1. Lecture 1: Introduction and review –Quiz 1 –Website: http://www.esf.edu/chemistry/nomura/fch530/ –Review of acid/base chemistry –Universal features of cells on Earth Cell types: Prokaryotes and Eukaryotes

2. Review of pH, acids, and bases pH is generally defined as the negative logarithm of the hydrogen ion activities (concentration) expressed over 14 orders of magnitude pH = -log 10 [H + ] The pH scale is a reciprocal relationship between [H + ] and [OH - ] Because the pH scale is based on negative logarithms, low pH values represent the highest [H + ] and thus the lowest [OH - ] At neutrality, pH 7, [H + ] = [OH - ]

3. Review of pH, acids, and bases Strong electrolytes dissociate completely in water –Electrolytes are substances capable of generating ions in solution –Increase the electrical conductivity of the solution The dissociation of a strong acid in water The equilibrium constant is [H 2 O] is constant in dilute aq. Solutions and is incorporated into the equilibrium constant. giving rise to a new term K a -the acid dissociation constant = K[H 2 O], [H 3 O + ] is expressed as [H + ] HCl + H 2 O H 3 O + + Cl - [H + ][Cl - ] Ka=Ka= [HCl] K = [H 3 O + ][Cl - ] [H 2 O][HCl] Because K a is large for HCl, [H + ] in solution = [HCl] added to solution. Thus, a 1M HCl solution has a pH of 0, a 1 mM HCl solution has a pH of 3, and so on. Conversely, 0.1 M NaOH solution has a pH of 13.

4. [H + ][Cl - ] Ka=Ka= [HCl] For a strong acid K a will approach be large because the nearly all of the protons will be dissociated. The [H+] at equilibrium is equal to the initial concentration of the acid. Calculate the pH of a 1M HCl solution HCl + H 2 O H 3 O + + Cl - 0.0004% at equilibrium 99.996% at equilibrium Since we are at equilibrium, H 3 O + is equal to the initial concentration of acid. [H + ] = [H 3 O + ] = [HCl] = 1M We know that pH is the -log of [H+], therefore for 1M HCl at equilibrium pH = -log 10 [H + ] pH = -log 10 (1) pH = 0

5. 0(10 0 )1.00.00000000000001(10 -14 ) 1(10 -1 )0.10.0000000000001(10 -13 ) 2(10 -2 )0.010.000000000001(10 -12 ) 3(10 -3 )0.0010.00000000001(10 -11 ) 4(10 -4 )0.00010.0000000001(10 -10 ) 5(10 -5 )0.000010.000000001(10 -9 ) 6(10 -6 )0.0000010.00000001(10 -8 ) 7(10 -7 )0.0000001 (10 -7 ) 8(10 -8 )0.000000010.000001(10 -6 ) 9(10 -9 )0.0000000010.00001(10 -5 ) 10(10 -10 )0.00000000010.0001(10 -4 ) 11(10 -11 )0.000000000010.001(10 -3 ) 12(10 -12 )0.0000000000010.01(10 -2 ) 13(10 -13 )0.00000000000010.1(10 -1 ) 14(10 -14 )0.000000000000011.0(10 0 ) pH [H + ][OH - ]

6. Review of pH, acids, and bases Weak electrolytes only slightly dissociate in water –Acetic acid, CH 3 COOH The dissociation of a weak acid in water The acid dissociation constant is K a is also called the ionization constant, because K a is small, most of the acetic acid is not ionized. CH 3 COOH + H 2 O H 3 O + + CH 3 COO - K a = [H + ][CH 3 COO - ] [CH 3 COOH] = 1.74 X 10 -5 M

7. Acid dissociation constant –The general ionization of an acid is as follows: So the acid dissociation constant is as follows: HA H + + A - K a = [H + ][A - ] [HA] There are many orders of magnitude spanned by K a values, so pK a is used instead: pK a = - log 10 K a The larger the value of the pK a, the smaller the extent of dissociation. pK a <2 is a strong acid

8. Henderson-Hasselbalch Equation Describes the dissociation of a weak acid in the presence of its conjugate base –The general ionization of a weak acid is as follows: So the acid dissociation constant is as follows: Rearranging this expression in terms of the parameter of interest [H+] gives the following: HA H + + A - K a = [H + ][A - ] [HA] KaKa [H + ] = [HA] [A - ]

9. Henderson-Hasselbalch Equation Take the log of both sides: log K a + log log[H + ] = [HA] [A - ] Change the signs and define pK a as -log K a : pK a - log pH = [HA] [A - ] pK a + log pH = [A - ] [HA] or

10. Titration curves and buffers Titration curves can be calculated by the Henderson-Hasselbalch equation –As OH - is added to the reaction, it reacts completely with HA to form A - x = the equivalents of OH - added and V represents the volume of the solution. If we let c o represent HA equivalents initially present, then: We can reincorporate this into the Henderson- Hasselbalch eqn. [A - ] = x vol [HA] = (co-x)(co-x) vol pK a + log pH = x co-xco-x ( )