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Acid Base Equilibrium. Write equilibrium expressions for the following. HC 2 H 3 O 2(aq) + H 2 O (l) ⇌ C 2 H 3 O 2 - (aq) + H 3 O + (aq) NH 3(aq) + H 2 O (l) ⇌ NH 4 + (aq) + OH - (aq). Strong vs Weak Acid.

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Presentation Transcript
slide2
Write equilibrium expressions for the following.
  • HC2H3O2(aq) + H2O(l) ⇌ C2H3O2-(aq) + H3O+(aq)
  • NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
strong vs weak acid
Strong vs Weak Acid
  • Strong Acid: ionizes 100 % in water to form hydronium ions and the anion of the acid.
  • Examples are hydrochloric, nitric, sulfuric
  • Weak Acid: ionizes less than 50% in water to form hydronium ions and the anion of the acids.
  • Examples are acetic, formic, nitrous
  • Weak acids reach equilibrium with their anion and hydronium and therefore do not ionize any more.
  • The equilibrium constant for weak acids is called the Ka
strong vs weak base
Strong vs Weak Base
  • Strong Base: metal hydroxides (NaOH, Ca(OH)2) that dissociate completely in water to form hydroxide and metal ions.
  • Examples are sodium hydroxide, calcium hydroxide
  • Weak Base: substances that react with water to form hydroxide ions and a cation.
  • Examples are ammonia, carbonates
  • Weak bases reach equilibrium with their cation and hydroxide and therefore do not ionize any more.
  • The equilibrium constant for a weak base is called the Kb
sample problem
Sample Problem

A 0.25 M solution of acetic acid reacts with water to form hydronium and acetate ions. If the equilibrium constant (Ka) is 1.8x10-5

  • Calculate the concentration of hydronium ions at equilibrium.
  • Calculate the pH at equilibrium.
sample problem1
Sample Problem

A 0.15 M solution of ammonia reacts with water to form hydroxide and ammonium ions. If the equilibrium constant (Kb) is 1.8x10-5

  • Calculate the concentration of hydroxide ions at equilibrium.
  • Calculate the pH at equilibrium.
autoionization of water
Autoionization of Water
  • Water reacts with itself to form H3O+ and OH- ions.

H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq)

  • The equilibrium constant for this reaction is referred to as Kw and is equal to 1.0 x10-14.
  • For any acid-base equilibrium, the Ka x Kb for any conjugate acid-base pair must equal Kw.
percent ionization
Percent Ionization
  • The percent ionization indicates the degree to which a weak acid or base ionizes in water.
  • It is determined using the following formula:

% ionization = [H3O+]eq/ [HA]initial x 100%

acid base properties of salt solutions
Acid-Base Properties of Salt Solutions
  • Salts are solids at room temperature composed of cations and anions arranged in a crystal lattice (i.e. ionic compounds).
  • When salts dissolve in water, they dissociate into aqueous solutions of ions that may or may not affect the pH of a solution.

Salts that Form Neutral Solutions

  • Salts with the cation of a strong base and the anion of a strong acid. (eg NaCl, K2SO4)
slide10
Salts that Form Acidic Solutions
  • Salts where the cation is the conjugate acid of a weak base and the anion of a strong acid. (eg NH4Cl)
  • Non-metal oxides (eg CO2)
  • Salts where the cation is a highly charged metal ion and the anion is from a strong acid. (eg AlCl3)
slide11
Salts that Form Basic Solutions
  • Salts with the cation of strong base and the anion of a weak acid. (eg NaC2H3O2)
  • Metal oxides (eg CuO)
  • The pH of a salt where the cation is the conjugate acid of a weak base and the anion is the conjugate base of the weak acid will depend on the Ka value of the conjugate acid and the Kb value of the conjugate base.

If Ka> Kb, pH will be acidic

Ka< Kb, pH will be basic

Ka= Kb, pH will be neutral

acid base titration
Acid-Base Titration
  • Acid-Base titration, involves a reaction between an acid and base resulting in the production of water and a salt (neutralization).
  • An indicator is used to determine the end point.
  • At the end point, the indicator will change colour.
  • The equivalence point or stoichiometric point occurs when equal moles of H3O+ and OH- ions have reacted to produce water.
slide13
The equivalence point is not always equal to the end point and does not always occur at pH 7.
  • When choosing an indicator for a titration, one tries to chose an indicator that will change when the equivalence point is reached.
  • The rule of thumb is to choose an indicator whose pKa is equal to (or close to) the pH at the equivalence point.
  • Titration curves plot pH vs volume of acid or base added and the equivalence point occurs at the inflection point on the curve.
slide14
Strong Acid and Base Titration
  • Equivalence point will occur at a pH of 7.
  • These problems are done as a simple limiting factor problem since there is complete ionization/dissociation.
slide16
Weak Acid and Strong Base Titration
  • Equivalence point will occur at a pH above 7 as the conjugate base of the weak acid determines the pH of the solution.
  • These problems are done using an ICE table.
slide18
Strong Acid and Weak Base Titration
  • Equivalence point will occur at a pH below 7 as the conjugate acid from the weak base determines pH of the solution.
  • These problems are done using an ICE table.
titration questions
Titration Questions

A 275 mL sample of 0.25 M methyl amine is

titrated with 0.35 M nitric acid. Calculate

  • the volume of acid required to reach equivalence.
  • The pH at equivalence point.
challenge question
Challenge Question

A 120 mL sample of 0.50 M formic acid is titrated with 0.15 M calcium hydroxide. Determine

  • The pH after the addition of 100.0 mL of calcium hydroxide
  • Determine the volume of calcium hydroxide needed to reach equivalence.
  • Determine the pH at equivalence.
buffers
Buffers
  • Buffers are mixtures of conjugate acid-base pairs that allow a solution to resist changes in pH.
  • Buffers have common ions in them that act as a reservoir and help maintain a relatively constant pH.
  • Buffer problems should be treated as common ion problems.
slide24
Examples of buffers
  • Acetic acid and sodium acetate
  • Formic acid and sodium formate
  • Ammonia and ammonium chloride
  • Methylamine and methylammonium nitrate
slide25
A buffer is made by mixing 0.60 mol of acetic acid

and 0.50 mol of sodium acetate in 2.0 L of water.

Calculate a) the pH of the original buffer.

b) the pH after the addition of 0.10 mol of OH- ignoring any volume changes.

c) the pH after the addition of 0.10 mol of H3O+ ignoring any volume changes.

polyprotic acids
Polyprotic Acids
  • Polyprotic acids have more than 1 acidic hydrogen.
  • Each ionization will have its own Ka value and in general, Ka1> Ka2> Ka3 etc.
  • Since the first ionization is the largest, only the first ionization is used to determine the pH.