Outline:2/26/07 è è Today: Start Chapter 17 è CAPA 10 & 11 – due today è Pick up CAPA 12 & 13 – outside è Turn in seminar reports – to me Ù Acid-base Equilibria: Acid & Bases: definitions Defining/Calculating pH Simple pH Calculations from K a

Any questions on equilibria? Lots of suggested problems to do…. 16.1, 16.9, 16.11, 16.17, 16.19, 16.21, 16.23, 16.25, 16.27, 16.29, 16.31, 16.33, 16.35, 16.37, 16.43, 16.45, 16.47, 16.49, 16.51, 16.57, 16.59, , 16.67, 16.69, 16.71, 16.73, 16.75, 16.79, 16.81

Quiz #5 Please put books/papers away… Remember: lots of partial credit for setting up the problem correctly… Solving it just gets you the last points

Quiz #5 Please turn your quizzes over and pass them to your right…

Acids & Bases: These are all examples of equilibria big K eq = strong acid (dissociation) little K eq = weak acid  HSO 3  + H 2 O  H 3 O + + SO 3  K eq =  HCl + H 2 O  H 3 O + + Cl  K eq = 6.4  10 

Chapter 17: More Equilibrium! What exactly is in an aqueous solution? HNO 3 H3O+H3O+ H2OH2O NO 3  p. 140

Does this differ from a weak acid? HOAc H2OH2O p. 141

A combination of previous chapters: n n Solubility rules: Highly soluble:   ionic salts of Cl , Br , I ,    ionic salts of NO 3 , SO 4 ,   ionic salts of ClO 4 , etc. n Equilibrium rules: K eq = [products]/[reactants] Big K eq  large extent of reaction = Strong Acids

Just a few strong acids & bases, the rest are weak:

The extent of dissociation:K a and K b n Define acid dissociation constant:  K a = [H + ][A  ]/[ HA] n Define base dissociation constant:  K b = [B + ][OH  ]/[BOH] n Called water dissociation constant or “autoionization” = 1  10  14 (by definition) n OK, what exactly is water anyway?  H 2 O + H 2 O  H 3 O + + OH   K eq = [H 3 O + ][OH  ] = K w

Define water as neutral…. K w = 10  (arbitrary scale definition)   then: [H 3 O + ] = [OH  ] = x   or: x 2 = 10    or x = 10   n What’s the pH of neutral water?  =  log(10   Define pH =  log [H + ]

Same thing for bases…. Define pOH =  log(OH  ) Since K w = [H 3 O + ][OH  ] = 10   log[H 3 O + ] + log[OH  ] = log (10  ) or pH + pOH = 14 None of this is really new….using acids and bases in lab all year! e.g. pHydron paper, pH electrodes, Concentration vs. strength!

Calculation Examples: Calculate the pH of a solution of 2.5  10  2 M HClO 4 ? (perchloric acid) n Step #1: identify the acid If strong acid  complete dissociation Step #2: pH =  log [H + ]  log (2.5  10  2 ) = 1.60  HClO 4  H + + ClO 4   2.5  10  2 M 0 0 init  0 equil  2.5  10  2 M 2.5  10  2 M

Examples: Calculate the pH of a solution of 2.5  10  2 M HClO? (hypochlorous acid) n Step #1: identify the acid  if weak acid  look up K a = 3.5  10  8 Step #2: pH =  log [H + ]  log (2.7  10  5 ) = 4.56  HClO  H + + ClO   2.5  10  2 M 0 0 init  (2.5  10  2  x )M x M x M equil

Weak Base Example: Calculate the pH of a solution of 2.5  10  2 M trimethylamine (CH 3 ) 3 N ? n Step #1: identify it as a weak base  look up K b = 6.5  10  5  (CH 3 ) 3 N  H 2 O  OH  + (CH 3 ) 3 N H +  2.5  10  2 M 0 0 init  (2.5  10  2  x )M x M x M equil Step #2: pOH =  log [OH  ]  log (1.3  10  3 ) = 2.89 Step #3: 14  pOH = pH = 11.11

Problem solving overview: Skills needed to solve these problems: 1) Identify strong/weak acid/bases 2) Identify conjugate acid/bases & salts 3) Solving equilibrium problems 4) Identifying & making the correct K eq Know the 6 strong acids / 4 strong bases… Know solubility rules & acid/base definitions Solve for x…forward or backwards... Add equations to make new one & K’s multiply

And Practice, Practice, Practice…