8 Weak Acid Base Equilibria
Bronsted-Lowry Theory <ul><li>An acid is a species that can donate a proton to a base. </li></ul><ul><li>A base is a speci...
<ul><li>Conjugate acid base pairs </li></ul><ul><li>CH 3 COOH     CH 3 COO -   +  H + </li></ul><ul><li>acid  conjugate b...
<ul><li>Strong and Weak Acids </li></ul><ul><li>A strong acid completely dissociates in water </li></ul><ul><li>A weak aci...
<ul><li>Ionic Product of Water </li></ul><ul><li>Water only slightly dissociates into ions </li></ul><ul><li>H 2 O     H ...
<ul><li>K w  = [H +  ][OH - ] </li></ul><ul><li>At 25 0 C K w  = 1 x 10  -14  mol 2 /L 2 </li></ul><ul><li>and pK w  = pH ...
<ul><li>Weak Acids and Bases </li></ul><ul><li>General Equation for a Weak Acid </li></ul><ul><li>HA     A -  +  H +   </...
<ul><li>Some examples of weak acids </li></ul><ul><li>Phenol  K a  = 1 x 10  -10  mol/L </li></ul><ul><li>Ethanoic acid K ...
<ul><li>Finding the pH of weak acids </li></ul><ul><li>HA     A -  +  H +   </li></ul><ul><li>K a  =  [A - ][H + ]  </li>...
Example Find the pH of 0.1M ethanoic acid. K a  = 1.8 x 10  -5  mol/L H + = √KaC =  √  1.8 x 10 -5  x 0.1 = 1.34 x 10 -3  ...
<ul><li>Weak Bases </li></ul><ul><li>BOH     B +   +  OH - </li></ul><ul><li>K b  =  [B + ][OH - ] </li></ul><ul><li>[BOH...
<ul><li>Example </li></ul><ul><li>Find the pH of a solution of 0.1m ammonium hydroxide </li></ul><ul><li>if K b  = 1.7 x 1...
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8 Acid Base Equilibria

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8 Acid Base Equilibria

  1. 1. 8 Weak Acid Base Equilibria
  2. 2. Bronsted-Lowry Theory <ul><li>An acid is a species that can donate a proton to a base. </li></ul><ul><li>A base is a species that can accept a proton. </li></ul><ul><li>A  B + H + </li></ul><ul><li>acid base </li></ul><ul><li>(proton donator) (proton acceptor) </li></ul><ul><li>Note: A free proton does not exist. It is always attached by a coordinate bond to a lone pair of electrons on the solvent molecule </li></ul>
  3. 3. <ul><li>Conjugate acid base pairs </li></ul><ul><li>CH 3 COOH  CH 3 COO - + H + </li></ul><ul><li>acid conjugate base </li></ul><ul><li>H 2 O + H +  H 3 O + </li></ul><ul><li>base conjugate acid </li></ul><ul><li>Add 1) and 2) </li></ul><ul><li>CH 3 COOH + H 2 O  H 3 O + + CH 3 COO - </li></ul><ul><li>acid 1 base 2 acid 2 base 1 </li></ul><ul><li>There are 2 conjugate acid/base pairs in this reaction </li></ul>
  4. 4. <ul><li>Strong and Weak Acids </li></ul><ul><li>A strong acid completely dissociates in water </li></ul><ul><li>A weak acid only partially dissociates </li></ul><ul><li>A strong base completely dissociates in water </li></ul><ul><li>A weak base only partially dissociates </li></ul><ul><li>Strong Acids </li></ul><ul><li>HCl H 2 SO 4 HNO 3 </li></ul><ul><li>Weak Acids </li></ul><ul><li>All organic acids e.g. Ethanoic acid (acetic acid) </li></ul><ul><li>CH 3 COOH </li></ul><ul><li>Strong Bases </li></ul><ul><li>NaOH KOH </li></ul><ul><li>Weak Bases </li></ul><ul><li>NH 3 </li></ul>
  5. 5. <ul><li>Ionic Product of Water </li></ul><ul><li>Water only slightly dissociates into ions </li></ul><ul><li>H 2 O  H + + OH - </li></ul><ul><li>K c = [H + ][OH - ] </li></ul><ul><li>[H 2 O] </li></ul><ul><li>As water is only slightly ionised [H 2 O] is much bigger then [H + ] or [OH - ] </li></ul><ul><li> we can regard [H 2 O] as constant </li></ul><ul><li>K c [H 2 O] = [H + ][OH - ] </li></ul><ul><li>As K c is a constant and [H 2 O] is a constant we can replace both with a new constant called K w the ionic product of water </li></ul><ul><li> K w = [H + ][OH - ] </li></ul>
  6. 6. <ul><li>K w = [H + ][OH - ] </li></ul><ul><li>At 25 0 C K w = 1 x 10 -14 mol 2 /L 2 </li></ul><ul><li>and pK w = pH + pOH </li></ul><ul><li>p K w = 14 </li></ul><ul><li> pH + pOH = 14 </li></ul><ul><li>So if we know pOH we can find pH </li></ul>
  7. 7. <ul><li>Weak Acids and Bases </li></ul><ul><li>General Equation for a Weak Acid </li></ul><ul><li>HA  A - + H + </li></ul><ul><li>acid conjugate proton </li></ul><ul><li>base </li></ul><ul><li>K c = [A - ][H + ] for weak acids we use K a (acid dissociation constant) </li></ul><ul><li>[HA] </li></ul><ul><li> K a = [A - ][H + ] </li></ul><ul><li>[HA] </li></ul>
  8. 8. <ul><li>Some examples of weak acids </li></ul><ul><li>Phenol K a = 1 x 10 -10 mol/L </li></ul><ul><li>Ethanoic acid K a = 2.1 x 10 -4 mol/L </li></ul><ul><li>pK a = -log K a </li></ul><ul><li> pK a phenol = 10 </li></ul><ul><li>pK a ethanoic acid = 3.67 </li></ul><ul><li>Acid strength increases as pK a decreases (compare with pH) </li></ul>Increasing strength Increasing strength
  9. 9. <ul><li>Finding the pH of weak acids </li></ul><ul><li>HA  A - + H + </li></ul><ul><li>K a = [A - ][H + ] </li></ul><ul><li>[HA] </li></ul><ul><li>As there is very little dissociation of the weak acid we can assume that [HA] at equilibrium = initial concentration of the acid. </li></ul><ul><li>Let this conc = c </li></ul><ul><li>Then K a = [A - ][H + ] </li></ul><ul><li>c </li></ul><ul><li>Also A - = H +  K a = [ H + ] 2 </li></ul><ul><li>c </li></ul><ul><li>Then H + = √K a C </li></ul>
  10. 10. Example Find the pH of 0.1M ethanoic acid. K a = 1.8 x 10 -5 mol/L H + = √KaC = √ 1.8 x 10 -5 x 0.1 = 1.34 x 10 -3 mol/L pH = - log H +  pH = 2.87
  11. 11. <ul><li>Weak Bases </li></ul><ul><li>BOH  B + + OH - </li></ul><ul><li>K b = [B + ][OH - ] </li></ul><ul><li>[BOH] </li></ul><ul><li>We can assume [BOH] = c and [B + ] = [OH - ] </li></ul><ul><li>K b = [OH - ] 2 </li></ul><ul><li>c </li></ul><ul><li>And [OH - ] = √ K b x c </li></ul>
  12. 12. <ul><li>Example </li></ul><ul><li>Find the pH of a solution of 0.1m ammonium hydroxide </li></ul><ul><li>if K b = 1.7 x 10 -5 mol/L </li></ul><ul><li>[OH - ] = √ K b x c </li></ul><ul><li>= √ 1.7 x 10 -5 x 0.1mol/L </li></ul><ul><li>[OH - ] = 1.3 x 10 -3 mol/L </li></ul><ul><li>pOH = 2.88 </li></ul><ul><li> pH = 14 – 2.88 = 11.1 </li></ul>
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