Buffer Effectiveness, Titrations & ph curves. Section

Transcription

1 Buffer Effectiveness, Titrations & ph curves Section

2 Buffer effectiveness Buffer effectiveness refers to the ability of a buffer to resist ph change Effective buffers only neutralize small to moderate amounts of acid or base Factors influencing buffer effectiveness: Ratio of the buffers acid-conjugate base concentrations Composition of buffer solution Overall absolute concentration Concentration of buffer solution Optimal conditions for buffer effectiveness can be derived from the Henderson-Hasselbalch equation Reference Table

3 Relative amounts of Acid and Base Buffers are most effective when their acid and conjugate base concentrations are equal Buffers become less effective as the difference between the concentrations of acid and conjugate base increase In order for a buffer to be effective the difference in concentration should not differ by more than a factor of 10

4 The math This can be illustrated by examining two solutions of a generic buffer, pk a =5.00, neutralizing 0.01 mol NaOH. Both solutions have a volume of 1.0 liter and 0.20 mol total acid and conjugate base Solution I has equal concentrations of acid and conjugate base while Solution II has 0.18 mol acid (HA) and 0.02 mol base (A - ).

5 Solution I Solution II Initial ph = log (1.00) = 5.00 Initial ph = log (0.02/0.18) = 4.05 OH - + HA H OH - + HA H 2 O + A - 2 O + A - Before mol 0.10mol Before mol 0.02 mol Addition 0.01 mol Addition 0.01 mol After 0.00 mol 0.09 mol.110 mol After 0.00 mol 0.17 mol.03 mol ph =pk a + log ( [base] / [acid] ) = log ( /.090 ) =5.09 % change=( ) / 5.00 X100% =1.8% ph =pk a + log ( [base] / [acid] ) = log ( 0.03 /.17 ) =4.25 % change=( ) / 4.05 X100% =5.0% Reference Table

6 In the context of the H.-H. equation ph=pk a + log ( [base] / [acid] ) [base] = [acid] [base] / [acid] = 1 log(1)=0 ph=pk a + 0 ph=pk a

7 Concentrations of Acid and Base Buffers are most effective at high concentrations of acid and conjugate base The more dilute the buffer components the less effective

8 Solution I Solution II Initial ph = log (0.50/0.50) = 5.00 Initial ph = log (0.050/0.050) = 5.00 OH - + HA H OH - + HA H 2 O + A - 2 O + A - Before mol 0.50mol Before mol 0.05 mol Addition 0.01 mol Addition 0.01 mol After 0.00 mol 0.49 mol.51 mol After 0.00 mol 0.04 mol.06 mol ph =pk a + log ( [base] / [acid] ) = log ( 0.51 / 0.49 ) =5.02 % change=( ) / 5.00 X100% =0.4% ph =pk a + log ( [base] / [acid] ) = log ( 0.06 / 0.04 ) =5.18 % change=( ) / 5.00 X100% =3.6%

9 Buffer Range Because a buffer should not differ by more than a factor of 10, we can use the Henderson-Hasselbalch equation to find a ph range: ph = pk a + log ( [base] / [acid] ) = pk a + log 0.10 = pk a - 1 ph = pk a + log ( [base] / [acid] ) = pk a + log 10 = pk a + 1 the effective ph range for a buffer solution is pk a ± 1 When making a buffer solution, use the pka then adjust acid-conjugate base ratio

10 Practice A: Which acid would one use (combined with its sodium salt) to make a solution buffered at a ph of 4.25? a.) HClO 2 pk a = 1.95 b.) HNO 2 pk a = 3.34 c.) HCHO 2 pk a = 3.75 d.)hclo pk a = 7.54

11 Practice B: Calculate the ratio of the conjugate base to the acid in formic acid (HCHO 2 ) required to attain the desired ph of (pk a = 3.74) ph = pk a + log ( [base] / [acid] ) 4.25 = log ( [base] / [acid] ) = log ( [base] / [acid] ) 0.51 = log ( [base] / [acid] ) [base] / [acid] = = 3.24

12 Buffer Capacity The amount of acid or base a buffer solution can neutralize without a significant change in ph Buffer capacity increases with increasing absolute concentration of the buffer solution (increasing concentration of components) Buffer capacity also increases as the concentration of acid and conjugate base reach similar levels (the [base] / [acid] approaches one) However, if a buffer neutralizes mainly acid or mainly base it may have a much higher concentration of one of the components

13 Titrations Acid-Base titrations: an acidic (or basic) solution of unknown concentration reacts with a basic (or acidic) solution of known concentration in order to determine the original solution s concentration

14 Titrating A ph indicator is mixed with the solution to monitor ph The solution of known acid or base concentration is slowly added in using a buret The acid and base neutralize each other as the titration continues When the equivalence point is reached the titration is complete (ph indicator- dye whose color depends on ph) (Equivalence point- point when the number of moles of acid and base are stoichiometrically equal)

15 Buret

16 Strong Acid-Strong Base Titration Ex: Titration of 25 ml 0.100M HCl with 0.100M NaOH Calculating equivalence point: HCl (aq) + NaOH (aq) H 2 O (l) + NaCl (aq) Initial mol HCl = 0.025L x (0.100mol / 1 L) = mol HCl NaOH solution = mol x (1L / 0.100mol) = L The ph at the equivalence point will always be 7 (all H 3 O + and OH - ions have neutralized each other) To find the concentration of the initial solution convert the volume added to moles and then moles to concentration

17 M a V a = M b V b (mol acid) x (volume acid) = (mol base) x (volume acid) ( mol / L ) x (L) = ( mol / L ) x (L) This equation is used to find the equivalence point

18 Strong Acid-Strong Base Titration Continued Initial ph= -log [H 3 O + ] = -log (0.100) = 1 ph at any given point before equivalence point: mol NaOH added = (L added) x ( [NaOH] / 1 L ) Because we are working with a strong acid and base H 3 O + = mol HCl - mol NaOH [H 3 O + ] = (mol H 3 O + / total volume) ph = -log [H 3 O + ] ph at any given point after equivalence point: [OH - ] = (mol OH - - mol H 3 O + ) / total volume [H 3 O + ] = / [OH - ] ph = -log [H 3 O + ]

19 Practice Convert ml mol, Use addition table to find mol H 3 O + mol H 3 O + [H 3 O + ], [H 3 O + ] ph Calculate the ph after adding 5 ml of M NaOH to 25 ml of M HCl OH - + H 3 O + 2 H 2 O Before Addition Addition After Addition mol mol 0.00 mol mol [H 3 O + ] = (.0020 mol H 3 O + ) / ( L L ) = ph = -log = 1.18 Total Volume

20 Titration curve/ph curve

21 Strong Base-Strong Acid Titration

22 College Board- Buffer

23 College Board- Titrations