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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 Kb 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 Kb for ammonia is 1.75 105. • The pKb = -log10(1.75 x 105) = 4.757 • pKa + pKb = 14.0, pKa for the acid form (NH4+) = 9.243 • The total volume = 200 + 100 = 300 mL • Using the H-H equation, pH = pKa + log10(base/acid) • pH = 9.243 + log10{(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 pKa 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 + log10{[base] / [acid]} Generally, the pH range that the buffer will work most effectively is pH = pKa ± 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 pKa 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 ([HPO42-] / [H2PO4-]) is calculated from the Henderson-Hasselbalch expression. • pH = pKa + log10{[HPO42-] / [H2PO4 -]} • 7.00 = 7.199 + log10{[HPO42-] / [H2PO4 -]} • log10{[HPO42-] / [H2PO4-]}= 7.00 – 7.199 = - 0.199 • {[HPO4 2-] / [H2PO4-]}= 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 pKa 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 nHBor 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 Awhose pKa = 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: • Addition of strong base (OH-) • HB(aq) + OH-(aq)  H2O + B- (aq) • 2. Addition of strong acid (H+ or H3O+) • 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 [H2PO4-] = 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 H2PO4- to increase by 0.100 mol and the HPO4-2 to decrease by 0.100 mol; the reaction is • HPO4-2 + H+ → H2PO4-

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 [H2PO4-] = 0.800 M. __________________________________________ The new mol of HPO4-2 = (1.00L)(0.506) – 0.100 = 0.406 mol; since in 1.00 L, [HPO4-2] = 0.406 M The new mol of H2PO4- = (1.00)(0.800) + 0.100 = 0.900 mol; since in 1.00 L, [H2PO4-] = 0.900M The new pH is found by substituting the new concentration values into the H-H equation: pH = pKa + log10{[base] / [acid]} pH = 7.199 + log10{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.