1. __________________________________________ Class Monday, Oct 11, 2004

2. __________________________________________ Another pH Buffer Problem What is the pH of a solution prepared by mixing together 100 mL of 0.2500 M ammonia and 200 mL of 0.1500 M ammonium chloride. The K b for ammonia is 1.75 10  5. Answer: pH = 9.16

3. __________________________________________ Another pH Buffer Problem What is the pH of a solution prepared by mixing together 100 mL of 0.2500 M ammonia and 200 mL of 0.1500 M ammonium chloride. The K b for ammonia is 1.75 10  5. The pK b = -log 10 (1.75 x 10  5 ) = 4.757 pK a + pK b = 14.0, pK a for the acid form (NH 4 + ) = 9.243 The total volume = 200 + 100 = 300 mL Using the H-H equation, pH = pK a + log 10 (base/acid) pH = 9.243 + log 10 {(100 x 0.2500/300) ÷ (200 x 0.1500/300)} pH = 9.243 + (  0.0792) – 9.163 = 9.16 Note that the (no. mol of acid) > (no. mol of base); the pH will lie to the side of the pK a of which ever one is the larger, here acidic side of 9.243 Answer: pH = 9.16

4. __________________________________________ Class Monday, Oct 11, 2004 pH = pKa + log 10 {[base] / [acid]} Generally, the pH range that the buffer will work most effectively is pH = pK a ± 1.00

5. __________________________________________ Another Buffer Problem – Choose a system (conjugate pair) to make a buffer whose pH = 7.0

6. __________________________________________ Another Buffer Problem – Choose a system (conjugate pair) to make a buffer whose pH = 7.0 As mentioned in a previous slide, the pH of the buffer is roughly equal to pK a of the weak acid. From Appendix B, pages 540ff there are several system whose pKa values are close to 7.0; I am going to choose the phosphate buffer with pKa = 7.199.

7. __________________________________________ Another Buffer Problem – Choose a system (conjugate pair) to make a buffer whose pH = 7.0 The ratio of ([HPO 4 2- ] / [H 2 PO4 - ]) is calculated from the Henderson-Hasselbalch expression. pH = pKa + log 10 {[HPO 4 2- ] / [H 2 PO4 - ]} 7.00 = 7.199 + log 10 {[HPO 4 2- ] / [H 2 PO4 - ]} log 10 {[HPO 4 2- ] / [H 2 PO4 - ]}= 7.00 – 7.199 = - 0.199 {[HPO 4 2- ] / [H 2 PO 4 - ]}= 10 -0.199 = 0.632 This means that the ratio of {[base] / [acid]} must be 0.632:1 to have a buffer with a pH of 7.00

8. __________________________________________ Another Buffer Problem – Choose a system (conjugate pair) to make a buffer whose pH = 7.0

9. __________________________________________ Buffer Capacity Buffer capacity measures the resistance the buffer solution has to changes in pH whenever an acid or a base is added. It is technically defined as the number of moles of acid or base one liter of the buffer solution can absorb with a change of pH not to exceed 1 pH unit. The greater the concentrations of the acid and base forms, the greater is the buffer capacity. The buffer capacity is also greatest near the pK a of the acid form of the system.

10. __________________________________________ Buffer Capacity Buffer capacity (  ) is the number of moles of OH– or H + that 1.00 Liter of a buffer can absorb without the pH change exceeding 1 pH unit. The buffer capacity depends on the concentrations of the weak acid and its conjugate base. For the addition of base: nOH- = nHB originally present For the addition of acid:nH+ = nB- originally present In practice, pH starts to change drastically as nHB or nB→ 0, as is shown in the next slide.

11. __________________________________________ Buffer Capacity The effect of adding increments of H + or OH  to a buffer system of HA and A whose pK a = 5.0 and the total concentration of = 1 M.

12. __________________________________________ Buffer Capacity Whenever a strong acid or a strong base is added to a buffer the following reactions occur: 1.Addition of strong base (OH - ) HB(aq) + OH - (aq)  H 2 O + B - (aq) 2. Addition of strong acid (H + or H 3 O + ) B - (aq) + H + (aq)  HB(aq)

13. __________________________________________ Buffer Capacity So long as the system has plenty of HB and B – to consume the H + or OH - ions that have been added there is not a drastic change in the pH. The actual pH will depend on the ratio of the base form : acid form as shown in the Henderson-Hasselbalch equation.

14. __________________________________________ Buffer Capacity Problem What is the new pH whenever 0.100 mol of HCl is added to 1 liter of the pH 7.0 phosphate buffer chosen earlier if the [H 2 PO 4 - ] = 0.800 M. Earlier we calculated that Base : Acid ratio needed to be 0.632, so if the [acid] = 0.800 M, the [base] = 0.632 x 0.800M = 0.506M The addition of 0.100 mol of HCl (H + ) will cause H 2 PO 4 - to increase by 0.100 mol and the HPO 4 -2 to decrease by 0.100 mol; the reaction is HPO 4 -2 + H + → H 2 PO 4 -

15. __________________________________________ What is the new pH whenever 0.100 mol of HCl is added to 1 liter of the pH 7.0 phosphate buffer chosen earlier if the [H 2 PO 4 - ] = 0.800 M. The new mol of HPO 4 -2 = (1.00L)(0.506) – 0.100 = 0.406 mol; since in 1.00 L, [HPO 4 -2 ] = 0.406 M The new mol of H 2 PO 4 - = (1.00)(0.800) + 0.100 = 0.900 mol; since in 1.00 L, [H 2 PO 4 - ] = 0.900M The new pH is found by substituting the new concentration values into the H-H equation: pH = pKa + log 10 {[base] / [acid]} pH = 7.199 + log 10 {0.406 / 0.900} pH = 7.199 + (- 0.346) = 6.853

16. __________________________________________ Buffer Solutions Note that the addition of strong acid causes the pH of the buffer to become more acidic (lower pH). Conversely, the addition of a strong base would cause the pH of the buffer to become more basic (higher pH).

17. __________________________________________ Buffer Solutions The Buffer capacity of the 0.500 M lactic acid/lactate buffer. Note the middle of the buffer range occurring at pH of = 3.85, the pKa of this system.