4-4-2011

Choosing a Buffer Pick a buffer whose acid has a pKa near the pH that you want Must consider pH and capacity Buffer Capacity: Not all buffers resist changes in pH by the same extent. The effectiveness of a buffer to resist changes in pH is called the buffer capacity. Buffer capacity is the measure of the ability of a buffer to absorb acid or base without significant change in pH.

Buffer capacity increases as the concentration of acid and conjugate base increases. Larger volumes of buffer solutions have a larger buffer capacity than smaller volumes with the same concentration. Buffer solutions of higher concentrations have a larger buffer capacity than a buffer solution of the same volume with smaller concentrations. Buffers with weak acid/conjugate base ratio close to 1 have larger capacities 3

You should observe that 0.2>>x Calculate the pH of a solution which is 0.40 M in acetic acid (HOAc) and 0.20 M in sodium acetate (NaOAc). Ka=1.7x10-5 HOAc H+ + OAc- Initial M: 0.40 0 0.20 Change -x +x +x Equil M: 0.40-x x 0.20+x 0.20 >> x 1.7x10-5 = [H+][OAc-] = [x][0.20] x = [H+] = 3.4x10-5 M [HOAc] [0.40] You should observe that 0.2>>x pH = - log 3.4x10-5 = 4.47

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOH (aq) H+ (aq) + HCOO- (aq) Initial (M) Change (M) Equilibrium (M) 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x 0.30 – x  0.30 0.52 + x  0.52

Ka = {0.52 * x}/0.30 1.7*10-4 = 0.52*x/0.3 X = 9.4*10-5 Again 0.3 is really much greater than x X = [H+] = 9.4*10-5 M pH = 4.02

Calculate the pH of a solution that is 0. 025 mol/L HCN and 0 Calculate the pH of a solution that is 0.025 mol/L HCN and 0.010 mol/L NaCN. (Ka of HCN = 4.9 x 10-10) HCN D H+ + CN- Even if you do not know it is a buffer you can still work it out easily: 7

+ - (all in mol/L) HCN (aq) + H O (l) D H O (aq) + CN (aq) 2 3 Initial conc. 0. 025 N/A 0. 0. 10 Conc. change - x N/A +x +x Equil. conc. 0. 025 – x N/A +x 0.010 + x Ka = {x*(0.01+x)}/(0.025 – x), assume 0.01>>x However if you observe it is a buffer you could work it without including the “x” as it is always very small. 4.9X10-10 = {x*(0.01)}/(0.025) X = [H+] = 1.2*10-9 M, and pH = 8.91

A diprotic acid, H2A, has Ka1 = 1. 1. 10-3 and Ka2 = 2. 5. 10-6 A diprotic acid, H2A, has Ka1 = 1.1*10-3 and Ka2 = 2.5*10-6. To make a buffer at pH = 5.8, which combination would you choose, H2A/HA- or HA-/A2-? What is the ratio between the two buffer components you chose that will give the required pH? The weak acid/conjugate base system should have a pKa as close to the required pH as possible. pKa1 = 2.96, while pKa2 = 5.6 It is clear that the second equilibrium should be used (HA-/A2-).

The ratio between HA- and A2- can easily be found from the equilibrium constant expression where: HA- D H+ + A2- Ka2 = {[H+][A2-]}/[HA-] Ka2/[H+] = [A2-]/[HA-] [H+] = 10-5.8 = 1.6*10-6 M {2.5*10-6/1.6*10-6} = [A2-]/[HA-] [A2-]/[HA-] = 1.56

First, get initial pH before addition of any base Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? Kb = 1.8*10-5 NH4+ (aq) H+ (aq) + NH3 (aq) First, get initial pH before addition of any base Ka = {[H+][NH3]}/[NH4+] (10-14/1.8*10-5) = {[H+] * 0.30}/0.36 [H+] = 6.7*10-10 pHinitial = 9.18

Now calculate the pH after addition of the base: Addition of the base will decrease NH4+ and will increase NH3 by the same amount due to a 1:1 stoichiometry, which is always the case. mmol of NH4+ = 0.36*80 – 0.05*20 = 27.8 mmol NH3 = 0.30*80 + 0.05 * 20 = 25 Can use mmoles instead of molarity, since both ammonia and ammonium are present in same solution NH4+ (aq) H+ (aq) + NH3 (aq)

DpH = pHfinal – pHinitial DpH = 9.21 – 9.18 = + 0.03 NH4+ (aq) H+ (aq) + NH3 (aq) Ka = {[H+][NH3]}/[NH4+] (10-14/1.8*10-5) = {[H+] * 25}/27.8 [H+] = 6.2*10-10 pHfinal = 9.21 DpH = pHfinal – pHinitial DpH = 9.21 – 9.18 = + 0.03

A 0. 10 M acetic / 0. 10 M acetate mixture has a pH of 4 A 0.10 M acetic / 0.10 M acetate mixture has a pH of 4.74 and is a buffer solution! What happens if we add 0.01 mol of NaOH (strong base) to 1.00 L of the acetic acid – acetate buffer solution? This reaction goes to completion since OH- is a strong base and keeps occurring until we run out of the limiting reagent OH- (all in mo les ) CH COOH (aq) + OH - - (aq) g H O (l) + CH COO (aq) 3 2 3 Initial 0.10 0.01 N/A 0. 10 Change - x - x N/A + x Final (where x = 0.01 0.1 – x 0.01 – x = N/A 0.10 + x due to limiting OH - ) = 0.09 0.00 = 0.11 14

(all in mol/L) CH 3 COOH (aq) + H 2 O (l) D H O + (aq) + CH COO - (aq) Initial conc. 0. 09 N/A 0.0 0.1 1 Conc. change x +x Equil. conc. – With the assumption that x is much smaller than 0.09 mol (an assumption we always need to check, after calculations are done!), we find pH = - log [H3O+] pH = - log 1.5 x 10-5 pH = 4.82 15

New [CH3COOH] = 0.11 M and new [CH3COO-] = 0.09 M What happens if we add 0.01 mol of HCl (strong acid) to 1.00 L of the acetic acid – acetate buffer solution? CH3COO- (aq) + H3O+ (aq) → H2O (l) + CH3COOH (aq) This reaction goes to completion and keeps occurring until we run out of the limiting reagent H3O+ New [CH3COOH] = 0.11 M and new [CH3COO-] = 0.09 M 16

Note we’ve made the assumption that x << 0.09 M! (all in mol/L) CH COOH (aq) + H O (l) D + - H O (aq) + CH COO (aq) 3 2 3 3 Initial conc. 0.11 N/A 0.0 0. 09 Conc. change - x N/A +x +x Equil. conc. 0.11 – x N/ A +x 0. 09 + x With the assumption that x is much smaller than 0.09 mol (an assumption we always need to check, after calculations are done!), we find Note we’ve made the assumption that x << 0.09 M! pH = - log [H3O+] pH = - log 2.2 x 10-5 pH = 4.66 17

Assume 0.25>>x, which is always safe to assume for buffers Calculate the pH of a 0.100 L buffer solution that is 0.25 mol/L in HF and 0.50 mol/L in NaF. (all in mol/L) H F (aq) + 2 O (l) D 3 O - Initial conc. 0.25 N/A 0.0 0. 5 Conc. change x +x Equil. conc. – 0 + x Assume 0.25>>x, which is always safe to assume for buffers 18

M 10 x 1.8 0.50 0.25 3.5 ] [F [HF] K O [H = pH = - log [H3O+] 4 a 3 - + = pH = - log [H3O+] pH = - log 1.8 x 10-4 pH = 3.76 No. mol HF = 0.25*0.1 = 0.025 No. mol F- = 0.50*0.1 = 0.050 19

Use moles: New HF = 0.027 mol and new F- = 0.048 mol a) What is the change in pH on addition of 0.002 mol of HNO3? It is easier here to use moles not molarity (all in mo les ) F - (aq) + H O + (aq) g H O (l) + H F (aq) 3 2 Initial 0. 050 0.0 2 N/A 0. 025 Change - x - x N/A + x Final (where x = 0.0 2 0. 50 – x 0.0 2 – x N/A 0. 25 + x due to limiting H O + ) = 0. 48 = 0.00 = 0. 27 3 Use moles: New HF = 0.027 mol and new F- = 0.048 mol (all in mol/L) HF (aq) + H O (l) D H O + (aq) + F - (aq) 2 3 Initial conc. 0. 02 7 N/A 0.00 0. 048 Conc. change - x N/A +x +x Equil. conc. 0. 02 7 – x N/A +x 0. 048 + x ) (0.027 (x)(0.048 10 x 3.5 [HF] ] ][F O [H K 4 3 a = - + 20

Notice we’ve made the assumption that x << 0.027 M. [HF] 0.027 [H O + ] = K = 3.5 x 10 - 4 = - 2.0 x 10 4 M 3 a [F - ] 0.048 Notice we’ve made the assumption that x << 0.027 M. pH = - log [H3O+] pH = - log 2.0 x 10-4 pH = 3.71 21

Use moles: New HF = 0.021 mol and new F- = 0.054 mol b) What is the change in pH on addition of 0.004 mol of KOH? (all in mo les ) HF (aq) + OH - (aq) g H O (l) + F - (aq) 2 Initial 0. 025 0.004 N/A 0.050 Change - x - x N/A + x Final (where x = 0.0 4 0. 025 – x 0.04 – x N/A 0.050 + x due to limiting OH - ) = 0. 021 = 0.00 = 0. 054 Use moles: New HF = 0.021 mol and new F- = 0.054 mol (all in mol/L) HF (aq) + H2O (l) D H + O (aq) + F - - (aq) 3 Initial conc. 0.021 N/A 0.00 0.054 Conc. change - x N/A +x +x Equil. conc. 0.021 – x N/A +x 0.054 + x [H O + ][F - ] (x)(0.054 ) - 4 K = 3.5x10 - 4 = 3 so 3.5x10 = a [HF] (0.021 ) 22

Notice we’ve made the assumption that [HF] 0.21 + - x - 4 [H O ] = K = 3.5 x 10 4 = 1.4 10 M 3 a - [F ] 0.54 Notice we’ve made the assumption that x << 0.21 M. We should check this! pH = - log [H3O+] pH = - log 1.4 x 10-4 pH = 3.87 23