00 at 25 °C

So when pH = 5, pOH = 14 -5 = 9.

[OH-] = 10-9

Solution is acidic if [H+] > [OH-]Solution is basic if [H+] < [OH-]

Brønsted-Lowry Acids and Bases

DefinitionsBrønsted-Lowry Acids: Proton DonorBrønsted-Lowry Bases: Proton Acceptor

acid

For the reminder of the text, we will use Brønsted-Lowry’s definitions when we speak of acids and bases.

SaltsAny ionic solids, such as ammonium chloride, is called a salt. In a formal sense, a a salt can be thought of as the product of an acid-base reactions.

The nature of H+ and OH-

Conjugated Acids and Bases

AutoprotolysisH2O + H2O H3O+ + OH-←→

(base) (acid) (acid) (base)

H2O H+ + OH-←→

(6-16)

(6-17)

(6-16) and (6-17) mean the same.

Other protic solvents undergo autoprotolysis

Strengths of Acids and BasesStrong Acids and BasesA strong acid or base is completely

dissociated in aqueous solution.HCl(aq) H+ + Cl- (~ 100%)KOH(ac) K+ + OH-(~ 100%)

Remember all the acids and bases in Table 6-2. Weak acids and bases

HA H+ + A- a weak acid

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

B + H2O BH+ + OH- a weak baseKb = [BH+][OH-]/[B]

←→

←→

←→

←→

Problem

(a)[H+] = 0.010 M = 1.0× 10-[Q1]pH = -log(1.0 × 10-[Q1])

= -log(1.0) + 2 = [Q2] (See p48)

(b) [OH-] = 3.5 × 10-2[H+] = Kw/[Q3] = 1.0 × 10-14/(3.5 × 10-2)

= 2.85 × 10-13

pH = - log[H+] = -log(2.85) –log10-13= -0.455 +13= 12.545 = 12.54

Lewis Acids and Bases (p108)

adduct

Ch. 7 Titration

Significant figures ? (a) 10 mL(b) 10.3 mL(c) 10.31 mL(d) 10.315 mL

Titrant

Analyte

7-1 Titration

The color changes indicates “end point”, which is close to the “equivalent point”

The titrant functions as an “indicator”

(HOCH2)3CNH2 + HCl (HOCH2)3CNH3+ + Cl-

“Tris”

Analyte Titrant Indicator•BB (Blue Yellow)•MR (Yellow Red)

• pH meter (pH is detected byvoltage; see P 323)

At “equivalent point”

Home Work

• Read Ch 6 (6-3, 6-4, 6-5, 6-6)• Read Ch 7 & Ch 12 for Exp7, 8

ProblemsCh6: 6-B, 6-C, 6-D, 6-6, 6-14, 6-

15, 6-17, 6-19, 6-21a, 6-21b, 6-41, 6-46, 6-47, 6-48, 6-49, 6-50, 6-51, 6-52, 6-53