Chapter 8 – pH Buffers Lect 2 Oct 18 Chapter 8 – pH Buffers Buffers are extremely important in the regulation of a variety of biochemical reactions. Often the rate of the reaction is dependent on the pH. In certain cases the enzyme is degraded and no longer functions correctly outside a narrow pH range. The next slide shows this for the rate of -chymotrysin as a catalyst for the cleavage of a C – N bond. Rate of cleavage of C – N bond with chymotrysin (catalyst) As a function of pH. Chapter 8 – pH Buffers This last expression may be rearranged to solve for [H+ ] as [H+ ] = Ka x {[HX] / [X- ]} (Note that the last term is the ratio of the acid : base forms.) Henderson and Hasselbalch first rewrote this last equation by taking the – log 10 of each term to obtain pH = pKa + log 10{[base] / [acid]} This last equation bear Henderson and Hasselbalch names and is widely used in biochemistry. Chapter 8 – pH Buffers pH = pKa + log 10{[base] / [acid]} It is important to note form the Henderson-Hasselbalch equation that the pH depends on 1) The pKa of the conjugate acid form of the acid-base pair Chapter 8 – pH Buffers pH = pKa + log 10{[base] / [acid]} It is important to note form the Henderson-Hasselbalch equation that the pH depends on 1) The pKa of the conjugate acid form of the acid-base pair 2) The ratio of Base : Acid Chapter 8 – pH Buffers pH = pKa + log 10{[base] / [acid]} Also note that whenever [Base] = [Acid], log 10 1.00 = 0, so the pH = pKa of the conjugate acid. • This provides us with a handle for selecting a suitable buffer; Choose a system whose pKa of the conjugate acid is close to the desired region that you want to control the pH. Chapter 8 – pH Buffers pH = pKa + log 10{[base] / [acid]} Farther, note that a 10-fold change in the ratio of [Base] : [Acid] leads to a change of 1.00 pH unit in the buffer. Chapter 8 – pH Buffers pH = pKa + log 10{[base] / [acid]} Generally, the pH range that the buffer will work effectively is pH = pKa ± 1.00 Problem 1 - Calculate the pH in a solution prepared by adding 0.200 mol of acetic acid and 0.200 mol of sodium acetate to one liter. Ka = 1.8 X 10-5 ; pKa = 4.74 [HC2H3O2] = 0.200 mol/L = 0.200 M [C2H3O2 -] = 0.200 mol/L = 0.200 M pH = pKa + log 10{[base] / [acid]} pH = 4.74 + log 10{0.200M/ 0.200M} pH = 4.74 + log 10{1.00} pH = 4.74 Problem 2 - Choose a system (conjugate pair) to give a buffer whose pH = 7.0 As mentioned above, the pH of the buffer is roughly equal to pKa of the weak acid. From Appendix B, pages 546ff there are several system whose pKa values are close to 7.0 . I am going to choose the phosphate buffer with pKa = 7.199. The ratio of ([HPO4 2-] / [H2PO4 -]) is calculated from the Henderson-Hasselbalch expression. pH = pKa + log 10{[HPO4 2-] / [H2PO4 -]} 2 7.00 = 7.199 + log 10{[HPO4 ] / [H2PO4 ]} log 10{[HPO4 2-] / [H2PO4 -]}= 7.00 – 7.199 = - 0.199 {[HPO4 2-] / [H2PO4 -]}= 10 -0.199 = 0.632 This means that any ratio of {[base] / [acid]} = 0.632 will have a pH of 7.00 Effect of adding strong acid or base to a buffer 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) Effect of adding strong acid or base to a buffer 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) • Addition of strong acid (H+ or H3O+) B-(aq) + H+(aq) HB(aq) Effect of adding strong acid or base to a buffer 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. 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 calculate 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 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 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- 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. HPO4-2 + H+ → H2PO4The 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 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 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 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. pH = 7.199 + (- 0.346) = 6.853 {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)} 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. 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. Next time, finish Buffers, Chapter 7, start Chapter 12 – EDTA