Download presentation
Presentation is loading. Please wait.
Published byramaic Ρέντης Modified over 6 years ago
1
Here’s the plan: Pre-activity Strong/Weak Acids/Bases
Writing equilibrium reactions for weak acids and weak bases Writing Acid/Base Reactions Titration calculations (weak/strong)
2
“Pre-Activity” According to the Bronsted-Lowry definition, an acid is something that will _________ __ ______ while a base will __________ __ _______ . According to the Arrhenius definition, an acid has ____ while a base has _____ . The strong acids are: __________ . The strong bases are: __________ . Conjugates are _____________ . Placing a conjugate acid and base in the same solution will create a _________ which resists ________ in ___.
3
1. Separate the cards given into 4 piles: Strong Acid Weak Acid Strong Base Weak Base
HBr CuOH Ca(OH)2 NH4+ HCOOH (which H is the one?) KOH HC2H3O2 H2SO4 OH-1 NH3 NaC2H3O2 HF NO2-1 CH3COOH (this is identical to which card?) HNO3 CH3NH2 + H2O CH3NH3+ + OH-1 NaOH HA KClO H+1
4
2. Write the equilibrium reactions each would form when dissolved in water.
HA (a general formula for a weak acid) B (a general formula for a weak base) HNO2 NH3 H2CO3 NH4+ CH3COO-1
5
3. Write NET IONIC EQUATIONS for the following reactants
3. Write NET IONIC EQUATIONS for the following reactants. If no reaction occurs, write the reactants and “No Reaction” for the products Formic acid (HCOOH) and KOH Ammonia and ammonium Pyradine (a base, C5H5N) and perchloric acid Propanoic acid (C3H5COOH) and ammonia The underlined H is the H+ that would be donated.
6
4. Perform the calculations below and sketch the titration curve
80. mL of 0.28 M chlorous acid (Ka=1.2 x 10-2) is titrated with 0.35 M potassium hydroxide. Calculate the pH at each point in the titration: 0 mL KOH has been added 32 mL KOH has been added 64 mL KOH has been added 96 mL KOH has been added Use a dashed line to sketch the titration curve if 0.35 M barium hydroxide had been used instead.
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.