Chem. 31 – 2/25 Lecture

Announcements I• Exam 1

– On Monday (3/2)– Will Cover the parts we have covered in Ch. 1,

3 and 4 plus parts of Ch. 6 (through Le Châtelier’s Principle)

– Some of HW1.3 postponed (see posted solutions)

– Help session (11:00 to 12:00 on Friday – after office hours)

Announcements II• Today’s Lecture

– Chapter 6 Material – Le Châtelier’s Principle (last part on Exam1)

– Review of Material on Exam 1 (including Equation)

– Chapter 6 Material not on Exam 1 (Sparingly soluble salts)

Le Châtelier’s Principle

Stess Number Two: Dilution

Side with more moles is favored at lower concentrations

Example: HNO2(aq) ↔ H+ + NO2-

If solution is diluted, reaction goes to products

If diluted to 2X the volume:

2

2

HNONOH

K

2

2

2

12

1

2

1

HNO

NOHQ

KQ2

1

So Q<K, products favored

Le Châtelier’s PrincipleStess Number Two: Dilution – Molecular Scale View

H+ NO2-

Concentrated Solution

H+ NO2-

H+

NO2-

H+ NO2-

H+ NO2-

Diluted Solution – dissociation allows ions to fill more space

H+ NO2-

H+ NO2-

H+

NO2-

H+ NO2-

H+ NO2-

Le Châtelier’s Principle

Stress Number 3: TemperatureIf ΔH>0, as T increases, products

favoredIf ΔH<0, as T increases, reactants

favoredEasiest to remember by considering

heat a reactant or productExample:

OH- + H+ ↔ H2O(l) + heatIncrease in T

Some Le Chatelier’s Principle Examples

• Looking at the reaction below, that is initially at equilibrium,

AgCl(s) ↔ Ag+(aq) + Cl-(aq) (ΔH°>0)determine the direction (toward products or reactants) each of the following changes will result ina) increasing the temperatureb) addition of waterc) addition of AgCl(s) d) addition of NaCl

Review for Exam

• Know the following (from Ch. 1)– Common base units (m, kg, s, mol, K) +

common multipliers (nano to kilo)– How to convert between different units*– Definitions of main concentration units (M,

weight fractions including % and ppm, and mass/volume)

– How to convert between concentration units*– Steps to make solutions of known

concentration + calculations for solution preparation*

– How to do stoichiometry problems (involving solids or solutions)*

Note: *means need quantitative knowledge

Review for Exam – cont.• Know the following (from Ch. 1 – cont.)

– Titration definitions (titrant, equivalence point, end point, standardization titrations, analyte titrations, back titrations)

– Practical titration requirements– How to solve normal and back titration

problems*

• Know the following (from Ch. 3)– Rules for significant figures (including for

calculations with +, -, *, or / and when uncertainties are given)*

– Definitions for: systematic and random error, accuracy and precision, uncertainty, relative error and relative uncertainty

Review for Exam

• Know the following (from Ch. 3 – cont.)-How to do propagation of uncertainty problems (+, -, *, /, exponent, and mixed operations) and to convert between absolute and relative uncertainty*

•Know the following (from Ch. 4)-What a Gaussian distribution represents-How to calculate mean values and standard deviations (can use calculators)*-The differences between populations and samples-How to calculate Z values*

Review for Exam – Ch. 4 (cont.)

•How to use Table 4-1 and Z values to calculate probabilities between limits*•How to determine confidence intervals* + factors which influence confidence intervals•Difference between Z and t based confidence intervals (lecture only)•What confidence intervals tell you•How to perform t-test (case 1)*•How to recognize and select a proper test (3 t tests, F test and Grubbs test)•How to deal with poor data points (including use of Grubbs test)*

Review for Exam• Chapter 4 – cont.

- How method of least squares works (qualitatively)

- Steps to the calibration process- Assumptions required for least squares analysis- How to determine concentrations of unknowns*

+ limitations in this

• Chapter 6- Be able to write equilibrium equations from

given equilibrium reactions- Manipulate equilibrium reactions/equations*- Definitions of changes in Enthalpy, Entropy and

Free Energy plus predictions given reaction

Review for Exam – Ch. 6 (cont.)

•Chapter 6•Be able to determine K from DG° or visa versa•Be able to calculate DG from DH and DS•Be able to predicts shift in equilibrium due to changes in conditions (Le Châtelier’s Principle)

Review for Exam• Equations I will give

- Basic propagation of uncertainty equations (e.g. for Y = a + b, Y = a·b, and Y = xn)

- Equation for calculation of standard deviation- Grubb’s test equation- Equation for converting K to DGº

Ch. 6 – Solubility Problems

•Why Solubility is Important• Use in gravimetric analysis (predict if precipitation is

complete enough)• Use in precipitation titrations (not covered)• Use in separations (e.g. separation of Mg2+ from Ca2+

in tap water for separate analysis)• Understand phase in which analytes will exist

•Problem Overview• Dissolution of sparingly soluble salts in water• Dissolution of sparingly soluble salts in common ion• Precipitation problems (and selective precipitation

problems)

Solubility Product Problems - Solubility in Water

Example: solubility of Mg(OH)2 in waterSolubility defined as mol Mg(OH)2

dissolved/L sol’n or g Mg(OH)2 dissolved/L sol’n or other units

Use ICE approach:Mg(OH)2(s) ↔ Mg2+ + 2OH-

Initial 0 0Change +x +2xEquilibrium x 2xNote: x = [Mg2+] = solubility

Solubility Product Problems

- Solubility of Mg(OH)2 in water

Equilibrium Equation: Ksp = [Mg2+][OH-]2

Ksp = 7.1 x 10-12 = x(2x)2 = 4x3 (see Appendix F for Ksp)

x = (7.1 x 10-12/4)1/3 = 1.2 x 10-4 MSolubility = 1.2 x 10-4 M = [Mg2+]Conc. [OH-] = 2x = 2.4 x 10-4 M

Solubility Product Problems

- Solubility of Mg(OH)2 in Common Ion

If we dissolve Mg(OH)2 in a common ion (OH- or Mg2+), from Le Châtelier’s principle, we know the solubility will be reduced

Example 1) What is the solubility of Mg(OH)2 in a pH = 11.0 buffer?

No ICE table needed because, from pH, we know [OH-]eq and buffer means dissolution of Mg(OH)2 doesn’t affect pH.

Solubility Product Problems

- Solubility of Mg(OH)2 at pH 11 – cont.

[H+] = 10-pH = 10-11 Mand [OH-] = Kw/[H+] = 10-3 M

Ksp = [Mg2+][OH-]2

Moles Mg(OH)2 dissolved = moles Mg2+

[Mg2+] = Ksp/[OH-]2 = 7.1 x 10-12/(10-3)2

[Mg2+] = 7 x 10-6 M

Solubility Product Problems

- Solubility of Mg(OH)2 in Common Ion

Example 2) Solubility of Mg(OH)2 in 5.0 x 10-3 M MgCl2.

Solubility Product Problems Precipitation Problems

What occurs if we mix 50 mL of 0.020 M BaCl2 with 50 mL of 3.0 x 10-4 M (NH4)2SO4?

Does any solid form from the mixing of ions?

What are the concentrations of ions remaining?