Acid­base Chemistry.notebook
February 20, 2014
Acid­Base Chemistry
Arrhenius:
acid: generates H+ in solution
base: generates OH­ in solution
Bronstead­Lowry:
acid: donates a H+ (proton donor)
base: accepts a H+ (proton acceptor)
Lewis:
acid: accepts an electron pair
base: donates and electron pair
Acid­base Chemistry.notebook
February 20, 2014
Stength:
Strong Acids: HCl, HBr, HI, H2SO4, HNO3, HClO3, HClO4
­Dissociate completely into ions
­ 0.50M HCl = 0.50M H+
Weak acids: H3PO4, HF, Acetic Acid
­ Dissociates depending on equilibrium constant, Ka
Ka = Ex. Determine the pH of 0.30M acetic acid (HC2H3O2) with a Ka of 1.8x10­5.
Strong Bases: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2
­ Like strong acids, these bases completely ionize in solution.
­ 0.50 M LiOH = 0.50 M OH­
­ 0.50 M Ba(OH)2 = 1.0 M OH­
Weak bases: NH3
­ dissociation depends on the equilibrium constant, Kb
Acid­base Chemistry.notebook
February 20, 2014
Find the pH of a solution formed by dissolving 0.100 mol of HC2H3O2 with a Ka of 1.8x10­8 and 0.200 mol of NaC2H3O2 in a total volume of 1.00 L.
Acid­base Chemistry.notebook
February 20, 2014
Acids and bases react with one another to yield two products: water, and an ionic compound known as a salt. This kind of reaction is called a neutralization reaction.
Ex. HCl + NaOH ­­> NaCl + H2O
Ex. H2SO4 + 2 NH4OH ­­> (NH4)2SO4 + 2 H2O
What mass of NaHCO3 (M = 84.0) is required to completely neutralize 25.0 mL of 0.125 M H2SO4? 0.525g
Acid­base Chemistry.notebook
February 20, 2014
Titrations
• An acid­base titration is when you add a base to an acid until the equivalence point is reached which is where the number of moles of acid equals the number of moles of base. For the titration of a strong base and a strong acid, this equivalence point is reached when the pH of the solution is seven (7) as seen on the following titration curve:
• For the titration of a strong base with a weak acid, the equivalence point is reached when the pH is greater than seven.
Acid­base Chemistry.notebook
February 20, 2014
Henderson­Hasselbach Equation
pH = pKa + log ([Base]/[Acid])
25.0 mL of 0.400 M KOH is added to 100. mL of 0.150 M benzoic acid, HC7H5O2 (Ka=6.3x10­5).