Chemistry*12* Name:* Acid0Base*Equilibrium*V Date:* Block:*
Transcript of Chemistry*12* Name:* Acid0Base*Equilibrium*V Date:* Block:*
Chemistry 12 Acid-‐Base Equilibrium V
Name: Date: Block:
1. Buffers 2. Hydrolysis
Buffers
An acid-‐base buffer is a solution that resists changes in pH following the addition of relatively small amounts of a strong acid or base. Example: Consider a 1.0 L solution of 0.10 M acetic acid.
CH3COOH (aq) + H2O (l) ⇋ CH3COO-‐ (aq) + H3O+ (aq)
• Acetic acid is a weak acid – only a small percent of the weak acid is ionized Visually… If a strong base was added to a solution, acetic acid will be there to neutralize the base. If a strong acid was added, there would be no species to neutralize it.
In order for a buffer solution to be effective, equivalent concentrations of a weak acid and a conjugate base must be in solution.
Circle the pairs of chemical species below that could be used to prepare a buffer solution:
HNO3 and NaNO3 KF and HF HNO2 and HNO3 HCOOH and LiHCOO
NaHSO4 and Na2SO4 K2CO3 and K2C2O4 HCl and NaCl KH2PO4 and K2HPO4
Acidic Buffer: Because … Then … Or in general… The hydronium ion concentration (and therefore the pH) or a buffer solution depends on:
1. the Ka value 2. the ratio of the concentration of the weak acid to its conjugate base
Using our acetic acid example…
[H3O+] = pH =
We now add 0.10 mol HCl to 1.0 L buffer solution with no volume change.
• The 0.10 mol H3O+ will be consumed
• [CH3COO-‐] will _________________________________________
• [CH3COOH] will _________________________________________
[H3O+] = pH =
We now add 0.10 mol OH-‐ to 1.0 L buffer solution with no volume change. • The 0.10 mol OH-‐ will be consumed
• [CH3COO-‐] will _________________________________________
• [CH3COOH] will _________________________________________
[H3O+] = pH =
Basic Buffer: Consider a NH3/NH4+ buffer solution: Because … Then … Or in general… (Remember: Kb for NH3 = Kw / Ka for NH4+) Using our NH3/NH4+ example…
[OH-‐] = [H3O+] = pH =
We now add 0.010 mol HCl to 1.0 L buffer solution with no volume change. • The 0.010 mol H3O+ will be consumed
• [NH3] will _________________________________________
• [NH4+] will _________________________________________
[OH-‐] = [H3O+] = pH =
We now add 0.010 mol OH-‐ to 1.0 L buffer solution with no volume change.
• The 0.010 mol OH-‐ will be consumed
• [NH3] will _________________________________________
• [NH4+] will _________________________________________
[OH-‐] = [H3O+] = pH =
Hebden Workbook Pg. 181 # 131 – 138
Hydrolysis In previous Chemistry courses, you have learned about neutralization reactions where:
Acid + Base à Salt + Water
The “salt” produced in neutralization reactions are actually acidic or basic. The ions that make up the salt behave as weak acids or bases.
Hydrolysis is the reaction of an ion with water to produce either: • The conjugate base of the ion and hydronium ions or… • The conjugate acid of the ion and hydroxide ions.
Circle the ions in the following list that represent cations of strong bases:
Al3+ Rb+ Fe3+ Cr3+ Ca2+ Sn4+ Cs+ Ba2+ Circle the ions in the following list that represent the conjugate bases of strong acids:
F-‐ ClO2-‐ ClO4-‐ SO42-‐ Cl-‐ NO2-‐ CH3COO-‐ CN-‐ NO3-‐ Circle the following salts whose ions will not hydrolyze when dissociated in water. NH4Cl Na2CO3 RbClO4 Li2SO3 BaI2 NH4HCOO KIO3 CsF CaBr2
For the above list, for the salts that will hydrolyze:
a) Write out the balanced dissociation equation. b) For the ion that will hydrolyze, write out the equation that will occur when it reacts with water. c) Write out the Ka or Kb expression. (the larger Ka will “win”!)
Example: A 9.54g sample of Mg(CN)2 is dissolved in enough water to make 500.0 mL of solution. Calculate the pH of this solution.
• What is the concentration of Mg(CN)2?
• What is initial concentration of each ion? (*Hint – dissociation equation required)
• What are the 2 ions produced? Which will hydrolyze?
• What is the equation when it reacts with water? What is the base ionization expression (Kb)?
• Make an ICE table.
• What is the [OH-‐] at equilibrium?
• Calculate pH and pOH.
Sodium carbonate is used in the manufacture of glass and also as a laundry additive to soften water. A 200.0 mL aqueous solution of 0.50 M Na2CO3 is diluted to 500.0 mL. Calculate the pH of the resulting solution.
The Kb for pyridine, C5H5N, a weak base, is 4.7 x 10-‐9. Calculate the pH of a 0.10 M solution of C5H5NHNO3.
Hebden Workbook Pg 148 #69-‐73