Reference no: EM13163890

19.02 Phosphate Buffer Capacity Buffer capacity refers to the amount of acid or base a buffer can "absorb" without a significant pH change. It is governed by the concentrations of the conjugate acid and base forms of the buffer. A 0.5 M buffer will require five times as much acid or base as a 0.1 M buffer for a given pH change. In this problem you begin with a buffer of known pH and concentration and calculate the new pH after a particular quantity of acid or base is added. In the laboratory you will carry out some stepwise additions of acid or base and measure the resulting pH values. Reaction: HPO4 2? + H3O+ H2PO4 ? Initial nin x nin Change -x -x +x Equilibrium nin - x small nin + x Starting with 37 mL of 0.50 M phosphate buffer, pH = 6.83, you add 4.9 mL of 1.00 M HCl. Using the Henderson Hasselbalch equation with a pK2 for phosphate of 6.64, calculate the following values to complete the ICE table.(Carry out your intermediate calculations to at least one more significant figure than needed. See "Phosphate Buffer Issues" to explain why you do not use the value of 7.21 for pK2 that is given in your textbook.) 1.) What is the composition of the buffer to begin with in terms of the concentration of the two major species? 2.) What is the millimolar quantity of the two major phosphate species? 3.) What is the millimolar quantity of H3O+ added as HCl (x in the ICE table)? 4.) What is the final millimolar quantity of HPO4 2? and H2PO4 ? at equilibrium? 5.) Calculate the new HPO4 2?/H2PO4 ? ratio, the log(HPO4 2?/H2PO4 ? ) and the new pH of the solution. (Note: You can use the molar ratio rather than the concentration ratio because both species are in the same volume.) 6.) log(HPO4 2?/H2PO4 ?) 7.) pH = 8.) Now take another 37 mL of the 0.50 M pH 6.83 buffer and add 1.3 mL of 1.00 M NaOH. Using steps similar to those above, calculate the new pH of the solution.