Acid Equilibrium and pH

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Presentation transcript:

Acid Equilibrium and pH Søren Sørensen

Acid/Base Definitions Arrhenius Model Acids produce hydrogen ions in aqueous solutions Bases produce hydroxide ions in aqueous solutions Bronsted-Lowry Model Acids are proton donors Bases are proton acceptors Lewis Acid Model Acids are electron pair acceptors Bases are electron pair donors

Acid Dissociation Constant HA  H+ + A- Acid Proton Conjugate base Alternately, H+ may be written in its hydrated form, H3O+ (hydronium ion). HA(aq) + H2O(l)  H3O+(aq) + A-(aq) acid base conj. Acid conj. base

Practice: write the simple dissociation reaction for the following acids HCl HC2H3O2 NH4+ C6H5NH3+ [Al(H2O)6]3+

Dissociation of Strong Acids Strong acids are assumed to dissociate completely in solution. Large Ka or small Ka? Reactant favored or product favored?

Dissociation Constants: Strong Acids Formula Conjugate Base Ka  Perchloric   HClO4   ClO4-   Very large   Hydriodic   HI   I-   Hydrobromic   HBr   Br-   Hydrochloric   HCl   Cl-   Nitric   HNO3   NO3-   Sulfuric   H2SO4   HSO4-   Hydronium ion   H3O+   H2O   1.0 

Dissociation of Weak Acids Weak acids are assumed to dissociate only slightly (less than 5%) in solution. Large Ka or small Ka? Reactant favored or product favored?

Dissociation Constants: Weak Acids Formula Conjugate Base Ka  Iodic   HIO3   IO3-   1.7 x 10-1   Oxalic   H2C2O4   HC2O4-   5.9 x 10-2   Sulfurous   H2SO3   HSO3-   1.5 x 10-2   Phosphoric   H3PO4   H2PO4-   7.5 x 10-3   Citric   H3C6H5O7   H2C6H5O7-   7.1 x 10-4   Nitrous   HNO2   NO2-   4.6 x 10-4   Hydrofluoric  HF   F-   3.5 x 10-4   Formic   HCOOH   HCOO-   1.8 x 10-4   Benzoic   C6H5COOH   C6H5COO-   6.5 x 10-5   Acetic   CH3COOH   CH3COO-   1.8 x 10-5   Carbonic   H2CO3   HCO3-   4.3 x 10-7   Hypochlorous   HClO   ClO-   3.0 x 10-8   Hydrocyanic   HCN   CN-   4.9 x 10-10 

Autoionization of Water: Amphoteric H2O + H2O  H3O+ + OH- At 25C, [H3O+] = [OH-] = 1 x 10-7 Kw (ion-product constant) is a constant at 25 C: Kw = [H3O+][OH-] = [H+][OH-] Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

Relationship between pH and pOH Calculating pH, pOH pH = -log10[H+] pOH = -log10[OH-] Relationship between pH and pOH pH + pOH = 14 Finding [H3O+], [OH-] from pH, pOH [H3O+] = 10-pH [OH-] = 10-pOH

pH and pOH Calculations

The pH Scale Graphic: Wikimedia Commons user Slower

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #1: Write the dissociation equation HC2H3O2  C2H3O2- + H+

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #2: ICE it! HC2H3O2  C2H3O2- + H+ I C E 0.50 - x +x +x 0.50 - x x x

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #3: Set up the law of mass action HC2H3O2  C2H3O2- + H+ E 0.50 - x x x

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #4: Solve for x, which is also [H+] HC2H3O2  C2H3O2- + H+ E 0.50 - x x x [H+] = 3.0 x 10-3 M

A Weak Acid Equilibrium Problem What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-5 ? Step #5: Convert [H+] to pH HC2H3O2  C2H3O2- + H+ E 0.50 - x x x

Your turn P. 691 #49, 61, 63, 65 Refer to table 14.2 for Ka’s of common monoprotic acids on page 644

Bases Sodium hydroxide Preferred IUPAC name Systematic name Sodium oxidanide Other names Caustic soda Lye Bases Graphics Source: Wikipedia

Acid/Base Definitions Arrhenius Model Acids produce hydrogen ions in aqueous solutions Bases produce hydroxide ions in aqueous solutions Bronsted-Lowry Model Acids are proton donors Bases are proton acceptors Lewis Acid Model Acids are electron pair acceptors Bases are electron pair donors

Dissociation of Strong Bases MOH(s)  M+(aq) + OH-(aq) Strong bases are metallic hydroxides Group I hydroxides (NaOH, KOH) are very soluble Group II hydroxides (Ca, Ba, Mg, Sr) are less soluble pH of strong bases is calculated directly from the concentration of the base in solution

Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH3NH2 + H2O  CH3NH3+ + OH- Kb = 4.38 x 10-4

Kb for Some Common Weak Bases Many students struggle with identifying weak bases and their conjugate acids. What patterns do you see that may help you? Base Formula Conjugate Acid Kb Ammonia   NH3  NH4+  1.8 x 10-5   Methylamine  CH3NH2  CH3NH3+  4.38 x 10-4   Ethylamine  C2H5NH2  C2H5NH3+  5.6 x 10-4   Diethylamine  (C2H5)2NH  (C2H5)2NH2+  1.3 x 10-3   Triethylamine   (C2H5)3N   (C2H5)3NH+  4.0 x 10-4   Hydroxylamine  HONH2   HONH3+    1.1 x 10-8   Hydrazine H2NNH2  H2NNH3+    3.0 x 10-6   Aniline  C6H5NH2   C6H5NH3+    3.8 x 10-10   Pyridine  C5H5N   C5H5NH+    1.7 x 10-9 

Reaction of Weak Bases with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H2O  BH+ + OH- (Yes, all weak bases do this – DO NOT endeavor to make this complicated!)

A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #1: Write the equation for the reaction NH3 + H2O  NH4+ + OH-

A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #2: ICE it! NH3 + H2O  NH4+ + OH- I C E 0.50 - x +x +x 0.50 - x x x

A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #3: Set up the law of mass action NH3 + H2O  NH4+ + OH- E 0.50 - x x x

A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #4: Solve for x, which is also [OH-] NH3 + H2O  NH4+ + OH- E 0.50 - x x x [OH-] = 3.0 x 10-3 M

A Weak Base Equilibrium Problem What is the pH of a 0.50 M solution of ammonia, NH3, Kb = 1.8 x 10-5 ? Step #5: Convert [OH-] to pH NH3 + H2O  NH4+ + OH- E 0.50 - x x x

Reference Table