1. 16.2: Buffers http://acidsandbases-101.weebly.com/buffers-- indicators.html

2. A buffer is a solution which resists pH change Buffers are especially important in biological processes, many of which must occur in a narrow pH range (ex: carbonic acid & bicarbonate in blood maintain a pH between 7.36 and 7.42) A buffer contains either: a weak acid and its conjugate base a weak base and its conjugate acid What is a Buffer?

3. Antifreeze (HOCH 2 CH 2 OH) first stage resembles intoxication antifreeze is metabolized in the liver where it is oxidized to glycolic acid; this acid then enters the bloodstream If consumed antifreeze quantities are large enough, the glycolic acid will overwhelm the capacity of the buffer, lowering pH dangerously leads to acidosis HbH + (aq) + O 2 (g) HbO 2 (aq) + H + condition in which the acid affects the equilibrium between hemoglobin and oxygen

4. Calculating the pH of a Buffer Solution As a buffer solution has both a weak acid and its conjugate base, in order to calculate pH we need the initial concentrations of both Due to Lechatelier’s principle, the presence of the conjugate base will suppress the ionization of the acid, called the common ion effect Then, we use ICE tables to solve in order to find pH

5. Practice Problem #1 HC 2 H 3 O 2 MH 3 O + MC 2 H 3 O 2 - M Initial0.1000.000.100 Change-x+x Equilibrium.1-xx.1+x Calculate the pH of a buffer solution that is.100 M HC 2 H 3 O 2 and.100 M NaC 2 H 3 O 2 : (K a =1.8 x 10 -5 ) HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 - (aq)

6. Practice Problem #1 continued K a = ([H 3 O + ][C 2 H 3 O 2 - ])/[HC 2 H 3 O 2 ] 1.8 x 10 -5 = (x(.100 + x))/(.100-x) 1.8 x 10 -5 = (x(.100)/(.100) x = [H 3 O + ] = 1.8 x 10 -5 pH = -log [H 3 O + ] = -log ( 1.8 x 10 -5 ) = 4.74

7. The Henderson-Hasselbach Equation The Henderson-Hasselbach Equation is a simplified equation that relates the pH of a buffer solution to the initial concentrations of the buffer components pH = pK a + log [base]/[acid] This equation assumes the x is small!!! It is recommended to use the equilibrium approach when first working with buffer calculations until you are comfortable with them & able to see when x will be negligible. Some basic guidelines for when x is small: the initial concentration of acids (or bases) is not too dilute, and the equilibrium constant is relatively small (generally, initial concentrations of acids/bases should be at least 10 2 -10 3 times greater than the equilibrium constant)

8. Derivation of Henderson-Hasselbach Equation Given generic weak acid equation: HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) solve equilibrium equation for H 3 O + : K a = ([H 3 O + ][A - ])/[HA] [H 3 O + ] = K a [HA]/[A - ] to get pH: -log[H 3 O + ] = -log (K a [HA]/[A - ]) -log[H 3 O + ] = -log K a - log ([HA]/[A - ]) -log[H 3 O + ] = -log K a + log ([A - ]/[HA]) pH = pK a + log ([A - ]/[HA]) pH = pK a + log ([base]/[acid]) pH = -log[H 3 O + ] pK a = -log K a

9. Practice Problem #2 Calculate the pH of a buffer solution that is 0.050 M benzoic acid (HC 7 H 5 O 2 ) and 0.150 M sodium benzoate (NaC 7 H 5 O 2 ). K a = 6.5 x 10 -5 pH = pK a + log ([base]/[acid]) pH = -log (6.5 x 10 -5 ) + log (0.150/0.050) pH = 4.187 + 0.477 pH = 4.66

10. Calculating pH Changes in a Buffer Solution Though buffers resist pH change, the addition of an acid or base to a buffer solution causes a small amount of pH change To calculate the pH, we break the problem into two pieces The stoichiometry calculation is where we determine how the addition changes the relative amounts of acid and conjugate base (in moles) based on the neutralization reaction The equilibrium calculation is where we calculate the pH based on the new amounts of acid and conjugate base (can use equilibrium equations or Henderson-Hasselbach Equation) To try this calculation, we have a 1 L buffer solution composed of 0.100 M generic acid HA and 0.100 M conjugate base A -. 0.025 mol of strong acid H is added.

11. Stoichiometric Calculation The neutralization of the acid converts a stoichiometric amount of the base into its conjugate acid: H + (aq) + A - (aq) HA (aq) neutralizing 0.025 mol of the strong acid requires 0.025 mol of the conjugate base, so the amount of A - (aq) decreases by 0.025 mol and the amount HA (aq) increases by 0.025 mol H + (aq) A - (aq) HA (aq) Before addition0 mol0.100 mol Addition+0.025 mol----- After addition0 mol0.075 mol0.125 mol

12. Equilibrium Calculation Our solution is 1 L, so our new concentrations are 0.075 M A - (aq) and 0.125 M HA (aq) With these new initial concentrations, we can solve for pH as we would any other buffer HA MH 3 O + MA - M Initial0.1250.000.075 Change-x+x Equilibrium.125-xx.075+x

13. Equilibrium Calculation continued K a = ([H 3 O + ][A - ])/[HA] K a = (x(0.075+x))/(0.125 - x) K a = (x(0.075))/0.125 x = [H 3 O + ] = K a (0.125/0.075) pH = -log [H 3 O + ] pH = -log K a (0.125/0.075)

14. Calculating pH Changes in a Buffer Solution Alternatively, if 0.025 mol of a strong base were added to our initial solution, the stoichiometric calculation would have been based on the neutralization reaction: OH - (aq) + HA (aq) H 2 O (l) + A - (aq) OH - MHA MA - M Initial0.1250.000.075 Change-x+x Equilibrium.125-xx.075+x

15. C When calculating the pH of a buffer after adding small amounts of acid or base, remember: Adding a small amount of strong acid to a buffer converts a stoichiometric amount of the base to the conjugate acid and decreases the pH of the buffer (adding acid decreases pH) Adding a small amount of strong base to a buffer converts a stoichiometric amount of the acid to the conjugate base and increases the pH of the buffer (adding base increases pH) Calculating pH Changes in a Buffer Solution

16. Buffer Containing a Base and its Conjugate Acid We calculate the pH this sort of a solution the same way as one one containing a weak acid & its conjugate base However, when using the Henderson-Hasselbach equation (pH = pK a + log [base]/[acid]), we must calculate pK a first for the conjugate acid of the weak base. We do this using the equation K a + K b = K w and pK a + pK b = 14 We subtract the pK b of the weak base from 14 to get the pK a and proceed

17. Helpful Links Crash Course Chemistry: Buffers https://www.youtube.com/watch?v=8Fdt5WnYn1k Bozeman Science: pH & Buffers https://www.youtube.com/watch?v=rIvEvwViJGk