Le Châtelier’s principle

The significance of Kc values If Kc is small (0.001 or lower), [products] must be small, thus forward reaction is weak If Kc is large (1000 or more), [products] must be large, thus forward reaction is strong [Products] Kc = [Reactants] Reactants  Products If Kc is about 1, then reactants and products are about equal but not exactly since they may be raise to different exponents

Stresses to equilibria  Changes in reactant or product concentrations is one type of “stress” on an equilibrium  Other stresses are temperature, and pressure.

 The response of equilibria to these stresses is explained by Le Chatelier’s principle: If an equilibrium in a system is upset, the system will tend to react in a direction that will reestablish equilibrium  Thus we have: 1) Equilibrium, 2) Disturbance of equilibrium, 3) Shift to restore equilibrium  Le Chatelier’s principle predicts how an equilibrium will shift (

N 2 + 3H 2  2NH kJ Summary of Le Chatelier’s principle E.g. N 2 + 3H 2  2 NH kJ Pressure (due to decreased volume): increase in pressure favors side with fewer molecules Amounts of products and reactants: equilibrium shifts to compensate  N 2  H 2 Temperature: equilibrium shifts to compensate:  Heat shift right shift left

Part II. Equilibria involving sparingly soluble salts  Ag + + CO 3 -2 Ag 2 CO H + + CO 3 -2 H 2 O + CO 2

Part II. Equilibria involving sparingly soluble salts  Ag + + Cl - AgCl Ag + + 2NH 3 Ag(NH 3 ) 2 + heat NH 3 + H + (NH4)

Questions  Omit part 1  Pg 256 omit part 3

Titration Lab 20- Acids & Bases

A. Neutralization  Chemical reaction between an acid and a base.  Products are a salt (ionic compound) and water.

Neutralization ACID + BASE  SALT + WATER HCl + NaOH  NaCl + H 2 O HC 2 H 3 O 2 + NaOH  NaC 2 H 3 O 2 + H 2 O Salts can be neutral, acidic, or basic. Neutralization does not mean pH = 7. weak strong neutral basic

Titration  Titration Analytical method in which a standard solution is used to determine the concentration of an unknown solution. standard solution unknown solution

 Equivalence point (endpoint) Point at which equal amounts of H 3 O + and OH - have been added. Determined by… indicator color change B. Titration dramatic change in pH

B. Titration moles H 3 O + = moles OH - M  V  n = M  V  n M:Molarity V:volume n:# of H + ions in the acid or OH - ions in the base

B. Titration  42.5 mL of 1.3M NaOH are required to neutralize 50.0 mL of KHP. Find the molarity of KHP. H3O+H3O+ M = ? V = 50.0 mL n = 1 OH - M = 1.3M V = 42.5 mL n = 1 MV = MV M(50.0mL) =(1.3M)(42.5mL) M = 1.11 M H 2 SO 4

Procedures  Data you need for part B Molarity of NaOH Trial 1 [.115 M] Trial 2 [.116 M] Trial 3 [.117 M]

Procedures Formulas M = mol/liters moles H 3 O + = moles OH - MV (H 3 O + ) = MV (OH - ) % KHP = mass of KHP/ Mass of mixture

Part B. Standardization of unknown acid Add 2.5 grams of unknown acid into 3 Erlenmeyer flasks Use analytical balance to 4 sig figs Add 100 ml of D.I water into each flask Add 2 drops of Phenolphthalein soln

 A. Fill a Burett with standarized NaOH Slowly add NaOH into 1 st flask, while swirling flask See fig 20-2 and read procedures for proper titration Record amount of NaOH used to titrate

calculations  Determine moles of base used  Determine moles of KHP  Determine mass of KHp  Determine % of KHP in unknown

Questions  223 and 224  Questions 1 -3