Buffer Solutions A buffer is a solution that

Buffer Solutions A buffer is a solution that resists changes in p. H upon addition of small quantities of acids or bases. In other words, a buffer is a solution that keeps its p. H almost constant. A buffer is a solution containing a weak acid and its conjugate base or a weak base and its conjugate acid. Below is an explanation of how buffers work: Let us look at a buffer formed from a weak acid (like acetic acid) and its conjugate base (e. g. sodium acetate); we have the following equilibria: 1

HOAc D H+ + OAc- + H 2 O D HOAc + OHIn the first equilibrium, acetate is produced from the dissociation of acetic acid. However, there is a lot of acetate added to solution from the sodium acetate. Therefore, the first equilibrium does not occur to any significant degree. We can fairly assume that HOAc will practically not dissociate. The same applies for the second equilibrium where acetic acid is produced. Since a great excess of acetic acid is present in solution from the first equilibrium, one can say that acetate will not practically associate in water. 2

Therefore, neither acetic acid nor acetate equilibria will proceed to any significant extent. The equilibrium constant for dissociation of acetic acid can be written as: Ka = [OAc-][H+]/[HOAc] Where the [OAc-] = COAc- and [HOAc] = CHOAc When an acid is added to the buffer above, H+ from the acid will combine with OAc- to form HOAc which is reflected by an increase in HOAc and a decrease in OAc-. 3

In case the amount of H+ added is not very large, the ratio [OAc-]/[HOAc] will only change slightly. A similar argument can be presented when a small amount of base is added to the buffer solution. However, we will treat the buffer problem as we described for the common ion effect 4

Buffer Capacity The degree of buffer resistance to changes in p. H is referred to as buffer capacity. The buffer capacity is defined as the concentration of acid or base ( in moles ) needed to cause a p. H change equal to dp. H where: b = d. Cbase/dp. H = - d. Cacid/dp. H The minus sign is because addition of an acid causes a decrease in p. H. A more practical relation to use for calculation of buffer capacity is: b = 2. 303 Cacid Cconjugate base / (Cacid + Cconjugate base) 5

Example Calculate the buffer capacity of a solution containing 0. 10 M acetic acid and 0. 10 M acetate. Find the p. H change when you add Na. OH so that the solution becomes 0. 005 M in Na. OH Solution b = 2. 303 Cacid Cconjugate base / (Cacid + Cconjugate base) b = 2. 303 * 0. 10/( 0. 10 + 0. 10) = 0. 115 M b = d. Cbase/dp. H 0. 115 = 0. 005/dp. H = 0. 005/0. 115 = 0. 043 6

Hasselbalch-Henderson Equation In the acetic acid/acetate buffer described above, we have: Ka = [OAc-][H+]/[HOAc] Pka = p. H – log [OAc-]/[HOAc] p. H = pka + log [OAc-]/[HOAc] The above equation is referred to as Hasselbalch. Henderson equation. 7

this equation can be useful to describe buffer limits where the maximum p. H limit for a buffer is when the salt to acid ratio is 10 and the minimum p. H limit of the buffer is when the acid to salt ratio is 10. Inserting these values, one in a time, in the Hasselbalch equation gives p. H = pka + 1 Therefore, the buffer acts well within two p. H units and the midpoint p. H value is equal to pka. One should first look at the pka or pkb to choose the correct buffer system which can be used within a specific p. H range. 8

Example Calculate the p. H of the buffer containing 0. 50 M formic acid (HA, ka = 1. 8 x 10 -4) and 0. 25 M sodium formate (Na. A). Solution First, write the acid equilibrium equation HA D H+ + A- 9

Ka = x (0. 25 + x)/(0. 50 – x) Assume that 0. 25 >> x 1. 8 x 10 -4 = 0. 25 x /0. 50 x = 3. 6 x 10 -4 Relative error = (3. 6 x 10 -4/0. 25) x 100 = 0. 14% The assumption is therefore valid and we have [H+] = 3. 6 x 10 -4 M p. H = 3. 44 10

Example Calculate the p. H of the buffer solution prepared by mixing 10 m. L 0. 10 M HOAc (ka = 1. 75 x 10 -5) with 20 m. L of 0. 20 M sodium acetate. Solution Let us first calculate the concentrations after mixing (final concentrations of the acid and its conjugate base) mmol HOAc = 0. 10 x 10 = 1. 0 mmol [HOAc] = 1. 0/30 mmol OAc- = 0. 20 x 20 = 4. 0 mmol 11

[OAc-] = 4. 0/30 The equilibrium equation is HOAc = H+ + OAc- 12

Ka = x (4. 0/30 + x)/ (1. 0/30 – x) Assume 1. 0/30 >> x 1. 75 x 10 -5 = x (4. 0/30)/1. 0/30 x = 1. 75 x 10 -5 /4. 0 x = 4. 38 x 10 -6 Relative error = {4. 38 x 10 -6/(1. 0/30)} x 100 = 0. 013% The assumption is valid therefore: [H+] = 4. 38 x 10 -6 M p. H = 5. 36 13

Example Calculate the p. H of the solution resulting from adding 25 m. L of 0. 10 M Na. OH to 30 m. L of 0. 20 M acetic acid. Solution Let us find what happens when we mix the two solutions. Definitely the hydroxide will react with the acid to form acetate which also results in a decrease in the acid concentration. mmol OH- = 0. 10 x 25 = 2. 5 mmol HOAc = 0. 20 x 30 = 6. 0 mmol 14

Now the mmol base added will react in a 1: 1 mole ratio with the acid. Therefore we have mmol OAc- formed = 2. 5 mmol [OAc-] = 2. 5/55 M mmol HOAc left = 6. 0 – 2. 5 = 3. 5 mmol [HOAc] = 3. 5/55 M 15

Ka = x (2. 5/55 + x)/ (3. 5/55 – x) Assume 2. 5/55 >> x 1. 75 x 10 -5 = x (2. 5/55)/3. 5/55 x = 1. 75 x 10 -5 x 2. 5/3. 5 x = 2. 45 x 10 -5 Relative error = {2. 45 x 10 -5/(2. 5/55)} x 100 = 0. 054% The assumption is valid p. H = 4. 61 16

Example A buffer solution is 0. 20 M in HOAc and in Na. OAc. Find the change in p. H after addition of 0. 10 mmol of HCl to 10 m. L of the buffer without change in volume. Solution We should find the initial p. H (before addition of HCl) and then calculate the p. H after addition. HOAc D H+ + OAc- 17

Initial p. H Ka = x (0. 20 + x)/ (0. 20 – x) Assume 0. 20 >> x 1. 75 x 10 -5 = 0. 20 x/ 0. 20 x = 1. 75 x 10 -5 Relative error = (1. 75 x 10 -5/0. 20) x 100 = 8. 8 x 10 -3% [H+] = 1. 75 x 10 -5 M p. H = 4. 76 18

After addition of HCl, the acetate concentration will decrease while the acetic acid concentration will increase. mmol HOAc = 0. 20 x 10 + 0. 10 = 2. 1 mmol [HOAc] = 2. 1/10 = 0. 21 M mmol OAc- left = 0. 20 x 10 – 0. 10 = 1. 9 mmol [OAc-] = 1. 9/10 = 0. 19 M 19

Ka = x (0. 19 + x)/ (0. 21 – x) Assume 0. 19 >> x 1. 75 x 10 -5 = 0. 19 x/ 0. 21 x = 1. 93 x 10 -5 Relative error = (1. 93 x 10 -5/0. 19) x 100 = 0. 01% [H+] = 1. 93 x 10 -5 M p. H = 4. 71 Dp. H = 4. 71 – 4. 76 = - 0. 05 Try the same problem replacing Na. OH for the HCl and get the change in p. H using the same procedure. You should realize that in this case the acetic acid will decrease while acetate will increase. 20